Slides by JOHN LOUCKS St. Edward’s University

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© 2009 Thomson South-Western. All Rights Reserved© 2009 Thomson South-Western. All Rights Reserved

Slides by

JOHNLOUCKSSt. Edward’sUniversity

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Chapter 5Chapter 5 Discrete Probability Distributions Discrete Probability Distributions

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Random VariablesRandom Variables Discrete Probability DistributionsDiscrete Probability Distributions Expected Value and VarianceExpected Value and Variance Binomial Probability DistributionBinomial Probability Distribution Poisson Probability DistributionPoisson Probability Distribution Hypergeometric Probability Hypergeometric Probability

DistributionDistribution

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A A random variablerandom variable is a numerical description of the is a numerical description of the outcome of an experiment.outcome of an experiment. A A random variablerandom variable is a numerical description of the is a numerical description of the outcome of an experiment.outcome of an experiment.

Random VariablesRandom Variables

A A discrete random variablediscrete random variable may assume either a may assume either a finite number of values or an infinite sequence offinite number of values or an infinite sequence of values.values.

A A discrete random variablediscrete random variable may assume either a may assume either a finite number of values or an infinite sequence offinite number of values or an infinite sequence of values.values.

A A continuous random variablecontinuous random variable may assume any may assume any numerical value in an interval or collection ofnumerical value in an interval or collection of intervals.intervals.

A A continuous random variablecontinuous random variable may assume any may assume any numerical value in an interval or collection ofnumerical value in an interval or collection of intervals.intervals.

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Let Let xx = number of TVs sold at the store in one day, = number of TVs sold at the store in one day,

where where xx can take on 5 values (0, 1, 2, 3, 4) can take on 5 values (0, 1, 2, 3, 4)

Let Let xx = number of TVs sold at the store in one day, = number of TVs sold at the store in one day,

where where xx can take on 5 values (0, 1, 2, 3, 4) can take on 5 values (0, 1, 2, 3, 4)

Example: JSL AppliancesExample: JSL Appliances

Discrete Random VariableDiscrete Random Variablewith a Finite Number of Valueswith a Finite Number of Values

We can count the TVs sold, and there is a finiteWe can count the TVs sold, and there is a finiteupper limit on the number that might be sold (whichupper limit on the number that might be sold (whichis the number of TVs in stock).is the number of TVs in stock).

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Let Let xx = number of customers arriving in one day, = number of customers arriving in one day,

where where xx can take on the values 0, 1, 2, . . . can take on the values 0, 1, 2, . . .

Let Let xx = number of customers arriving in one day, = number of customers arriving in one day,

where where xx can take on the values 0, 1, 2, . . . can take on the values 0, 1, 2, . . .

Discrete Random VariableDiscrete Random Variablewith an Infinite Sequence of Valueswith an Infinite Sequence of Values

We can count the customers arriving, but there isWe can count the customers arriving, but there isno finite upper limit on the number that might arrive.no finite upper limit on the number that might arrive.


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Random VariablesRandom Variables

QuestionQuestion Random Variable Random Variable xx TypeType

FamilyFamilysizesize

xx = Number of dependents = Number of dependents reported on tax returnreported on tax return

DiscreteDiscrete

Distance fromDistance fromhome to storehome to store

xx = Distance in miles from = Distance in miles from home to the store sitehome to the store site

ContinuousContinuous

Own dogOwn dogor cator cat

xx = 1 if own no pet; = 1 if own no pet; = 2 if own dog(s) only; = 2 if own dog(s) only; = 3 if own cat(s) only; = 3 if own cat(s) only; = 4 if own dog(s) and cat(s)= 4 if own dog(s) and cat(s)

DiscreteDiscrete

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The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed overdescribes how probabilities are distributed over the values of the random variable.the values of the random variable.

The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed overdescribes how probabilities are distributed over the values of the random variable.the values of the random variable.

