1/71 Statistics Discrete Probability Distributions.
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Transcript of 1/71 Statistics Discrete Probability Distributions.
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1/71
Statistics
Discrete Probability Distributions
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Contents
.10.10
.20.20
.30.30
.40.40
0 1 2 3 4 0 1 2 3 4
Random Variables Discrete Probability Distributions Expected Value and Variance Binomial Distribution Poisson Distribution Hypergeometric Distribution
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STATISTICS in PRACTICE
Citibank makes available a wide range of financial services.
Citibanking’s automatic teller machines (ATMs) located in Citicard Banking Centers (CBCs), let customers do all their banking in one place with the touch of a finger.
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STATISTICS in PRACTICE Periodic CBC capacity studies are used to analyze customer waiting times and
to determine whether additional ATMs are needed.
Data collected by Citibank showed that the random customer arrivals followed a probability distribution known as the Poisson distribution.
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A random variable is a numerical description of the outcome of an experiment.
Random Variables
A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals.
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Example: Tossing a dicePossible outcomes: 1, 2, 3, 4, 5, and 6One can define random variables as 1 if outcome is greater than 3, and 0 if outcome is smaller or equal to 3.
Or1 if outcome is odd numbers0 if outcome is even numbers
Differences between outcomes and random variables
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Discrete Random Variables
Example 1.
The certified public accountant (CPA)
examination has four parts.
Define a random variable as x = the number of
parts of the CPA examination passed and It is
a discrete random variable because it may
assume the finite number of values 0, 1, 2, 3, or 4.
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Discrete Random Variables
Example 2.
An experiment of cars arriving at a tollbooth. The random variable is x = the number of cars arriving during a one-day period. The possible values for x come from the sequence of integers 0, 1, 2, and so on. x is a discrete random variable assuming one of the values in this infinite sequence.
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Discrete Random Variables
Examples of Discrete Random Variables
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Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4)
Let x = number of TVs sold at the store in one day, where x can take on 5 values (0, 1, 2, 3, 4)
Example: JSL Appliances Discrete random variable with a finite
number of values
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Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .
Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .
Example: JSL Appliances Discrete random variable with an
infinite sequence of values
We can count the customers arriving, but thereis no finite upper limit on the number that might arrive.
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Random Variables
Question Random Variable x Type
Familysize
x = Number of dependents reported on tax return
Discrete
Distance fromhome to store
x = Distance in miles from home to the store site
Continuous
Own dogor cat
x = 1 if own no pet; = 2 if own dog(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s)
Discrete
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Continuous Random VariablesExample 1.
Experimental outcomes based on measurement scales such as time, weight, distance, and temperature can be described by continuous random variables.
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Continuous Random Variables
Example 2.
An experiment of monitoring incoming telephone calls to the claims office of a major insurance company. Suppose the random variable of interest is x = the time between consecutive incoming calls in minutes. This random variable may assume any value in the interval x ≥ 0.
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Continuous Random Variables
Example of Continuous Random Variables
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The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. We can describe a discrete probability distribution with a table, graph, or equation.
Discrete Probability Distributions
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The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable.
The required conditions for a discrete probability function are:
Discrete Probability Distributions
f(x) > 0
f(x) = 1
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Discrete Probability Distributions Example: Probability Distribution for the
Number of Automobiles Sold During a Day at Dicarlo Motors.
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a tabular representation of the probability distribution for TV sales was developed.
Using past data on TV sales, …
Number Units Sold of Days
0 80 1 50 2 40 3 10 4 20
200
x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10 1.00
80/200
Discrete Probability Distributions
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.10.10
.20.20
.30.30
.40.40
.50.50
0 1 2 3 4Values of Random Variable x (TV sales)
Pro
babili
ty
Discrete Probability Distributions
Graphical Representation of Probability Distribution
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Discrete Uniform Probability Distribution
The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula.
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Discrete Uniform Probability Distribution
the values of the random variable are equally likely
The discrete uniform probability function is
f(x) = 1/nwhere:
n = the number of values the random variable may assume
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Discrete Uniform Probability Distribution
Example: An experiment of rolling a die we define
the random variable x to be the number of dots on the upward face. There are n = 6 possible values for the random variable;
x = 1, 2, 3, 4, 5, 6. The probability function for this discrete uniform random variable is f (x) = 1/6 x = 1, 2, 3, 4, 5, 6.
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Discrete Uniform Probability Distribution
x f (x)1 1/62 1/63 1/64 1/65 1/66 1/6
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Example: Consider the random variable x with the
following discrete probability distribution.
This probability distribution can be defined by
the formula f (x) = x/ 10 for x = 1, 2, 3, or 4.
Discrete Uniform Probability Distribution
x f (x)1 1/102 2/103 3/104 4/10
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Expected Value and Variance The expected value, or mean, of a random variable is a measure of its central location.
The variance summarizes the variability in the values of a random variable.
