Chapter 6 Random Variables and the Normal Distribution

61 Discrete Random Variables

62 Binomial Probability Distribution

63 Continuous Random Variables and the Normal Probability Distribution

61 Discrete Random Variables

Objectives

By the end of this section I will be

able tohellip

1) Identify random variables

2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs

3) Calculate the mean variance and standard deviation of a discrete random variable

Random Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation for random variables

Suppose our experiment is to toss a single fair

die and we are interested in the number

rolled We define our random variable X to be

the outcome of a single die roll

a Why is the variable X a random variable

b What are the possible values that the

random variable X can take

c What is the notation used for rolling a 5

d Use random variable notation to express the probability of rolling a 5

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

61 Discrete Random Variables

Objectives

By the end of this section I will be

able tohellip

1) Identify random variables

2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs

3) Calculate the mean variance and standard deviation of a discrete random variable

Random Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation for random variables

Suppose our experiment is to toss a single fair

die and we are interested in the number

rolled We define our random variable X to be

the outcome of a single die roll

a Why is the variable X a random variable

b What are the possible values that the

random variable X can take

c What is the notation used for rolling a 5

d Use random variable notation to express the probability of rolling a 5

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Random Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation for random variables

Suppose our experiment is to toss a single fair

die and we are interested in the number

rolled We define our random variable X to be

the outcome of a single die roll

a Why is the variable X a random variable

b What are the possible values that the

random variable X can take

c What is the notation used for rolling a 5

d Use random variable notation to express the probability of rolling a 5

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example 62 - Notation for random variables

Suppose our experiment is to toss a single fair

die and we are interested in the number

rolled We define our random variable X to be

the outcome of a single die roll

a Why is the variable X a random variable

b What are the possible values that the

random variable X can take

c What is the notation used for rolling a 5

d Use random variable notation to express the probability of rolling a 5

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example 62 continued

Solution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Types of Random Variables

Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process

Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Identify each as a discrete or continuous random variable

(a) Total amount in ounces of soft drinks you consumed in the past year

(b) The number of cans of soft drinks that you consumed in the past year

Example

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(a) continuous

(b) discrete

Example

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Identify each as a discrete or continuous random variable

(a) The number of movies currently playing in US theaters

(b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie

Example

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(a) discrete

(b) continuous

(c) continuous

Example

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Discrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

The probability distribution table of

the number of heads observed when

tossing a fair coin twice

Table 62 in your textbook

Example

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution

Example

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Determine the outcomes with a tree diagram

Example

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Total number of outcomes is 16

Total number of ways to have 0 girls is 1

Total number of ways to have 1 girl is 4

Total number of ways to have 2 girls is 6

06250161girls) 0(P

25000164girl) 1(P

3750166girls) 2(P

Example

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Total number of ways to have 3 girls is 4

Total number of ways to have 4 girls is 1

25000164girls) 3(P

06250161girls) 4(P

Example

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

Distribution is

NOTE

x P(x)

0 00625

1 02500

2 03750

3 02500

4 00625

1)(xP

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Mean of a Discrete Random Variable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Finding the Mean of a Discrete Random Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Variability of a Discrete Random Variable Formulas for the Variance and Standard

Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

02)(xxP

)(xPx 2x )(2 xPx

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

x P(x)

0 00625 0 0 0

1 02500 025 1 02500

2 03750 075 4 15000

3 02500 075 9 22500

4 00625 025 16 10000

)(xPx 2x )(2 xPx

000010000400005)( 222 xPx

0100001

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Discrete Probability Distribution as a Graph Graphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

Page 270

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

Probability distribution (table)

Omit graph

x P(x)

0 025

1 035

2 025

3 015

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

Page 270

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

ANSWER

(a) Probability X is fewer than 3

850

250350250

)2P()1( )0(

)2 1 0(

XXPXP

XXXP

scored goals ofnumber X

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

ANSWER

(b) The most likely number of goals is the

expected value (or mean) of X

She will most likely score one goal

x P(x)

