Note 6 of 5E Statistics with Economics and Business Applications Chapter 4 Useful Discrete Probability Distributions Binomial, Poisson and Hypergeometric.

Note 6 of 5E

Statistics with Economics and Statistics with Economics and Business ApplicationsBusiness Applications

Chapter 4 Useful Discrete Probability Distributions

Binomial, Poisson and Hypergeometric Probability Distributions

Note 6 of 5E

ReviewReview

I. I. What’s in last two lectures?What’s in last two lectures?Experiment, Event, Sample space, Probability, Counting rules, Conditional probability, Bayes’s rule, random variables, mean, variance. Chapter 3.

II. II. What's in this lecture?What's in this lecture?Binomial, Poisson and Hypergeometric Probability Distributions. Read Chapter 4.

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IntroductionIntroduction

• Discrete random variables take on only a finite or countable number of values.

• There are several useful discrete probability distributions. We will learn Binomial and Poisson distributions.

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The Binomial Random VariableThe Binomial Random Variable The coin-tossing experimentcoin-tossing experiment is a

simple example of a binomial binomial random variablerandom variable. . Toss a fair coin n = 3 times and record x = number of heads.

x p(x)

0 1/8

1 3/8

2 3/8

3 1/8

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The Binomial Random VariableThe Binomial Random Variable• Many situations in real life resemble the coin

toss, but the coin is not necessarily fair, so that P(H) 1/2.

• Example: Example: A geneticist samples 10 people and counts the number who have a gene linked to Alzheimer’s disease.

Person• Coin:Coin:

• Head:Head:

• Tail:Tail:

• Number of tosses:Number of tosses:

• P(H):P(H):Has gene

Doesn’t have gene

n = 10

P(has gene) = proportion in the population who

have the gene.

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The Binomial ExperimentThe Binomial Experiment1. The experiment consists of nn identical trials identical trials..2. Each trial results in one of two outcomesone of two outcomes, success (S)

or failure (F).3. The probability of success on a single trial is p and

remains constantremains constant from trial to trial. The probability of failure is q = 1 – p.

4. The trials are independentindependent.5. We are interested in xx, the number of successes in , the number of successes in n n

trialstrials..

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Binomial or Not?Binomial or Not? The independence is a key assumption

that often violated in real life applications

• Select two people from the U.S. population, and suppose that 15% of the population has the Alzheimer’s gene.• For the first person, p = P(gene) = .15• For the second person, p P(gene) = .15,

even though one person has been removed from the population.

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Binomial or Not?Binomial or Not?

2 out of 20 PCs are defective. We randomly select 3 for testing. Is this a binomial experiment?

1. The experiment consists of n=3 identical trials2. Each trial result in one of two outcomes3. The probability of success (finding the defective) is

2/20 and remains the same4. The trials are not independent. For example, P( success on the 2nd trial | success on the 1st trial) =

1/19, not 2/20

Rule of thumb: if the sample size n is relatively large to the population size N, say n/N >= .05, the resulting experiment would not be binomial.

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The Binomial Probability The Binomial Probability DistributionDistribution

For a binomial experiment with n trials and probability p of success on a given trial, the probability of k successes in n trials is

.1!01)2)...(2)(1(!

)!(!

!

.,...2,1,0)!(!

!)(

and with

Recall

for

nnnn

knk

nC

nkqpknk

nqpCkxP

nk

knkknknk

.1!01)2)...(2)(1(!

)!(!

!

.,...2,1,0)!(!

!)(

and with

Recall

for

nnnn

knk

nC

nkqpknk

nqpCkxP

nk

knkknknk

SticiGuiSticiGui

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The Mean and Standard The Mean and Standard DeviationDeviation

For a binomial experiment with n trials and probability p of success on a given trial, the measures of center and spread are:

npq

npq

np

:deviation Standard

:Variance

:Mean2

npq

npq

np

:deviation Standard

:Variance

:Mean2

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n = p = x =success =

ExampleExampleA marksman hits a target 80% of the time. He fires five shots at the target. What is the probability that exactly 3 shots hit the target?

333)3( nn qpCxP

5 .8hit # of hits

353 )2(.)8(.!2!3

!5

2048.)2(.)8(.10 23

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ExampleExample

What is the probability that more than 3 shots hit the target?

55555

45454)3( qpCqpCxP

0514 )2(.)8(.!0!5

!5)2(.)8(.

!1!4

!5

7373.)8(.)2(.)8(.5 54

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Cumulative Cumulative Probability TablesProbability Tables

You can use the cumulative probability cumulative probability tablestables to find probabilities for selected binomial distributions.

Find the table for the correct value of n.

Find the column for the correct value of p.

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Find the table for the correct value of n.

Find the column for the correct value of p.

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

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ExampleExample

k p = .80

0 .000

1 .007

2 .058

3 .263

4 .672

5 1.000

What is the probability that exactly 3 shots hit the target?

P(P(xx = 3) = 3) = P(x 3) – P(x 2)= .263 - .058= .205

P(P(xx = 3) = 3) = P(x 3) – P(x 2)= .263 - .058= .205 Check from formula:

P(x = 3) = .2048

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ExampleExample

k p = .80

0 .000

1 .007

2 .058

3 .263

4 .672

5 1.000

What is the probability that more than 3 shots hit the target?

