4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Business & Economics Discrete Random Variables...

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© 1998 Prentice-Hall, Inc.© 1998 Prentice-Hall, Inc.

Statistics forStatistics forBusiness & EconomicsBusiness & Economics

Discrete Random VariablesDiscrete Random VariablesChapter 4Chapter 4

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Learning ObjectivesLearning Objectives

1.1. Define random variableDefine random variable

2.2. Compute the expected value & variance Compute the expected value & variance of discrete random variablesof discrete random variables

3.3. Describe the binomial & Poisson Describe the binomial & Poisson probability distributionsprobability distributions

4.4. Calculate probabilities for binomial & Calculate probabilities for binomial & Poisson random variablesPoisson random variables

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Thinking ChallengeThinking Challenge

You’re taking a You’re taking a 3333 question multiple choice question multiple choice test. Each question has test. Each question has 4 4 choiceschoices. Clueless on . Clueless on 11 question, you decide to question, you decide to guess. What’s the chance guess. What’s the chance you’ll get it right?you’ll get it right?

If you guessed on all If you guessed on all 3333 questions, what would be questions, what would be your grade? Pass?your grade? Pass?

AloneAlone GroupGroup Class Class

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Random VariableRandom Variable

1.1. A numerical outcome of an experimentA numerical outcome of an experiment

2.2. May be discrete or continuousMay be discrete or continuous

3.3. Discrete random variableDiscrete random variable Countable number of valuesCountable number of values Example: Number of tails in 2 coin tossesExample: Number of tails in 2 coin tosses

4.4. Continuous random variableContinuous random variable Infinite number of values within an intervalInfinite number of values within an interval Example: Amount of soda in a 12 oz. canExample: Amount of soda in a 12 oz. can

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Discrete Random Discrete Random VariablesVariables

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Discrete Discrete Random VariableRandom Variable

1. Type of random variable1. Type of random variable

2.2. Whole number (0, 1, 2, 3 etc.)Whole number (0, 1, 2, 3 etc.)

3.3. Obtained by countingObtained by counting

4.4. Usually finite number of valuesUsually finite number of values Poisson random variable is exception (Poisson random variable is exception ())

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Discrete Random Discrete Random Variable ExamplesVariable Examples

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RandomRandomVariableVariable

PossiblePossibleValuesValues

ExperimentExperiment

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Make 100 sales callsMake 100 sales calls


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Make 100 sales callsMake 100 sales calls # Sales# Sales 0, 1, 2, ..., 1000, 1, 2, ..., 100


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Inspect 70 radiosInspect 70 radios


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Inspect 70 radiosInspect 70 radios # Defective# Defective 0, 1, 2, ..., 700, 1, 2, ..., 70


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Answer 33 questionsAnswer 33 questions


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Answer 33 questionsAnswer 33 questions # Correct# Correct 0, 1, 2, ..., 330, 1, 2, ..., 33


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Count cars at tollCount cars at tollbetween 11:00 & 1:00between 11:00 & 1:00


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Count cars at tollCount cars at tollbetween 11:00 & 1:00between 11:00 & 1:00

# Cars# Carsarrivingarriving

0, 1, 2, ..., 0, 1, 2, ...,


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Discrete Discrete Probability Probability DistributionDistribution

1.1. List of all possible [List of all possible [xx, , pp((xx)] pairs)] pairs xx = Value of random variable (outcome) = Value of random variable (outcome) pp((xx) = Probability associated with value) = Probability associated with value

2.2. Mutually exclusive (no overlap)Mutually exclusive (no overlap)

3.3. Collectively exhaustive (nothing left out)Collectively exhaustive (nothing left out)

4. 4. 0 0 pp((xx) ) 1 (or 1 (or pp((xx) ) 0) 0)

5. 5. pp((xx) = 1) = 1

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Discrete Probability Discrete Probability Distribution Distribution

ExampleExample

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ExampleExampleExperiment: Toss 2 coins. Count # tails.Experiment: Toss 2 coins. Count # tails.