0 025 0

1 035 035

2 025 050

3 015 045

)(xPx

131)(xxP

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Example

ANSWER

(c) Probability X is at least one

750

150250350

)3P()2( )1(

)3 2 1(

XXPXP

XXXP

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

62 Binomial Probability Distribution

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

62 Binomial Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Explain what constitutes a binomial experiment

2) Compute probabilities using the binomial probability formula

3) Find probabilities using the binomial tables

4) Calculate and interpret the mean variance and standard deviation of the binomial random variable

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Factorial symbol

For any integer n ge 0 the factorial symbol n is defined as follows

0 = 1

1 = 1

n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Find each of the following

1 4

2 7

Example

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

504012345677 2

2412344 1

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Factorial on Calculator

Calculator

7 MATH PRB 4

which is

Enter gives the result 5040

7

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Combinations

An arrangement of items in which

r items are chosen from n distinct items

repetition of items is not allowed (each item is distinct)

the order of the items is not important

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

The number of different possible 5 card

poker hands Verify this is a combination

by checking each of the three properties

Identify r and n

Example of a Combination

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Five cards will be drawn at random from a deck of cards is depicted below

Example

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

An arrangement of items in which

5 cards are chosen from 52 distinct items

repetition of cards is not allowed (each card is distinct)

the order of the cards is not important

Example

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Combination Formula

The number of combinations of r items chosen

from n different items is denoted as nCr and

given by the formula

n r

nC

r n r

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Find the value of

Example

47 C

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

35

57

123

567

)123()1234(

1234567

34

7

)47(4

747 C

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Combinations on Calculator

Calculator

7 MATH PRB 3nCr 4

To get

Then Enter gives 35

47 C

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Determine the number of different

possible 5 card poker hands

Example of a Combination

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

9605982552C

Example

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

Genetics bull In mice an allele A for agouti (gray-brown

grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

bull A single offspring has genotypes

Sample Space

A a

A AA Aa

a aA aa

aaaAAaAA

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

bull Agouti genotype is dominant bull Event that offspring is agouti

bull Therefore for any one birth

aAAaAA

43genotype) agouti(P

41genotype) agoutinot (P

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti

bull Exactly three agouti offspring may occur in four different ways (in order of birth)

NGGG GNGG GGNG GGGN

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

bull Consecutive events (birth of a mouse) are independent and using multiplication rule

256

27

4

3

4

3

4

3

4

1

)()()()()( GPGPGPNPGGGNP

256

27

4

3

4

3

4

1

4

3

)()()()()( GPGPNPGPGGNGP

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

256

27

4

3

4

1

4

3

4

3

)()()()()( GPNPGPGPGNGGP

256

27

4

1

4

3

4

3

4

3

)()()()()( NPGPGPGPNGGGP

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Motivational Example

bull P(exactly 3 offspring has agouti fur)

4220256

108

4

1

4

34

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

4

3

4

3

4

3

4

3

4

1

N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (

3

P

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Experiment

Agouti fur example may be considered a binomial experiment

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Experiment

Four Requirements

1) Each trial of the experiment has only two possible outcomes (success or failure)

2) Fixed number of trials

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Experiment

Agouti fur example may be considered a binomial experiment

1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2) Fixed number of trials (4 births)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (frac34)

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Probability Distribution

When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Probability Distribution Formula

For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

where

XnX

Xn ppCXP )1()(

XnX

nCXn

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Probability Distribution Formula

Let q=1-p

XnX

Xn qpCXP )(

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Rationale for the Binomial

Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

The number of outcomes with exactly x successes among n

trials

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Probability Formula

P(x) = bull px bull qn-x n

(n ndash x )x

Number of outcomes with exactly x successes among n

trials

The probability of x successes among n

trials for any one particular order

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Agouti Fur Genotype Example

4220256

108

4

1

64

274

4

1

4

34

4

1

4

3

3)34(

4)(

13

13

XP

fur agouti birth with a ofevent X

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

2ND VARS

Abinompdf(4 75 3)

