Chapter 6 Discrete Probability Distributions

36
Chapter 6 Discrete Probability Distributions

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Chapter 6 Discrete Probability Distributions. North Seattle Community College BUS210 Business Statistics. Learning Objectives. In this section, you will learn about:. Probability distributions and their properties Important discrete probability distributions: The Binomial Distribution - PowerPoint PPT Presentation

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Page 1: Chapter 6 Discrete Probability Distributions

Chapter 6

Discrete

Probability Distributions

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Learning Objectives

In this section, you will learn about: Probability distributions and their properties Important discrete probability distributions:

The Binomial Distribution The Poisson Distribution

Calculating the expected value and variance Using the Binomial and Poisson distributions

to solve business problems

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Definitions

Random variable: representation of a possible numerical value from an uncertain event. Discrete random variables: “How many”

produce outcomes that come from counting (e.g. number of courses you are taking).

Continuous random variables: “How much”

produce outcomes that come from a measurement (e.g. your annual salary, or your weight).

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Random

Variables

Discrete Random Variable

ContinuousRandom Variable

Ch. 6 Ch. 7

Random Variables

Discrete Random Variable

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Can only assume a countable number of values

Discrete Random Variables

If X is the number of times that a 6 shows up, then X could be equal to 0, 1, 2, 3, 4, or 5

If X is the number of times that heads comes up, then X could be equal to 0, 1, 2, or 3

Flipping a coin 3 times.

Examples:

Rolling five dice (as in Yahtzee)

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A probability distribution is…

a mutually exclusive listing of all possible numerical outcomes for a variable and the probability of an occurrence associated with each outcome.

Number of Classes Taken Probability

1 0.20

2 0.10

3 0.40

4 0.24

5 or more 0.06

Discrete Random Variables

Notice that the total probability adds up to 1.00

Example for a discrete random variable:

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X Freq Probability

0 1 1/4 = 0.25

1 2 2/4 = 0.50

2 1 1/4 = 0.25

Experiment: Toss 2 Coins. Let X = # heads.

Probability Distribution

0 1 2 X

0.50

0.25

Pro

bab

ility

Discrete Random Variable

Probability Distribution

T T

T H

TH

H HTotal = 1.00

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Example: Experiment of 2 Tossed Coins

Expected Value is the mean of a discrete random variable

Discrete Random Variable

Expected Value

i X P(X)X1 0 1/4 = 0.25

X2 1 2/4 = 0.50

X3 2 1/4 = 0.25

Also known as the Weighted Average

XiP(Xi)

(0)(0.25) = 0.00

(1)(0.50) = 0.50

(2)(0.25) = 0.50

E(X) = 1.00

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Variance of a discrete random variable

Standard Deviation of a discrete random variable

Discrete Random Variable

Measuring Dispersion

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Example: standard deviation of 2 tossed coins

(continued)

Discrete Random Variable

Measuring Dispersion

X E(X) X-E(X) (X-E(X))2 P(X) (X-E(X))2P(X)

X1 0 1 -1 1 0.25 0.25

X2 1 1 0 0 0.50 0.00

X3 2 1 +1 1 0.25 0.25

∑= 0.50

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Continuous Probability

Distributions

Probability Distributions

Binomial

Poisson

Discrete Probability

Distributions

Normal

Ch. 6 Ch. 7

Probability Distributions

Binomial

Exponential

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A finite number of observations, n 20 tosses of a coin 12 auto batteries purchased from a wholesaler

Each observation is categorized as… Success: “event of interest” has occurred, or

i.e. - heads (or tails) in each toss of a coin; i.e. - defective (or not defective) battery

Failure: “event of interest” has not occurred The complement of a success

These two categories, success & failure, are mutually exclusive and collectively exhaustive

Probability Distributions

Binomial

Note: Note: A random A random experiment with experiment with only 2 outcomes is only 2 outcomes is known as a known as a Bernoulli Bernoulli experimentexperiment..