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ExampleExample

Probability DistributionProbability Distribution

Values, Values, xx Probabilities, Probabilities, pp((xx))

Experiment: Toss 2 coins. Count # tails.Experiment: Toss 2 coins. Count # tails.

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ExampleExample




© 1984-1994 T/Maker Co.© 1984-1994 T/Maker Co.

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ExampleExample



00



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ExampleExample



00

11



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ExampleExample



00

11



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ExampleExample



00

11

22



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ExampleExample



00 1/4 = .251/4 = .25

11

22



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ExampleExample



00 1/4 = .251/4 = .25

11 2/4 = .502/4 = .50

22



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ExampleExample



00 1/4 = .251/4 = .25

11 2/4 = .502/4 = .50

22 1/4 = .25 1/4 = .25



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Visualizing Discrete Visualizing Discrete Probability Probability

DistributionsDistributions

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{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }

ListingListing

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{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }

ListingListing TableTable# Tails# Tails f(xf(x))

CountCountp(xp(x))

00 11 .25.2511 22 .50.5022 11 .25.25

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{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }

ListingListing TableTable

GraphGraph

# Tails# Tails f(xf(x))CountCount

p(xp(x))

00 11 .25.2511 22 .50.5022 11 .25.25

.00.00

.25.25

.50.50

00 11 22xx

p(x)p(x)

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{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }{ (0, .25), (1, .50), (2, .25) }

ListingListing TableTable

GraphGraph EquationEquation

# Tails# Tails f(xf(x))CountCount

p(xp(x))

00 11 .25.2511 22 .50.5022 11 .25.25

pp xxnn

xx nn xxpp ppxx nn xx(( ))

!!

!! (( )) !!(( ))

11

.00.00

.25.25

.50.50

00 11 22xx

p(x)p(x)

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Summary MeasuresSummary Measures

1.1. Expected valueExpected value Mean of probability distributionMean of probability distribution Weighted average of all possible valuesWeighted average of all possible values = = EE((XX)) = = xx pp((xx))

2.2. VarianceVariance Weighted average squared deviation about Weighted average squared deviation about

mean mean 22 = = EE[ ([ (xx ((xx pp((xx))

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Summary Measures Summary Measures Calculation TableCalculation Table

xx p(xp(x)) xx p(xp(x )) xx - - (x(x -- ))22((xx -- ))22 p(p(xx ))

TotalTotal ((xx -- ))22 p(p(xx ))xx p(xp(x ))

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You toss 2 coins. You’re You toss 2 coins. You’re interested in the numberinterested in the number of tails. What are the of tails. What are the expected valueexpected value & & standard deviationstandard deviation ofof this random variable, this random variable, number of tails?number of tails?

© 1984-1994 T/Maker Co.


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Expected Value & Expected Value & Variance Solution*Variance Solution*

00 .25.25 00 -1.00-1.00 1.001.00 .25.25

11 .50.50 .50.50 00 00 00

22 .25.25 .50.50 1.001.00 1.001.00 .25.25

= 1.0= 1.0 22 = .50= .50

xx p(xp(x)) xx p(xp(x )) xx - - (x(x -- ))22((xx -- ))22 p(p(xx ))

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Discrete Probability Discrete Probability Distribution FunctionDistribution Function

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FunctionFunction

1.1. Type of modelType of model Representation of some Representation of some

underlying phenomenonunderlying phenomenon

2.2. Mathematical formula Mathematical formula

3.3. Represents discrete Represents discrete random variablerandom variable

4.4. Used to get exact Used to get exact probabilitiesprobabilities

P X x

x

( )

!

x e-

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Discrete Probability Discrete Probability Distribution ModelsDistribution Models

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DiscreteProbabilityDistribution

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Binomial

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Binomial Poisson

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Binomial Poisson Other



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Binomial DistributionBinomial Distribution

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Binomial Binomial Random VariableRandom Variable

1.1. Number of ‘successes’ in a Number of ‘successes’ in a samplesample of of nn observations (trials)observations (trials)