Var(x) = 2 = (x - )2f(x)Var(x) = 2 = (x - )2f(x)
E(x) = = x f(x)E(x) = = x f(x)
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Expected Value and Variance
The standard deviation, , is defined as the positive square root of the variance.
Here, the expected value and variance are computed from random variables instead of outcomes
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Expected Value and Variance
Example: Calculation of the Expected Value for the Number of Automobiles Sold During A Day at Dicarlo Motors.
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Expected Value and Variance Example: Calculation of the Variance for the Number of Automobiles Sold During A Day at Dicarlo Motors.
The standard deviation is 118.125.1
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Expected Value
expected number of TVs sold in a day
x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40
E(x) = 1.20
Expected Value and Variance
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Variance and Standard Deviation
01234
-1.2-0.2 0.8 1.8 2.8
1.440.040.643.247.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
x - (x - )2 f(x) (x - )2f(x)
Variance of daily sales = s 2 = 1.660
x
TVssquared
Standard deviation of daily sales = 1.2884 TVs
Expected Value and Variance
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Binomial Distribution Four Properties of a Binomial Experiment
3. The probability of a success, denoted by p, does not change from trial to trial.
4. The trials are independent.
2. Two outcomes, success and failure, are possible on each trial.
1. The experiment consists of a sequence of n identical trials.
stationarityassumption
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Binomial Distribution
Our interest is in the number of successes occurring in the n trials. We let x denote the number of successes occurring in the n trials.
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where: f(x) = the probability of x successes in n trials, n = the number of trials, p = the probability of success on any one trial.
( )!( ) (1 )
!( )!x n xn
f x p px n x
Binomial Distribution
Binomial Probability Function
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Binomial Distribution
!!( )!
nx n x
( )(1 )x n xp p Probability of a particular sequence of trial outcomes with x successes in n trials
Number of experimental outcomes providing exactlyx successes in n trials
=
=
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( )!( ) (1 )
!( )!x n xn
f x p px n x
Binomial Distribution
Binomial Probability Function
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Binomial Distribution Example:
The experiment of tossing a coin five times
and on each toss observing whether the coin lands with a head or a tail on its
upward face. we want to count the number
of heads appearing over the five tosses. Does this experiment show the properties of
a binomial experiment?
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Binomial Distribution
Note that:
1. The experiment consists of five identical
trials; each trial involves the tossing of one
coin.
2. Two outcomes are possible for each trial: a
head or a tail. We can designate head a
success and tail a failure.
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Binomial Distribution Note that:
3. The probability of a head and the probability of
a tail are the same for each trial, with p = .5
and 1- p = .5.
4. The trials or tosses are independent because the
outcome on any one trial is not affected by
what happens on other trials or tosses.
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Example: Evans Electronics
Evans is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
Binomial Distribution
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Binomial Distribution Using the Binomial Probability Function
Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year?
f xn
x n xp px n x( )
!!( )!
( )( )
1
1 23!(1) (0.1) (0.9) 3(.1)(.81) .243
1!(3 1)!f
Let: p = .10, n = 3, x = 1
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Binomial Distribution
1st Worker 1st Worker 2nd Worker2nd Worker 3rd Worker3rd Worker xx Prob.Prob.
Leaves (.1)Leaves (.1)
Stays (.9)Stays (.9)
33
22
00
22
22
Leaves (.1)Leaves (.1)
Leaves (.1)Leaves (.1)
S (.9)S (.9)
Stays (.9)Stays (.9)
Stays (.9)Stays (.9)
S (.9)S (.9)
S (.9)S (.9)
S (.9)S (.9)
L (.1)L (.1)
L (.1)L (.1)
L (.1)L (.1)
L (.1)L (.1) .0010.0010
.0090.0090
.0090.0090
.7290.7290
.0090.0090
11
11
.0810.0810
.0810.0810
.0810.0810
11
Tree Diagram
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Using Tables of Binomial Probabilities
n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50
3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .12501 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250
p
Binomial Distribution
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Binomial Distribution
(1 )np p
E(x) = = np
Var(x) = 2 = np(1 - p)
Expected Value
Variance
Standard Deviation
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Binomial Distribution
3(.1)(.9) .52 employees
E(x) = = 3(.1) = .3 employees out of 3
Var(x) = 2 = 3(.1)(.9) = .27
Expected Value
Variance
Standard Deviation
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A Poisson distributed random variable is often useful in estimating the number of occurrences over a specified interval of time or space
It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2, . . . ).
Poisson Distribution
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Examples of a Poisson distributed random variable: the number of knotholes in 14 linear feet of pine board
the number of vehicles arriving at a toll booth in one hour
Poisson Distribution
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Poisson Distribution Two Properties of a Poisson Experiment
2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.
1. The probability of an occurrence is the same for any two intervals of equal length.
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Poisson Probability Function
Poisson Distribution
f xex
x( )
!
where:f(x) = probability of x occurrences in
an interval,
= mean number of occurrences in
an interval,
e = 2.71828.