We can describe a discrete probability distributionWe can describe a discrete probability distribution with a table, graph, or equation.with a table, graph, or equation. We can describe a discrete probability distributionWe can describe a discrete probability distribution with a table, graph, or equation.with a table, graph, or equation.

Discrete Probability DistributionsDiscrete Probability Distributions

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The probability distribution is defined by aThe probability distribution is defined by a probability functionprobability function, denoted by , denoted by ff((xx), which provides), which provides the probability for each value of the random variable.the probability for each value of the random variable.

The probability distribution is defined by aThe probability distribution is defined by a probability functionprobability function, denoted by , denoted by ff((xx), which provides), which provides the probability for each value of the random variable.the probability for each value of the random variable.

The required conditions for a discrete probabilityThe required conditions for a discrete probability function are:function are: The required conditions for a discrete probabilityThe required conditions for a discrete probability function are:function are:


ff((xx) ) >> 0 0ff((xx) ) >> 0 0

ff((xx) = 1) = 1ff((xx) = 1) = 1

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• a a tabular representationtabular representation of the probability of the probability distribution for TV sales was developed.distribution for TV sales was developed.

• Using past data on TV sales, …Using past data on TV sales, …

NumberNumber Units SoldUnits Sold of Daysof Days

00 80 80 11 50 50 22 40 40 33 10 10 44 20 20

200200

xx ff((xx)) 00 .40 .40 11 .25 .25 22 .20 .20 33 .05 .05 44 .10 .10

1.001.00

80/20080/200



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0 1 2 3 40 1 2 3 4Values of Random Variable Values of Random Variable xx (TV sales) (TV sales)Values of Random Variable Values of Random Variable xx (TV sales) (TV sales)

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Example: JSL AppliancesExample: JSL AppliancesGraphicalGraphical

representationrepresentationof probabilityof probabilitydistributiondistribution

GraphicalGraphicalrepresentationrepresentationof probabilityof probabilitydistributiondistribution

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Discrete Uniform Probability DistributionDiscrete Uniform Probability Distribution

The The discrete uniform probability distributiondiscrete uniform probability distribution is the is the simplest example of a discrete probabilitysimplest example of a discrete probability distribution given by a formula.distribution given by a formula.

The The discrete uniform probability distributiondiscrete uniform probability distribution is the is the simplest example of a discrete probabilitysimplest example of a discrete probability distribution given by a formula.distribution given by a formula.

The The discrete uniform probability functiondiscrete uniform probability function is is The The discrete uniform probability functiondiscrete uniform probability function is is

ff((xx) = 1/) = 1/nnff((xx) = 1/) = 1/nn

where:where:nn = the number of values the random = the number of values the random variable may assumevariable may assume

the values of the values of thethe

random random variablevariable

are equally are equally likelylikely

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Expected Value and VarianceExpected Value and Variance

The The expected valueexpected value, or mean, of a random variable, or mean, of a random variable is a measure of its central location.is a measure of its central location. The The expected valueexpected value, or mean, of a random variable, or mean, of a random variable is a measure of its central location.is a measure of its central location.

The The variancevariance summarizes the variability in the summarizes the variability in the values of a random variable.values of a random variable. The The variancevariance summarizes the variability in the summarizes the variability in the values of a random variable.values of a random variable.

The The standard deviationstandard deviation, , , is defined as the positive, is defined as the positive square root of the variance.square root of the variance. The The standard deviationstandard deviation, , , is defined as the positive, is defined as the positive square root of the variance.square root of the variance.

Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx))Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx))

EE((xx) = ) = = = xfxf((xx))EE((xx) = ) = = = xfxf((xx))

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expected number expected number of TVs sold in a of TVs sold in a

dayday

expected number expected number of TVs sold in a of TVs sold in a

dayday

xx ff((xx)) xfxf((xx))

00 .40 .40 .00 .00

11 .25 .25 .25 .25

22 .20 .20 .40 .40

33 .05 .05 .15 .15

44 .10 .10 .40.40

EE((xx) = 1.20) = 1.20

Expected ValueExpected Value


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00

11

22

33

44

-1.2-1.2

-0.2-0.2

0.80.8

1.81.8

2.82.8

1.441.44

0.040.04

0.640.64

3.243.24

7.847.84

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.576.576

.010.010

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.162.162

.784.784

x - x - ((x - x - ))22 ff((xx)) ((xx - - ))22ff((xx))

Variance of daily sales = Variance of daily sales = 22 = 1.660 = 1.660

xx

TVsTVssquaresquare

dd

Standard deviation of daily sales = 1.2884 TVsStandard deviation of daily sales = 1.2884 TVs

VarianceVariance


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Excel Formula Excel Formula WorksheetWorksheet

Using Excel to Compute the ExpectedUsing Excel to Compute the ExpectedValue, Variance, and Standard DeviationValue, Variance, and Standard Deviation

A B C

1 Sales Probability Sq.Dev.from Mean2 0 0.40 =(A2-$B$8)^23 1 0.25 =(A3-$B$8)^24 2 0.20 =(A4-$B$8)^25 3 0.05 =(A5-$B$8)^26 4 0.10 =(A6-$B$8)^278 Mean =SUMPRODUCT(A2:A6,B2:B6)9 Variance =SUMPRODUCT(C2:C6,B2:B6)

10 Std.Dev. =SQRT(B9)

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Excel Value Excel Value WorksheetWorksheetA B C

1 Sales Probability Sq.Dev.from Mean2 0 0.40 1.443 1 0.25 0.044 2 0.20 0.645 3 0.05 3.246 4 0.10 7.8478 Mean 1.29 Variance 1.66

10 Std.Dev. 1.2884

Using Excel to Compute the ExpectedUsing Excel to Compute the ExpectedValue, Variance, and Standard DeviationValue, Variance, and Standard Deviation

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Binomial Probability DistributionBinomial Probability Distribution

Four Properties of a Binomial ExperimentFour Properties of a Binomial Experiment

3. The probability of a success, denoted by 3. The probability of a success, denoted by pp, does, does not change from trial to trial.not change from trial to trial.3. The probability of a success, denoted by 3. The probability of a success, denoted by pp, does, does not change from trial to trial.not change from trial to trial.

4. The trials are independent.4. The trials are independent.4. The trials are independent.4. The trials are independent.

2. Two outcomes, 2. Two outcomes, successsuccess and and failurefailure, are possible, are possible on each trial.on each trial.2. Two outcomes, 2. Two outcomes, successsuccess and and failurefailure, are possible, are possible on each trial.on each trial.

1. The experiment consists of a sequence of 1. The experiment consists of a sequence of nn identical trials.identical trials.1. The experiment consists of a sequence of 1. The experiment consists of a sequence of nn identical trials.identical trials.

stationaritstationarityy

assumptioassumptionn

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Our interest is in the Our interest is in the number of successesnumber of successes occurring in the occurring in the nn trials. trials. Our interest is in the Our interest is in the number of successesnumber of successes occurring in the occurring in the nn trials. trials.

We let We let xx denote the number of successes denote the number of successes occurring in the occurring in the nn trials. trials. We let We let xx denote the number of successes denote the number of successes occurring in the occurring in the nn trials. trials.

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where:where: ff((xx) = the probability of ) = the probability of xx successes in successes in nn trials trials nn = the number of trials = the number of trials pp = the probability of success on any one trial = the probability of success on any one trial

( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability FunctionBinomial Probability Function

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( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x


Binomial Probability Binomial Probability FunctionFunction

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with x successes in with x successes in nn trials trials

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with x successes in with x successes in nn trials trials

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

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Example: Evans ElectronicsExample: Evans Electronics

Evans is concerned about a low retention Evans is concerned about a low retention rate forrate for

employees. In recent years, management has employees. In recent years, management has seen aseen a

turnover of 10% of the hourly employees turnover of 10% of the hourly employees annually.annually.

