slides prepared by john s. loucks st. edward’s university

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© 2003 South-Western/Thomson Learning© 2003 South-Western/Thomson LearningTMTM

Slides Prepared bySlides Prepared byJOHN S. LOUCKSJOHN S. LOUCKS

St. Edward’s UniversitySt. Edward’s University

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Chapter 5Chapter 5 Discrete Probability Distributions Discrete Probability Distributions

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 DistributionHypergeometric Probability Distribution

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Random VariablesRandom Variables

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

A random variable can be classified as being A random variable can be classified as being either discrete or continuous depending on the either discrete or continuous depending on the numerical values it assumes.numerical values it assumes.

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

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

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Example: JSL AppliancesExample: JSL Appliances

Discrete random variable with a finite number of Discrete random variable with a finite number of valuesvalues

Let Let xx = number of TV sets sold at the store in one = number of TV sets sold at the store in one dayday

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

Discrete random variable with an infinite sequence Discrete random variable with an infinite sequence of valuesof values

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, . . .

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

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Discrete Probability DistributionsDiscrete Probability Distributions

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

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

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

ff((xx) ) >> 0 0

ff((xx) = 1) = 1 We can describe a discrete probability We can describe a discrete probability

distribution with a table, graph, or equation.distribution with a table, graph, or equation.

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Using past data on TV sales (below left), a Using past data on TV sales (below left), a tabular representation of the probability tabular representation of the probability distribution for TV sales (below right) was distribution for TV sales (below right) was developed.developed.

NumberNumber

Units SoldUnits Sold of Daysof Days xx ff((xx))

00 80 80 0 0 .40 .40

11 50 50 1 1 .25 .25

22 40 40 2 2 .20 .20

33 10 10 3 3 .05 .05

44 2020 4 4 .10 .10

200200 1.00 1.00


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Graphical Representation of the Probability Graphical Representation of the Probability DistributionDistribution

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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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babili

ty

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Discrete Uniform Probability DistributionDiscrete Uniform Probability Distribution

The The discrete uniform probability distributiondiscrete uniform probability distribution is is the simplest example of a discrete probability the simplest 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

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

where:where:

nn = the number of values the = the number of values the randomrandom

variable may assumevariable may assume Note that the values of the random variable Note that the values of the random variable

are equally likely.are equally likely.

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Expected Value and VarianceExpected Value and Variance

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

variable:variable:

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

The The variancevariance summarizes the variability in the summarizes the variability in the values of a random variable.values of a random variable.• Variance of a discrete random variable:Variance of a discrete random variable:

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

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

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Expected Value of a Discrete Random VariableExpected Value of a Discrete Random Variable

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

The expected number of TV sets sold in a day The expected number of TV sets sold in a day is 1.2is 1.2

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Variance and Standard DeviationVariance and Standard Deviation

of a Discrete Random Variableof a Discrete Random Variable

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

00 -1.2-1.2 1.44 1.44 .40.40 .576 .57611 -0.2-0.2 0.04 0.04 .25.25 .010 .01022 0.8 0.8 0.64 0.64 .20.20 .128 .12833 1.8 1.8 3.24 3.24 .05.05 .162 .16244 2.8 2.8 7.84 7.84 .10.10 .784 .784

1.660 = 1.660 =

The variance of daily sales is 1.66 TV sets The variance of daily sales is 1.66 TV sets squaredsquared.. The standard deviation of sales is 1.2884 TV The standard deviation of sales is 1.2884 TV sets.sets.


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Using Excel to Compute the Expected Using Excel to Compute the Expected Value, Variance, and Standard DeviationValue, Variance, and Standard Deviation

Formula WorksheetFormula Worksheet

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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Value WorksheetValue Worksheet

Using Excel to Compute the Expected Using Excel to Compute the Expected Value, Variance, and Standard DeviationValue, Variance, and Standard Deviation

A 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

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Binomial Probability DistributionBinomial Probability Distribution

Properties of a Binomial ExperimentProperties of a Binomial Experiment

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

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

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

• The trials are independent.The trials are independent.

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Example: Evans ElectronicsExample: Evans Electronics


Evans is concerned about a low retention Evans is concerned about a low retention rate for employees. On the basis of past rate for employees. On the basis of past experience, management has seen a turnover experience, management has seen a turnover of 10% of the hourly employees annually. of 10% of the hourly employees annually. Thus, for any hourly employees chosen at Thus, for any hourly employees chosen at random, management estimates a probability random, management estimates a probability of 0.1 that the person will not be with the of 0.1 that the person will not be with the company next year.company next year.

Choosing 3 hourly employees a random, Choosing 3 hourly employees a random, what is the probability that 1 of them will leave what is the probability that 1 of them will leave the company this year?the company this year?

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

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Binomial Probability FunctionBinomial Probability Function

where:where:

ff((xx) = the probability of ) = the probability of xx successes in successes in nn trialstrials

nn = the number of trials = the number of trials

pp = the probability of success on any = the probability of success on any one trialone trial

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

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Using the Binomial Probability FunctionUsing the Binomial Probability Function

= (3)(0.1)(0.81)= (3)(0.1)(0.81)

= .243 = .243

f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

f ( )!

!( )!( . ) ( . )1

31 3 1

0 1 0 91 2

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Using the Tables of Binomial ProbabilitiesUsing the Tables of Binomial Probabilities

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

pn x .10 .15 .20 .25 .30 .35 .40 .45 .503 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

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Using a Tree DiagramUsing a Tree Diagram


FirstWorker FirstWorker

SecondWorkerSecondWorker

ThirdWorkerThird

WorkerValueof x

Valueof x Probab.Probab.