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Binomial Binomial Random VariableRandom Variable

# Reds in 15 spins of roulette wheel# Reds in 15 spins of roulette wheel # Defective items in a batch of 5 items# Defective items in a batch of 5 items # Correct on a 33 question exam# Correct on a 33 question exam # Customers who purchase out of 100 # Customers who purchase out of 100

customers who enter storecustomers who enter store

1.1. Number of ‘successes’ in a Number of ‘successes’ in a samplesample of of nn observations (trials)observations (trials)

2.2. ExamplesExamples

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Binomial Binomial Distribution Distribution

CharacteristicsCharacteristics1.1. Sequence of Sequence of nn identical trials identical trials

2.2. Each trial has 2 outcomesEach trial has 2 outcomes ‘‘Success’ (desired outcome) or ‘failure’Success’ (desired outcome) or ‘failure’

3.3. Constant trial probability Constant trial probability

4.4. Trials are independent Trials are independent

5.5. Two different sampling methodsTwo different sampling methods InfiniteInfinite population population withwith replacement replacement FiniteFinite population population withoutwithout replacement replacement

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Binomial Probability Binomial Probability Distribution Distribution

FunctionFunction

p xn

xp q

nx n x

p px n x x n x( )!

!( )!( )FHG

IKJ

1p x

n

xp q

nx n x

p px n x x n x( )!

!( )!( )FHG

IKJ

1

pp((xx) = Probability of ) = Probability of x x ‘successes’‘successes’ in n trials in n trials

nn == SampleSample size size

pp == Probability of ‘success’Probability of ‘success’

xx == Number of ‘successes’ in Number of ‘successes’ in samplesample ( (xx = 0, 1, 2, ..., = 0, 1, 2, ..., n n))

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CharacteristicsCharacteristics

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MeanMean

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E x np

np p

( )

( )1

E x np

np p

( )

( )1

MeanMean

Standard DeviationStandard Deviation

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.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

n = 5 p = 0.1

n = 5 p = 0.5

E x np

np p

( )

( )1

E x np

np p

( )

( )1

MeanMean


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ExampleExample

Experiment: Toss 1 coin 5 times in a row. Experiment: Toss 1 coin 5 times in a row. Note # tails. What’s the probability of 3 tails?Note # tails. What’s the probability of 3 tails?

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ExampleExample

p xn

x n xp p

p

x n x( )!

!( )!( )

( )!

!( )!. ( . )

1

35

3 5 35 1 5

0

3 5 3

.3125

p xn

x n xp p

p

x n x( )!

!( )!( )

( )!

!( )!. ( . )

1

35

3 5 35 1 5

0

3 5 3

.3125

Experiment: Toss 1 coin 5 times in a row. Experiment: Toss 1 coin 5 times in a row. Note # tails. What’s the probability of 3 tails?Note # tails. What’s the probability of 3 tails?

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Using the Binomial Using the Binomial Probability TableProbability Table

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n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

Cumulative probabilities: Cumulative probabilities: pp(x (x kk) given) given n n & & pp

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n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049


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n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049


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n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049

n = 5 p

k .01 … 0.50 … .99

0 .951 … .031 … .000

1 .999 … .188 … .000

2 1.000 … .500 … .000

3 1.000 … .812 … .001

4 1.000 … .969 … .049


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Select table for Select table for n n = 5= 5

n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

Select row for Select row for k = 3k = 3

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

Select column for Select column for p p = 0.50= 0.50

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

Cumulative probability: Cumulative probability: pp(x (x 3) = .812 3) = .812

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

pp(x (x 3) = 3) = pp(x (x 3) - 3) - pp(x (x 2). Select row for 2). Select row for k = k = 22

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

Select column for Select column for p p = 0.50= 0.50

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

Cumulative probability: Cumulative probability: pp(x (x 2) = .500 2) = .500

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n = 5n = 5 pp

kk .01.01 …… 0.500.50 …… .99.99

00 .951.951 …… .031.031 …… .000.000

11 .999.999 …… .188.188 …… .000.000

22 1.0001.000 …… .500.500 …… .000.000

33 1.0001.000 …… .812.812 …… .001.001

44 1.0001.000 …… .969.969 …… .049.049

pp(x (x 3) = 3) = pp(x (x 3) - 3) - pp(x (x 2) = .812 - .500 = .312 2) = .812 - .500 = .312