Thus, for any hourly employee chosen at Thus, for any hourly employee chosen at random,random,

management estimates a probability of 0.1 management estimates a probability of 0.1 that thethat the

person will not be with the company next year.person will not be with the company next year.

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f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

1 23!(1) (0.1) (0.9) 3(.1)(.81) .243

1!(3 1)!f

1 23!

(1) (0.1) (0.9) 3(.1)(.81) .2431!(3 1)!

f

LetLet: p: p = .10, = .10, nn = 3, = 3, xx = 1 = 1

Choosing 3 hourly employees at random, Choosing 3 hourly employees at random, what iswhat is

the probability that 1 of them will leave the the probability that 1 of them will leave the companycompany

this year?this year?


Using theUsing theprobabilityprobabilityfunctionfunction


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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



S (.9)S (.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

1111

Example: Evans ElectronicsExample: Evans ElectronicsUsing a tree diagramUsing a tree diagramUsing a tree diagramUsing a tree diagram

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Using Excel to ComputeUsing Excel to ComputeBinomial ProbabilitiesBinomial Probabilities

Excel Formula WorksheetExcel Formula Worksheet

A B

1 3 = Number of Trials (n ) 2 0.1 = Probability of Success (p ) 34 x f (x )5 0 =BINOMDIST(A5,$A$1,$A$2,FALSE)6 1 =BINOMDIST(A6,$A$1,$A$2,FALSE)7 2 =BINOMDIST(A7,$A$1,$A$2,FALSE)8 3 =BINOMDIST(A8,$A$1,$A$2,FALSE)9

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Excel Value WorksheetExcel Value Worksheet

Using Excel to ComputeUsing Excel to ComputeBinomial ProbabilitiesBinomial Probabilities

A B

1 3 = Number of Trials (n ) 2 0.1 = Probability of Success (p ) 34 x f (x )5 0 0.7296 1 0.2437 2 0.0278 3 0.0019

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Using Excel to ComputeUsing Excel to ComputeCumulative Binomial ProbabilitiesCumulative Binomial Probabilities

Excel Formula WorksheetExcel Formula Worksheet

A B

1 3 = Number of Trials (n ) 2 0.1 = Probability of Success (p ) 34 x Cumulative Probability5 0 =BINOMDIST(A5,$A$1,$A$2,TRUE)6 1 =BINOMDIST(A6,$A$1,$A$2,TRUE)7 2 =BINOMDIST(A7,$A$1,$A$2,TRUE)8 3 =BINOMDIST(A8,$A$1,$A$2,TRUE)9

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Excel Value WorksheetExcel Value Worksheet

Using Excel to ComputeUsing Excel to ComputeCumulative Binomial ProbabilitiesCumulative Binomial Probabilities

A B

1 3 = Number of Trials (n ) 2 0.1 = Probability of Success (p ) 34 x Cumulative Probability5 0 0.7296 1 0.9727 2 0.9998 3 1.0009

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(1 )np p (1 )np p

EE((xx) = ) = = = npnp

Var(Var(xx) = ) = 22 = = npnp(1 (1 pp))

Expected ValueExpected Value

VarianceVariance

Standard DeviationStandard Deviation

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3(.1)(.9) .52 employees 3(.1)(.9) .52 employees

EE((xx) = ) = = 3(.1) = .3 employees out of 3 = 3(.1) = .3 employees out of 3

Var(Var(xx) = ) = 22 = 3(.1)(.9) = .27 = 3(.1)(.9) = .27

• Expected ValueExpected Value

• VarianceVariance

• Standard DeviationStandard Deviation


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A Poisson distributed random variable is oftenA Poisson distributed random variable is often useful in estimating the number of occurrencesuseful in estimating the number of occurrences over a over a specified interval of time or spacespecified interval of time or space

A Poisson distributed random variable is oftenA Poisson distributed random variable is often useful in estimating the number of occurrencesuseful in estimating the number of occurrences over a over a specified interval of time or spacespecified interval of time or space

It is a discrete random variable that may assumeIt is a discrete random variable that may assume an an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ). (x = 0, 1, 2, . . . ). It is a discrete random variable that may assumeIt is a discrete random variable that may assume an an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ). (x = 0, 1, 2, . . . ).