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(continued)

The probability of… success (event occurring) is represented as π failure (event not occurring) is 1 – π

Probability of success (π) is constant Probability of getting tails is the same for each toss of the coin

Observations are independent Each event is unaffected by any other event Two sampling methods deliver independence

Infinite population without replacement Finite population with replacement

Probability Distributions

Binomial

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Applications: Situations where are there only two outcomes

Apple marks newly manufactured iPads as either defective or acceptable

Visitors to Amazon’s website will either buy an item or not buy an item.

Asking if voters will approve a referendum, a pollster receives responses of “yes” or “no”

Federal Express marks a delivery as either damaged or not damaged

Probability Distributions

Binomial (continued)

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You toss a coin three times.

In how many ways can you get two heads?

Probability Distributions

Binomial (continued)

Possible ways: HHT, HTH, THH, so there are three ways you can getting two heads.

This situation is fairly simple….but, what about 10 times?…a 100?...or even a thousand?

Counting Revisited:

For more complicated situations we need a methodto be able to count the number of ways (combinations).

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The number of combinations of

selecting x objects out of n objects is

where: n! means “n factorial”2! = (1)(2)4! = (1)(2)(3)(4)

Note: 0! = 1 (by definition)

Counting Techniques

Rule of Combinations

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combinations

Counting Techniques

Rule of Combinations

You visit Baskin & Robbins. How many different possible 3 scoop combinations could you decide on for your ice cream cone if you select from their 31 flavors?

The total choices is n = 31, and we select x = 3.

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π = probability of “event of interest”

n = sample size (number of trials or observations)x = number of “events of interest” in sample

Probability Distributions

Binomial

n!(n-x)!n!

P(x) = π x (1-π) n-x

Note: other sources may show this equation as or similar.

for the number of heads in 3 coin flips, we would use…

π = 0.5 and (1 – π) = 0.5

n = 4

X = can be 0, or 1, or 2, or 3

Each value of X will have a different probability

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What is the probability of 2 successes in 7 observations if the probability of the event of interest is 0.4?

Probability Distributions

Binomial Example

x = 1, n = 7, and π = 0.4

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The probability of a single customer purchasing an extended warranty is 0.35.

What is the probability of having 3 of the next 10 customers purchase an extended warranty?

Probability Distributions

Binomial Example

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n x … π=.20 π=.25 π=.30 π=.35 π=.40 π=.45 π=.50

10 0

1

2

3

4

5

6

7

8

9

10

0.1074

0.2684

0.3020

0.2013

0.0881

0.0264

0.0055

0.0008

0.0001

0.0000

0.0000

0.0563

0.1877

0.2816

0.2503

0.1460

0.0584

0.0162

0.0031

0.0004

0.0000

0.0000

0.0282

0.1211

0.2335

0.2668

0.2001

0.1029

0.0368

0.0090

0.0014

0.0001

0.0000

0.0135

0.0725

0.1757

0.2522

0.2377

0.1536

0.0689

0.0212

0.0043

0.0005

0.0000

0.0060

0.0403

0.1209

0.2150

0.2508

0.2007

0.1115

0.0425

0.0106

0.0016

0.0001

0.0025

0.0207

0.0763

0.1665

0.2384

0.2340

0.1596

0.0746

0.0229

0.0042

0.0003

0.0010

0.0098

0.0439

0.1172

0.2051

0.2461

0.2051

0.1172

0.0439

0.0098

0.0010

10

9

8

7

6

5

4

3

2

1

0

… π=.80 π=.75 π=.70 π=.65 π=.60 π=.55 π=.50 x

Examples:

Probability Distributions

Using the Binomial Table

n = 10, π = .35, x = 3

n = 10, π = .75, x = 2

P(x = 3) = .2522

P(x = 2) = .0004

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The shape of the binomial distribution…

depends on the values of π and n

Probability Distributions

Shape of the Binomial

n = 5 π = 0.5n = 5 π = 0.1

0 1 2 3 4 5X

P(X)

.2

.4

.6

0

0 1 2 3 4 5X

P(X)

.2

.4

.6

0

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Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