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Thinking ChallengeThinking ChallengeYou’re a telemarketer selling You’re a telemarketer selling service contracts for Macy’s. service contracts for Macy’s. You’ve sold 20 in your last You’ve sold 20 in your last 100 calls (100 calls (pp = .20 = .20). If you ). If you call call 1212 people tonight, people tonight, what’s the probability ofwhat’s the probability ofA. No sales?A. No sales?

B. Exactly 2 sales?B. Exactly 2 sales?

C. At most 2 sales? C. At most 2 sales?

D. At least 2 sales?D. At least 2 sales?


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Solution*Solution*Using the Binomial Formula:Using the Binomial Formula:

AA. . pp(0) = (0) = .0687.0687 BB. . pp(2) = (2) = .2835.2835

CC. . pp(at most 2)(at most 2) = = pp(0) + (0) + pp(1) + (1) + pp(2)(2)= .0687 + .2062 + .2835= .0687 + .2062 + .2835= = .5584.5584

DD. . pp(at least 2)(at least 2) = = pp(2) + (2) + pp(3)...+ (3)...+ pp(12)(12)= 1 - [= 1 - [pp(0) + (0) + pp(1)] (1)] = 1 - .0687 - .2062= 1 - .0687 - .2062= = .7251.7251

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Poisson DistributionPoisson Distribution

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Poisson Random Poisson Random VariableVariable

1.1. Number of events that occur in an Number of events that occur in an interval interval Events Events per unitper unit

Time, length, area, spaceTime, length, area, space

2.2. ExamplesExamples # Customers arriving in 20 minutes# Customers arriving in 20 minutes # Strikes per year in the U.S.# Strikes per year in the U.S. # Defects per lot (group) of VCR’s# Defects per lot (group) of VCR’s

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Poisson ProcessPoisson Process

1.1. Constant event probabilityConstant event probability Average of 60/hr is 1/min Average of 60/hr is 1/min

for 60 1-minute intervalsfor 60 1-minute intervals

2.2. One event per intervalOne event per interval Don’t arrive togetherDon’t arrive together

3.3. Independent eventsIndependent events Arrival of 1 person does Arrival of 1 person does

not affect another’s arrivalnot affect another’s arrival © 1984-1994 T/Maker Co.

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Poisson Probability Poisson Probability Distribution Distribution

FunctionFunction

pp((xx) = Probability of ) = Probability of x x given given == Expected (mean) number of ‘successes’Expected (mean) number of ‘successes’

ee == 2.71828 (base of natural logs)2.71828 (base of natural logs)

xx == Number of ‘successes’ Number of ‘successes’ per unitper unit

pp xxxx

(( ))!!

xx ee--

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Poisson Distribution Poisson Distribution CharacteristicsCharacteristics

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MeanMean

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E x

x p xi

N

( )

( )1

E x

x p xi

N

( )

( )1

MeanMean


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.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 1 2 3 4 5

X

P(X)

.0

.2

.4

.6

0 2 4 6 8 10

X

P(X)

.0

.2

.4

.6

0 2 4 6 8 10

X

P(X)

= 0.5= 0.5

= 6= 6

E x

x p xi

N

( )

( )1

E x

x p xi

N

( )

( )1

MeanMean


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Poisson Distribution Poisson Distribution ExampleExample

Patients arrive at a Patients arrive at a hospital clinic at a rate hospital clinic at a rate of of 7272 per hour. What per hour. What is the probability of is the probability of 44 patients arriving in patients arriving in 33 minutes? minutes? © 1995 Corel Corp.