Poisson Probability DistributionPoisson Probability Distribution

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Examples of a Poisson distributed random variable:Examples of a Poisson distributed random variable: Examples of a Poisson distributed random variable:Examples of a Poisson distributed random variable:

the number of knotholes in 14 linear feet ofthe number of knotholes in 14 linear feet of pine boardpine board the number of knotholes in 14 linear feet ofthe number of knotholes in 14 linear feet of pine boardpine board

the number of vehicles arriving at athe number of vehicles arriving at a toll booth in one hourtoll booth in one hour the number of vehicles arriving at athe number of vehicles arriving at a toll booth in one hourtoll booth in one hour


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Two Properties of a Poisson ExperimentTwo Properties of a Poisson Experiment

2.2. The occurrence or nonoccurrence in anyThe occurrence or nonoccurrence in any interval is independent of the occurrence orinterval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

2.2. The occurrence or nonoccurrence in anyThe occurrence or nonoccurrence in any interval is independent of the occurrence orinterval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

1.1. The probability of an occurrence is the sameThe probability of an occurrence is the same for any two intervals of equal length.for any two intervals of equal length.1.1. The probability of an occurrence is the sameThe probability of an occurrence is the same for any two intervals of equal length.for any two intervals of equal length.

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Poisson Probability FunctionPoisson Probability Function


f xex

x( )

!

f x

ex

x( )

!

where:where:

f(x) f(x) = probability of = probability of xx occurrences in an interval occurrences in an interval

= mean number of occurrences in an interval= mean number of occurrences in an interval

ee = 2.71828 In Excel use EXP(1) = 2.71828 In Excel use EXP(1)

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Example: Mercy HospitalExample: Mercy Hospital

Patients arrive at the emergency room of Patients arrive at the emergency room of MercyMercy

Hospital at the average rate of 6 per hour onHospital at the average rate of 6 per hour on

weekend evenings.weekend evenings. What is the probability of 4 arrivals in 30 What is the probability of 4 arrivals in 30 minutesminutes

on a weekend evening?on a weekend evening?

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4 33 (2.71828)(4) .1680

4!f

4 33 (2.71828)

(4) .16804!

f

= 6/hour = 3/half-hour, = 6/hour = 3/half-hour, xx = 4 = 4

Example: Mercy HospitalExample: Mercy Hospital Using theUsing theprobabilityprobabilityfunctionfunction


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Poisson Probabilities

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Number of Arrivals in 30 Minutes

Pro

bab

ilit

y

actually, actually, the the

sequencesequencecontinues:continues:11, 12, …11, 12, …


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A property of the Poisson distribution is thatA property of the Poisson distribution is thatthe mean and variance are equal.the mean and variance are equal.

A property of the Poisson distribution is thatA property of the Poisson distribution is thatthe mean and variance are equal.the mean and variance are equal.

= = 22

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Variance for Number of ArrivalsVariance for Number of Arrivals

During 30-Minute PeriodsDuring 30-Minute Periods

= = 22 = 3 = 3 = = 22 = 3 = 3


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Poisson Probability Distributions

The limiting form of the binomial distribution where the probability of success is small and n is large is called the Poisson probability distribution

The binomial distribution becomes more skewed to the right (positive) as the probability of success become smaller

This distribution describes the number of times some event occurs during a specified interval (time, distance, area, or volume)

Examples of use: Distribution of errors is data entry Number of scratches in car panels

Poisson Probability Experiment1. The Random Variable is the number of times (counting) some event occurs during a

Defined Interval Random Variable:

The Random Variable can assume an infinite number of values, however the probability becomes very small after the first few occurrences (successes)

Defined Interval: The Defined Interval some kind of “Continuum” such as:

Misspelled words per page ( continuum = per page) Calls per two hour period ( continuum = per two hour period) Vehicles per day ( continuum = per day) Goals per game ( continuum = per game) Lost bags per flight ( continuum = per flight) Defaults per 30 year mortgage period ( continuum = per 30 year mortgage

period ) Interval = Continuum

2. The probability of the event is proportional to the size of the interval (the longer the interval, the larger the probability)

3. The intervals do not overlap and are independent (the number of occurrences in 1 interval does not affect the other intervals

4. When the probability of success is very small and n is large, the Poisson distribution is a limiting form of the binomial probability distribution (“Law of Improbable Events”)

4040

Poisson Probability

4141

Variable Descriptionmu mean = pi*npi Probability of successn Sample Size = # of TrialsX Random Variable = # of successes

e2.71828182845905 use the formula =EXP(1) to generate the number e

variance variance = mustandard deviation Square Root of Variance or (pi*n)^(1/2)

In Excel use the POISSON functionIn Excel use the POISSON function

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Poisson Probability Distributions

Variance = Mu Always positive skew (most of the successes are in the

first few counts (0,1,2,3…) As mu becomes larger, Poisson becomes more

symmetrical We can calculate Probability with only knowledge of Mu

and X. Although using the Binomial Probability Distribution is

still technically correct, we can use the Poisson Probability Distribution to estimate the Binomial Probability Distribution when n is large and pi is small Why? Because as n gets large and pi gets small, the Poisson

and Binomial converge

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Hypergeometric Probability DistributionHypergeometric Probability Distribution

The The hypergeometric distributionhypergeometric distribution is closely related is closely related to the binomial distribution. to the binomial distribution. The The hypergeometric distributionhypergeometric distribution is closely related is closely related to the binomial distribution. to the binomial distribution.

However, for the hypergeometric distribution:However, for the hypergeometric distribution: However, for the hypergeometric distribution:However, for the hypergeometric distribution:

the trials are not independent, andthe trials are not independent, and the trials are not independent, andthe trials are not independent, and

the probability of success changes from trialthe probability of success changes from trial to trial. to trial. the probability of success changes from trialthe probability of success changes from trial to trial. to trial.

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Hypergeometric Probability FunctionHypergeometric Probability Function


n

N

xn

rN

x

r

xf )(

n

N

xn

rN

x

r

xf )( for 0 for 0 << xx << rr

where: where: ff((xx) = probability of ) = probability of xx successes in successes in nn trials trials nn = number of trials = number of trials NN = number of elements in the population = number of elements in the population rr = number of elements in the population = number of elements in the population

labeled successlabeled success

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Hypergeometric Probability FunctionHypergeometric Probability Function


( )

r N r

x n xf x

N

n

( )

r N r

x n xf x

N

n

for 0 for 0 << xx << rr

number of waysnumber of waysxx successes can be selected successes can be selectedfrom a total of from a total of rr successes successes

in the populationin the population

number of waysnumber of waysnn – – x x failures can be selectedfailures can be selectedfrom a total of from a total of NN – – rr failures failures

in the populationin the population

number of waysnumber of waysa sample of size a sample of size n n can be selectedcan be selected

from a population of size from a population of size NN

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Bob Neveready has removed two dead batteriesBob Neveready has removed two dead batteriesfrom a flashlight and inadvertently mingled themfrom a flashlight and inadvertently mingled themwith the two good batteries he intended aswith the two good batteries he intended asreplacements. The four batteries look identical.replacements. The four batteries look identical.

Example: Neveready’s Example: Neveready’s BatteriesBatteries

Bob now randomly selects two of the fourBob now randomly selects two of the fourbatteries. What is the probability he selects batteries. What is the probability he selects

the twothe twogood batteries?good batteries?