Where n = sample size

π = probability of the event of interest for any trial

(1 – π) = probability of no event of interest for any trial

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n = 5 π = 0.1

n = 5 π = 0.5

Examples

Binomial Distribution Characteristics

0 1 2 3 4 5X

P(X)

.2

.4

.6

0

0 1 2 3 4 5X

P(X)

.2

.4

.6

0

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ExamplesExamples: : •The probability that The probability that less than less than 3 of the next 10 customers will purchase an 3 of the next 10 customers will purchase an extended warranty isextended warranty is

Binomial Distribution Characteristics

Compound EventsCompound Events

P(x<3) = P(x=0) + P(x=1) + P(x=2)

P(x<3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)

Individual probabilities can be added to obtain any desired combined Individual probabilities can be added to obtain any desired combined event probability.event probability.

• The probability that 3 The probability that 3 or fewer or fewer of the next 10 customers will purchase an of the next 10 customers will purchase an extended warranty isextended warranty is

= 0.0135 + 0.0725 + 0.1757 = 0.2617

= 0.0135 + 0.0725 + 0.1757 + 0.2522 = 0.5139

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Binomial DistributionUsing Excel

Cell Formulas

n x π

x n π cum

n x π x (1-π)

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Continuous Probability

Distributions

Binomial

Poisson

Probability Distributions

Discrete Probability

Distributions

Normal

Ch. 6 Ch. 7

Probability Distributions

Poisson Exponential

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Used when you are interested in…the number of times an event occurswithin a continuous unit or interval.

Such as time, distance, volume, or area.

Examples: The number of… potholes in mile of road (distance) computer crashes in a day (time) chocolate chips in a cookie (volume) mosquito bites on a person (area)

Probability Distributions

The Poisson Distribution

Was actually Was actually used in a study used in a study of malariaof malaria

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Defining features: Independent: The occurrence of any single event

does not impact the occurrence of any other event. Constant: The average probability of an event

occurring in one area of opportunity remains the same for all other similar areas. The average number of events per unit is represented by (lambda)

The average probability of an event decreases or increases proportionally to any decrease or increase in the size of the area of opportunity.

Probability Distributions

The Poisson Distribution

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where:

x = number of events in an area of opportunity

= expected value (mean) of number of events

e = base of the natural logarithm system (2.71828...)

Probability Distributions

The Poisson Distribution

Note: other sources may show this equation as or similar.

e -λ λx P(x) =

x!

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Mean

Variance and Standard Deviation

where = expected number of events

Poisson DistributionCharacteristics

Isn’t that interesting?

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X

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

0

1

2

3

4

5

6

7

0.9048

0.0905

0.0045

0.0002

0.0000

0.0000

0.0000

0.0000

0.8187

0.1637

0.0164

0.0011

0.0001

0.0000

0.0000

0.0000

0.7408

0.2222

0.0333

0.0033

0.0003

0.0000

0.0000

0.0000

0.6703

0.2681

0.0536

0.0072

0.0007

0.0001

0.0000

0.0000

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000

0.5488

0.3293

0.0988

0.0198

0.0030

0.0004

0.0000

0.0000

0.4966

0.3476

0.1217

0.0284

0.0050

0.0007

0.0001

0.0000

0.4493

0.3595

0.1438

0.0383

0.0077

0.0012

0.0002

0.0000

0.4066

0.3659

0.1647

0.0494

0.0111

0.0020

0.0003

0.0000

Example: Find P(X = 2) if = 0.50

Probability Distributions

Using the Poisson Table

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Poisson DistributionUsing Excel

Cell Formulas

x λ cum

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X P(X)

0

1

2

3

4

5

6

7

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

P(X=2) = 0.0758

= 0.50

Probability Distributions

Graphing the Poisson

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The shape of the Poisson Distribution depends on the parameter :

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

= 0.50 = 3.00

Probability Distributions

Shape of the Poisson

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Chapter Summary

Probability distributions Discrete random variables The Binomial distribution The Poisson distribution