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Poisson Distribution Example: We are interested in the number of arrivals at the drive-up teller window of a bank during a 15-minute period on weekday mornings. Assume that the probability of a car arriving is the same for any two time periods of equal length and that the arrival or nonarrival of a car in any time period is independent of the arrival or nonarrival in any other time period.
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The Poisson probability function is applicable. Suppose that the average number of cars arriving in a 15-minute period of time is 10; in this case, the following probability function applies. The random variable here is x = number of cars arriving in any 15-minute
period.
Poisson Distribution
!
10)(
10
x
exf
x
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Poisson Distribution
Example: Mercy Hospital
Patients arrive at the emergency
room of Mercy Hospital at the average rate of
6 per hour on weekend evenings.
What is the probability of 4 arrivals in 30 minutes on a weekend evening?
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Poisson Distribution
Using the Poisson Probability Function
4 33 (2.71828)(4) .1680
4!f
MERCY
= 6/hour = 3/half-hour, x = 4
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Poisson Distribution
Using Poisson Probability Tables
x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081
MERCY
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Poisson Distribution of Arrivals
MERCY
Poisson Probabilities
0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5 6 7 8 9 10
Number of Arrivals in 30 Minutes
Pro
bab
ilit
y
actually, the sequencecontinues:11, 12, …
Poisson Distribution
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Poisson Distribution A property of the Poisson distribution is thatthe mean and variance are equal.
m = s 2
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Poisson Distribution
MERCY
Variance for Number of Arrivals
During 30-Minute Periods
m = s 2 = 3 m = s 2 = 3
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Hypergeometric Distribution
The hypergeometric distribution is closely related to the binomial distribution.
However, for the hypergeometric distribution:
the trials are not independent, and
the probability of success changes from trial to trial.
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Hypergeometric Probability Function
Hypergeometric Distribution
n
N
xn
rN
x
r
xf )( for 0 < x < r
where: f(x) = probability of x successes in n trials, n = number of trials, N = number of elements in the population, r = number of elements in the population
labeled success.
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Hypergeometric Probability Function
Hypergeometric Distribution
( )
r N r
x n xf x
N
n
for 0 < x < r
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Hypergeometric Probability Function
Hypergeometric Distribution
number of ways x successes can be selected from a total of r successes in the population
number of waysa sample of size n can be selectedfrom a population of size N
x
r=
xn
rN=
number of ways n – x failures can be selected from a total of N – r failuresin the population
n
N=
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Hypergeometric Distribution
Example: A Quality Control Application. Electric fuses produced by Ontario Electric are
packaged in boxes of 12 units each. Suppose an inspector randomly selects 3 of the 12 fuses in a box for testing.
If the box contains exactly 5 defective fuses, what is the probability that the inspector will find exactly 1 of the 3 fuses defective?
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Hypergeometric Distribution
In this application, n = 3 and N = 12.
With r = 5 defective fuses in the box.
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The probability of finding x = 1 defective fuse is
What is the probability of finding at least 1 defective fuse?
Hypergeometric Distribution
4773.220
)21)(5(
!9!3
!12!5!2
!7
!4!1
!5
3
12
2
7
1
5
)1(
f
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The probability of x = 0 is
we conclude that the probability of finding at least 1 defective fuse must be 1 - .1591 = .8409.
Hypergeometric Distribution
1591.220
)35)(1(
!9!3
!12!4!3
!7
!5!0
!5
3
12
3
7
0
5
)0(
f
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Hypergeometric Distribution Example: Neveready
Bob Neveready has removed twodead batteries from a flashlight
and inadvertently mingled them with the two good batteries he intended
as replacements. The four batteries look identical. Bob now randomly selects two of the four batteries. What is the probability he selects the two good batteries?
ZAP
ZA
P
ZAPZAP
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Hypergeometric Distribution
Using the Hypergeometric Function2 2 2! 2!
2 0 2!0! 0!2! 1( ) .167
4 4! 62 2!2!
r N r
x n xf x
N
n
where: x = 2 = number of good batteries selected
n = 2 = number of batteries selected N = 4 = number of batteries in total r = 2 = number of good batteries in total
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Hypergeometric Distribution
( )r
E x nN
2( ) 11
r r N nVar x n
N N N
Mean
Variance
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Hypergeometric Distribution
22 1
4
rn
N
2 2 2 4 2 12 1 .333
4 4 4 1 3
Mean
Variance
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Hypergeometric Distribution
Consider a hypergeometric distribution with n trials and let p = (r/n) denote the probability of a success on the first trial.
If the population size is large, the term (N – n)/(N – 1) approaches 1.
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Hypergeometric Distribution
The expected value and variance can be written E(x) = np and Var(x) = np(1 – p).
Note that these are the expressions for the expected value and variance of a binomial distribution.
continued
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Hypergeometric Distribution
When the population size is large, a hypergeometric distribution can be approximated by a binomial distribution with n trials and a probability of success p = (r/N).