4 - 1 © 2001 prentice-Hall, Inc. Behavioral Statistics Discrete Random Variables Chapter 4.

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4 - 4 - 1 1 © 2001 © 2001 prentice-Hall, prentice-Hall, Inc. Inc. Behavioral Statistics Behavioral Statistics Discrete Random Variables Discrete Random Variables Chapter 4 Chapter 4

Transcript of 4 - 1 © 2001 prentice-Hall, Inc. Behavioral Statistics Discrete Random Variables Chapter 4.

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Behavioral StatisticsBehavioral Statistics

Discrete Random VariablesDiscrete Random VariablesChapter 4Chapter 4

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Learning ObjectivesLearning Objectives

1.1. Distinguish Between the Two Types of Distinguish Between the Two Types of Random VariablesRandom Variables

2.2. Compute the Expected Value & Compute the Expected Value & Variance of Discrete Random VariablesVariance of Discrete Random Variables

3.3. Describe the Binomial and PoissonDescribe the Binomial and Poisson

4.4. Calculate Probabilities for Discrete Calculate Probabilities for Discrete Random VariablesRandom 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?

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Data TypesData Types

Data

Numerical Qualitative

Discrete Continuous

Data

Numerical Qualitative

Discrete Continuous

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Discrete Random Discrete Random VariablesVariables

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Discrete Discrete Random VariableRandom Variable

1. 1. Random VariableRandom Variable A Numerical Outcome of an ExperimentA Numerical Outcome of an Experiment Example: Number of Tails in 2 Coin TossesExample: Number of Tails in 2 Coin Tosses

2. Discrete Random Variable 2. Discrete Random Variable Whole Number (0, 1, 2, 3 etc.)Whole Number (0, 1, 2, 3 etc.) Obtained by CountingObtained by Counting 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

ExperimentExperiment RandomRandomVariableVariable

PossiblePossibleValuesValues

Make 100 Sales CallsMake 100 Sales Calls # # SalesSales 0, 1, 2, ..., 1000, 1, 2, ..., 100

Inspect 70 RadiosInspect 70 Radios # # DefectiveDefective 0, 1, 2, ..., 700, 1, 2, ..., 70

Answer 33 QuestionsAnswer 33 Questions # # CorrectCorrect 0, 1, 2, ..., 330, 1, 2, ..., 33

Count Cars at TollCount Cars at TollBetween 11:00 & 1:00Between 11:00 & 1:00

# # CarsCarsArrivingArriving

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 1

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

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Discrete Probability Discrete Probability Distribution Distribution

ExampleExample

Probability DistributionProbability Distribution

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

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

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

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

Experiment: Toss 2 Coins. Count # Tails.Experiment: Toss 2 Coins. Count # Tails.

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Visualizing Discrete Visualizing Discrete Probability Probability

DistributionsDistributions

{ (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

# # TailsTails 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 Weighted Average Squared Deviation

about Mean about 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 - - ((xx --))22((xx --))22 p(p(xx ))

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

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Thinking ChallengeThinking Challenge

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?

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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 - - ((xx --))22((xx --))22 p(p(xx ))

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Discrete Probability Discrete Probability Distribution FunctionDistribution Function

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Discrete Probability Discrete Probability Distribution Distribution

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

B in om ia l P o isson

D isc re teP rob ab ilityD is trib u tion

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Binomial DistributionBinomial Distribution

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Discrete Probability Discrete Probability Distribution ModelsDistribution Models

B in om ia l P o isson

D isc re teP rob ab ilityD is trib u tion

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Binomial Binomial DistributionDistribution

1.1. Number of ‘Successes’ in a Number of ‘Successes’ in a SampleSample of of n Observations (Trials)n Observations (Trials) # 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

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Binomial Binomial Distribution Distribution PropertiesProperties

1.1. Two Different Sampling MethodsTwo Different Sampling Methods InfiniteInfinite Population Population WithoutWithout Replacement Replacement FiniteFinite Population Population WithWith Replacement Replacement

2.2. Sequence of Sequence of nn Identical Trials Identical Trials

3.3. Each Trial Has 2 OutcomesEach Trial Has 2 Outcomes ‘‘Success’ (Desired Outcome) or ‘Failure’Success’ (Desired Outcome) or ‘Failure’

4.4. Constant Trial ProbabilityConstant Trial Probability

5.5. Trials Are Independent Trials Are Independent

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Binomial Probability Binomial Probability Distribution Distribution

FunctionFunction

xnxxnx ppxnx

nqp

x

nxp

)1(

)!(!

!)( xnxxnx pp

xnx

nqp

x

nxp

)1(

)!(!

!)(

pp((xx) = Probability of ) = Probability of x x ‘Successes’‘Successes’

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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Binomial Probability Binomial Probability Distribution Distribution

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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Binomial Probability Binomial Probability Table (Portion)Table (Portion)

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 ProbabilitiesCumulative Probabilities

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Binomial Binomial Distribution Distribution

CharacteristicsCharacteristics

.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

Standard DeviationStandard Deviation

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Binomial Binomial Distribution Distribution

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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Binomial Binomial Distribution Distribution

Solution*Solution*Using the Binomial Tables:Using the Binomial Tables:

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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Discrete Probability Discrete Probability Distribution ModelsDistribution Models

B in om ia l P o isson

D isc re teP rob ab ilityD is trib u tion

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Poisson DistributionPoisson Distribution

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 Constant Event ProbabilityProbability

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

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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

.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

Standard DeviationStandard Deviation

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Poisson Distribution Poisson Distribution ExampleExample

Customers arrive at a Customers arrive at a rate of rate of 7272 per hour. per hour. What is the What is the probability of probability of 44 customers arriving in customers arriving in 33 minutes? minutes? © 1995 Corel Corp.

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Poisson Distribution Poisson Distribution SolutionSolution

72 72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. IntervalPer Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

0.1912

!4

e6.3)4(

!

e)(

.63-4

-

p

xxp

x

0.1912

!4

e6.3)4(

!

e)(

.63-4

-

p

xxp

x

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Poisson Probability Poisson Probability Table (Portion)Table (Portion)

x 0 … 3 4 … 9

.02 .980 …: : : : : : :

3.4 .033 … .558 .744 … .9973.6 .027 … .515 .706 … .9963.8 .022 … .473 .668 … .994: : : : : : :

x 0 … 3 4 … 9

.02 .980 …: : : : : : :

3.4 .033 … .558 .744 … .9973.6 .027 … .515 .706 … .9963.8 .022 … .473 .668 … .994: : : : : : :

Cumulative ProbabilitiesCumulative Probabilities

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Thinking ChallengeThinking Challenge

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?

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

75 75 words/min = (75 words/min)(60 min/hr)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 )(255 255 words) words)

= .34 errors/255-word transaction = .34 errors/255-word transaction

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Poisson Distribution Poisson Distribution Solution: Finding Solution: Finding

p(0)*p(0)*

0.7118

!0

e34.)4(

!

e)(

-.340

-

p

xxp

x

0.7118

!0

e34.)4(

!

e)(

-.340

-

p

xxp

x

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ConclusionConclusion

1.1. Distinguished Between the Two Types of Distinguished Between the Two Types of Random VariablesRandom Variables

2.2. Computed the Expected Value & Variance Computed the Expected Value & Variance of Discrete Random Variablesof Discrete Random Variables

3.3. Described the Binomial and Poisson Described the Binomial and Poisson DistributionsDistributions

4.4. Calculated Probabilities for Discrete Calculated Probabilities for Discrete Random VariablesRandom Variables

End of Chapter

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