Enter gives the result 0421875

n p x

Binomial Probability Distribution Formula Calculator

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Distribution Tables

n is the number of trials

X is the number of successes

p is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 284

Example

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

00148

)50()50(15504

)50()50(

)5(

155

5205

520C

XP

heads ofnumber X

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Binomial Mean Variance and Standard Deviation

Mean (or expected value) μ = n p

Variance

Standard deviation

npqpq 2 then 1 Use

)1(2 pnp

npqpnp )1(

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

20 coin tosses

The expected number of heads

Variance and standard deviation

Example

10)500)(20(np

05)500)(500)(20(2 npq

2425

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 284

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2) Fixed number of trials (50)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(a)

Example

05490

)4160()5840(

)25(

2525

2550C

XP

baskets ofnumber X

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(b) The expected value of X

The most likely number of baskets is 29

Example

229)5840)(50(np

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them

Example

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 285

Example

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Is this a Binomial Distribution

Four Requirements

1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2) Fixed number of trials (120)

3) Experimental outcomes are independent of each other

4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(a) X=number of white males who contracted AIDS through injected drug use

Example

08160

)890()110(

)10(

11010

10120C

XP

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(b) At most 3 men is the same as less than or equal to 3 men

Why do probabilities add

Example

)3()2()1()0()3( XPXPXPXPXP

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Use TI-83+ calculator 2ND VARS to

get Abinompdf(n p X)

Example

= binompdf(120 11 0) + binompdf(120 11 1)

+ binompdf(120 11 2) + binompdf(120 113)

)3()2()1()0( XPXPXPXP

0000554 4E 5553517238

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(c) Most likely number of white males is the expected value of X

Example

213)110)(120(np

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 286

Example

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

(a)

Example

74811)890)(110)(120(2 npq

43374811

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

RECALL Outliers and z Scores

Data values are not unusual if

Otherwise they are moderately

unusual or an outlier (see page 131)

2score-2 z

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Sample Population

Z score Formulas

s

xxz x

z

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Z-score for 20 white males who contracted AIDS through injected drug use

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS

981433

21320z

Example

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

63 Continuous Random Variables and the Normal Probability Distribution

Objectives

By the end of this section I will be

able tohellip

1) Identify a continuous probability distribution

2) Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

(a) Relatively small sample (n = 100) with large class widths (05 lb)

FIGURE 615- Histograms

(b) Large sample (n = 200) with smaller class widths (02 lb)

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Figure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Continuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Requirements for Probability Distribution of a Continuous Random Variable

1) The total area under the density curve

must equal 1 (this is the Law of Total

Probability for Continuous Random

Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Probability for a Continuous Random Variable

Probability for Continuous Distributions

is represented by area under the curve

above an interval

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

The Normal Probability Distribution

Most important probability distribution in the world

Population is said to be normally distributed the data values follow a normal probability distribution

population mean is μ

population standard deviation is σ

μ and σ are parameters of the normal distribution

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

FIGURE 619

The normal distribution is symmetric about its mean μ (bell-shaped)

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Properties of the Normal Density Curve (Normal Curve)

1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Properties of the Normal Density Curve (Normal Curve) continued

5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

The Empirical Rule

For data sets having a distribution that is

approximately bell shaped the following

properties apply

About 68 of all values fall within 1

standard deviation of the mean

About 95 of all values fall within 2

standard deviations of the mean

About 997 of all values fall within 3

standard deviations of the mean

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

FIGURE 623 The Empirical Rule

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Drawing a Graph to Solve Normal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Steps in Drawing a Graph to Help You Solve Normal Probability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the

normal curve This part will depend on what values of X the problem is asking about

3) Proceed to find the desired area or probability using the empirical rule

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 295

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 296

Example

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Page 296

Example

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

ANSWER

Example

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

Summary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it