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Poisson Distribution Poisson Distribution SolutionSolution

72 per hr. = 1.2 per min. = 3.6 per 3 min. interval72 per hr. = 1.2 per min. = 3.6 per 3 min. interval

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Poisson Distribution Poisson Distribution SolutionSolution

72 per hr. = 1.2 per min. = 3.6 per 3 min. interval72 per hr. = 1.2 per min. = 3.6 per 3 min. interval

p xx

p

x

( )!

( ).

!

e

e

0.1912

-

-3.6

43 6

4

4a f

p xx

p

x

( )!

( ).

!

e

e

0.1912

-

-3.6

43 6

4

4a f

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Using the Poisson Using the Poisson Probability TableProbability Table

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Cumulative probabilitiesCumulative probabilities

xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

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xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

Select row with Select row with = 3.6 = 3.6

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xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

pp(x (x 4) = 4) = pp(x (x 4) - 4) - pp(x (x 3). 3).

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xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

pp(x (x 4) = 4) = pp(x (x 4) - 4) - pp(x (x 3). Select column 3). Select column xx = = 4.4.

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xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

pp(x (x 4) = 4) = pp(x (x 4) - 4) - pp(x (x 3) = .706 - 3) = .706 - pp(x (x 3) 3)

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xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515 .706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

Select column Select column xx = 3 = 3

pp(x (x 4) = 4) = pp(x (x 4) - 4) - pp(x (x 3) = .706 - 3) = .706 - pp(x (x 3) 3)

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pp(x (x 4) = 4) = pp(x (x 4) - 4) - pp(x (x 3) = .706 - .515 = .191 3) = .706 - .515 = .191

xx 00 …… 33 44 …… 99

.02.02 .980.980 ……:: :: :: :: :: :: ::

3.43.4 .033.033 …… .558.558 .744.744 …… .997.9973.63.6 .027.027 …… .515.515.515.515 .706.706.706.706 …… .996.9963.83.8 .022.022 …… .473.473 .668.668 …… .994.994:: :: :: :: :: :: ::

4 - 4 - 9292



You work in Quality You work in Quality Assurance for an Assurance for an investment firm. A investment firm. A clerk enters clerk enters 7575 words words per minute withper minute with 66 errors per hour. What errors per hour. What is the probability of is the probability of 00 errorserrors in a in a 255-word255-word bond transaction? bond transaction?

© 1984-1994 T/Maker Co.


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Poisson Distribution Poisson Distribution Solution: Finding Solution: Finding **

75 words/min = (75 words/min)(60 min/hr)75 words/min = (75 words/min)(60 min/hr)

= = 45004500 words/hr words/hr

6 errors/hr6 errors/hr = 6 errors/= 6 errors/45004500 words words

= = .00133.00133 errors/word errors/word

In a In a 255255-word transaction (interval):-word transaction (interval):

= (= (.00133.00133 errors/word )( errors/word )(255255 words) words)

= = .34.34 errors/255-word transaction errors/255-word transaction

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Poisson Distribution Poisson Distribution Solution: Finding Solution: Finding

p(0)*p(0)*

pp xxxx

pp

(( ))!!

(( ))!!

xx

== .7118.7118

ee--

..ee-- 3434

0000ff..3434 00aa

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ConclusionConclusion

1.1. Defined random variableDefined random variable

2.2. Computed the expected value & Computed the expected value & variance of discrete random variablesvariance of discrete random variables

3.3. Described the binomial & Poisson Described the binomial & Poisson probability distributionsprobability distributions

4.4. Calculated probabilities for binomial & Calculated probabilities for binomial & Poisson random variablesPoisson random variables

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This Class...This Class...

1.1. What was the most important thing you What was the most important thing you learned in class today?learned in class today?

2.2. What do you still have questions about?What do you still have questions about?

3.3. How can today’s class be improved?How can today’s class be improved?

Please take a moment to answer the following questions in writing:

End of Chapter

Any blank slides that follow are blank intentionally.

4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Business & Economics Discrete Random Variables...

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Transcript of 4 - 1 © 1998 Prentice-Hall, Inc. Statistics for Business & Economics Discrete Random Variables...