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2 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

2 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:where: xx = 2 = number of = 2 = number of goodgood batteries selected batteries selected

nn = 2 = number of batteries selected = 2 = number of batteries selected NN = 4 = number of batteries in total = 4 = number of batteries in total rr = 2 = number of = 2 = number of goodgood batteries in total batteries in total




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A B

1 2 Number of Successes (x ) 2 2 Number of Trials (n ) 3 2 Number of Elements in the Population Labeled Success (r ) 4 4 Number of Elements in the Population (N ) 5 6 f (x ) =HYPGEOMDIST(A1,A2,A3,A4)7

Using Excel to ComputeUsing Excel to ComputeHypergeometric ProbabilitiesHypergeometric Probabilities

Formula WorksheetFormula Worksheet

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A B

1 2 Number of Successes (x ) 2 2 Number of Trials (n ) 3 2 Number of Elements in the Population Labeled Success (r ) 4 4 Number of Elements in the Population (N ) 5 6 f (x ) 0.16677

Value WorksheetValue Worksheet

Using Excel to ComputeUsing Excel to ComputeHypergeometric ProbabilitiesHypergeometric Probabilities

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( )r

E x nN

( )r

E x nN

2( ) 11

r r N nVar x n

N N N

2( ) 11

r r N nVar x n

N N N

MeanMean

VarianceVariance

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22 1

4

rnN

22 1

4

rnN

2 2 2 4 2 12 1 .333

4 4 4 1 3

2 2 2 4 2 12 1 .333

4 4 4 1 3

• MeanMean

• VarianceVariance


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Consider a hypergeometric distribution with Consider a hypergeometric distribution with nn trials trials and let and let pp = ( = (rr//nn) denote the probability of a success) denote the probability of a success on the first trial.on the first trial.

Consider a hypergeometric distribution with Consider a hypergeometric distribution with nn trials trials and let and let pp = ( = (rr//nn) denote the probability of a success) denote the probability of a success on the first trial.on the first trial.

If the population size is large, the term (If the population size is large, the term (NN – – nn)/()/(NN – 1) – 1) approaches 1.approaches 1. If the population size is large, the term (If the population size is large, the term (NN – – nn)/()/(NN – 1) – 1) approaches 1.approaches 1.

The expected value and variance can be writtenhe expected value and variance can be written EE((xx) = ) = npnp and and VarVar((xx) = ) = npnp(1 – (1 – pp).). The expected value and variance can be writtenhe expected value and variance can be written EE((xx) = ) = npnp and and VarVar((xx) = ) = npnp(1 – (1 – pp).).

Note that these are the expressions for the expectedhe expected value and variance of a binomial distribution.value and variance of a binomial distribution. Note that these are the expressions for the expectedhe expected value and variance of a binomial distribution.value and variance of a binomial distribution.

continuedcontinued

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When the population size is large, a hypergeometricWhen the population size is large, a hypergeometric distribution can be approximated by a binomialdistribution can be approximated by a binomial distribution with distribution with nn trials and a probability of success trials and a probability of success pp = ( = (rr//NN). ).

When the population size is large, a hypergeometricWhen the population size is large, a hypergeometric distribution can be approximated by a binomialdistribution can be approximated by a binomial distribution with distribution with nn trials and a probability of success trials and a probability of success pp = ( = (rr//NN). ).

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Hypergeometric Probability Distribution Requirements for a Hypergeometric Experiment?

1. An outcome on each trial of an experiment is classified into 1 of 2 mutually exclusive categories: Success or Failure

2. The random variable is the number of counted successes in a fixed number of trials

3. The trials are not independent (For each new trial, the sample space changes

4. We assume that we sample from a finite population (number of items in population is known) without replacement and n/N > 0.05. So, the probability of a success changes for each trial

N = Number Of Items In Population n = Number Of Items In Sample = Number Of Trials S = Number Of Successes In Population X = Number Of Successes In Sample

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Hypergeometric Probability Distribution If the selected items are not returned to the population and n/N <0.05,

then the Binomial Distribution can be used as a close approximation In Excel use the HYPGEOMDIST function

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End of Chapter 5End of Chapter 5

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Slides by JOHN LOUCKS St. Edward’s University