P(P(xx > 3) > 3) = 1 - P(x 3)= 1 - .263 = .737P(P(xx > 3) > 3) = 1 - P(x 3)= 1 - .263 = .737

Check from formula: P(x > 3) = .7373

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ExampleExample Would it be unusual to find that none of the shots hit the target?

P(P(xx = 0) = 0) = P(x 0) = 0P(P(xx = 0) = 0) = P(x 0) = 0

What is the probability that less than 3 shots hit the target?

P(P(xx < 3) < 3) = P(x 2) = 0.058P(P(xx < 3) < 3) = P(x 2) = 0.058

What is the probability that less than 4 but more than 1 shots hit the target?

P(1<P(1<xx < 4) < 4) = P(x 3) - P(x 1) = .263-.007=.256P(1<P(1<xx < 4) < 4) = P(x 3) - P(x 1) = .263-.007=.256

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ExampleExample Here is the probability distribution

for x = number of hits. What are the mean and standard deviation for x?

89.)2)(.8(.5

4)8(.5

npq

np

:deviation Standard

:Mean

89.)2)(.8(.5

4)8(.5

npq

np

:deviation Standard

:Mean

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The Poisson Random VariableThe Poisson Random Variable• The Poisson random variable x is often a model for

data that represent the number of occurrences of a specified event in a given unit of time or space.

• Examples:Examples:

• The number of calls received by a switchboard during a given period of time.

• The number of machine breakdowns in a day

• The number of traffic accidents at a given intersection during a given time period.

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The Poisson Probability The Poisson Probability DistributionDistribution

LetLet x x a Poisson random variable. The probability of k occurrences of this event is

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are

Mean:

Standard deviation:

!)(

k

ekxP

k

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ExampleExampleThe average number of traffic accidents on a certain section of highway is two per week. Find the probability of exactly one accident during a one-week period.

!)1(

k

exP

k

2707.2!1

2 221

ee

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Cumulative Cumulative Probability TablesProbability Tables

You can use the cumulative probability cumulative probability tablestables to find probabilities for selected Poisson distributions.

Find the column for the correct value of .

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Find the column for the correct value of .

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

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ExampleExample

k = 2

0 .135

1 .406

2 .677

3 .857

4 .947

5 .983

6 .995

7 .999

8 1.000

What is the probability that there is exactly 1 accident?

P(P(xx = 1) = 1) = P(x 1) – P(x 0)= .406 - .135= .271

P(P(xx = 1) = 1) = P(x 1) – P(x 0)= .406 - .135= .271 Check from formula:

P(x = 1) = .2707

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ExampleExample

What is the probability that 8 or more accidents happen?

P(P(xx 8) 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001

P(P(xx 8) 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001

k = 2

0 .135

1 .406

2 .677

3 .857

4 .947

5 .983

6 .995

7 .999

8 1.000

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The Hypergeometric The Hypergeometric Probability DistributionProbability Distribution

A bowl contains M red M&M® candies and N-M blue M&M® candies. Select n candies from the bowl and record x the number of red candies selected. Define a “red M&M®” to be a “success”.

The probability of exactly k successes in n trials is

Nn

NMkn

Mk

C

CCkxP

)(

mm

mmm

mm

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The Mean and VarianceThe Mean and Variance

1 :Variance

:Mean

2

N

nN

N

MN

N

Mn

N

Mn

1 :Variance

:Mean

2

N

nN

N

MN

N

Mn

N

Mn

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ExampleExampleA package of 8 AA batteries contains 2 batteries that are defective. A student randomly selects four batteries and replaces the batteries in his calculator. What is the probability that all four batteries work?

84

20

64)4(C

CCxP Success = working battery

N = 8

M = 6

n = 4 70

15

)1)(2)(3(4/)5)(6)(7(8

)1(2/)5(6

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38

64

N

Mn

ExampleExample

What are the mean and variance for the number of batteries that work?

4286.7

4

8

2

8

64

12

N

nN

N

MN

N

Mn

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Key ConceptsKey Concepts The Binomial Random VariableThe Binomial Random Variable

1. Five characteristics: n identical trials, each resulting in either success S or failure F; probability of success is p and remains constant from trial to trial; trials are independent; and x is the number of successes in n trials.

2. Calculating binomial probabilities

a. Formula:

b. Cumulative binomial tables

3. Mean of the binomial random variable: np

4. Variance and standard deviation: 2 npq and

knknk qpCkxP )(

npq npq

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Key ConceptsKey ConceptsII. The Poisson Random VariableII. The Poisson Random Variable

1. The number of events that occur in a period of time or space, during which an average of such events are expected to occur

2. Calculating Poisson probabilities

a. Formula:b. Cumulative Poisson tables

3. Mean of the Poisson random variable: E(x) 4. Variance and standard deviation: 2 and

!)(

k

ekxP

k

!)(

k

ekxP

k

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Key ConceptsKey ConceptsIII. The Hypergeometric Random VariableIII. The Hypergeometric Random Variable

1. The number of successes in a sample of size n from a finite population containing M successes and N M failures2. Formula for the probability of k successes in n trials:

3. Mean of the hypergeometric random variable:

4. Variance and standard deviation:

N

Mn

N

Mn

12

N

nN

N

MN

N

Mn

12

N

nN

N

MN

N

Mn

Nn

NMkn

Mk

C

CCkxP

)( N

n

NMkn

Mk

C

CCkxP

)(