Leaves (.1)Leaves (.1)

Stays (.9)Stays (.9)

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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

11

.0810.0810

.0810.0810

.0810.0810

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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 =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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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


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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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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Expected ValueExpected Value

EE((xx) = ) = = = npnp

VarianceVariance

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

Standard DeviationStandard Deviation

SD( ) ( )x np p 1SD( ) ( )x np p 1

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• Expected ValueExpected Value

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

• VarianceVariance

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

• Standard DeviationStandard Deviationemployees 52.)9)(.1(.3)(SD x employees 52.)9)(.1(.3)(SD x

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Poisson Probability DistributionPoisson Probability Distribution

Properties of a Poisson ExperimentProperties of a Poisson Experiment

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

• The occurrence or nonoccurrence in any The occurrence or nonoccurrence in any interval is independent of the occurrence or interval is independent of the occurrence or nonoccurrence in any other interval.nonoccurrence in any other interval.

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Poisson Probability DistributionPoisson Probability Distribution

Poisson Probability FunctionPoisson Probability Function

where:where:

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

= mean number of occurrences in an = mean number of occurrences in an intervalinterval

ee = 2.71828 = 2.71828

f xex

x( )

!

f x

ex

x( )

!

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Example: Mercy HospitalExample: Mercy Hospital

Using the Poisson Probability FunctionUsing the Poisson Probability Function

Patients arrive at the emergency room of Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a probability of 4 arrivals in 30 minutes on a weekend evening?weekend evening?

= 6/hour = 3/half-hour, = 6/hour = 3/half-hour, xx = 4 = 4f ( )

( . )!

.43 2 71828

41680

4 3

f ( )( . )

!.4

3 2 718284

16804 3

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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

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

Example: Mercy HospitalExample: Mercy Hospital

Using the Tables of Poisson ProbabilitiesUsing the Tables of Poisson Probabilities

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Using Excel to ComputeUsing Excel to ComputePoisson ProbabilitiesPoisson Probabilities


A B

1 3 = Mean No. of Occurrences ( ) 2

3Number of Arrivals (x ) Probability f (x )

4 0 =POISSON(A4,$A$1,FALSE)5 1 =POISSON(A5,$A$1,FALSE)6 2 =POISSON(A6,$A$1,FALSE)7 3 =POISSON(A7,$A$1,FALSE)8 4 =POISSON(A8,$A$1,FALSE)9 5 =POISSON(A9,$A$1,FALSE)

10 6 =POISSON(A10,$A$1,FALSE)

Etc. Etc.

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Using Excel to ComputeUsing Excel to ComputePoisson ProbabilitiesPoisson Probabilities

A B


3Number of Arrivals (x ) Probability f (x )

4 0 0.04985 1 0.14946 2 0.22407 3 0.22408 4 0.16809 5 0.1008

10 6 0.0504

Etc. Etc.

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Using Excel to ComputeUsing Excel to ComputeCumulative Poisson ProbabilitiesCumulative Poisson Probabilities


Etc. Etc.

A B


3Number of Arrivals (x ) Cumulative Probability

4 0 =POISSON(A4,$A$1,TRUE)5 1 =POISSON(A5,$A$1,TRUE)6 2 =POISSON(A6,$A$1,TRUE)7 3 =POISSON(A7,$A$1,TRUE)8 4 =POISSON(A8,$A$1,TRUE)9 5 =POISSON(A9,$A$1,TRUE)

10 6 =POISSON(A10,$A$1,TRUE)

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Using Excel to ComputeUsing Excel to ComputeCumulative Poisson ProbabilitiesCumulative Poisson Probabilities

Etc. Etc.

A B


3Number of Arrivals (x ) Cumulative Probability

4 0 0.04985 1 0.19916 2 0.42327 3 0.64728 4 0.81539 5 0.9161

10 6 0.9665

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Hypergeometric Probability DistributionHypergeometric Probability Distribution

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

With the hypergeometric distribution, the trials With the hypergeometric distribution, the trials are not independent, and the probability of are not independent, and the probability of success changes from trial to trial.success changes from trial to trial.

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n

N

xn

rN

x

r

xf )(

n

N

xn

rN

x

r

xf )(

Hypergeometric Probability FunctionHypergeometric Probability Function

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 = number of elements in the populationpopulation

rr = number of elements in the = number of elements in the populationpopulation

labeled successlabeled success

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Example: NevereadyExample: Neveready


Bob Neveready has removed two dead Bob Neveready has removed two dead batteries from a flashlight and inadvertently batteries from a flashlight and inadvertently mingled them with the two good batteries he mingled them with the two good batteries he intended as replacements. The four batteries intended as replacements. The four batteries look identical.look identical.

Bob now randomly selects two of the Bob now randomly selects two of the four batteries. What is the probability he four batteries. What is the probability he selects the two good batteries?selects the two good batteries?

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Example: NevereadyExample: Neveready


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

167.6

1

!2!2

!4

!2!0

!2

!0!2

!2

2

4

0

2

2

2

)(

n

N

xn

rN

x

r

xf 167.6

1

!2!2

!4

!2!0

!2

!0!2

!2

2

4

0

2

2

2

)(

n

N

xn

rN

x

r

xf

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Using Excel to ComputeUsing Excel to ComputeHypergeometric ProbabilitiesHypergeometric Probabilities


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

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Using Excel to ComputeUsing Excel to ComputeHypergeometric ProbabilitiesHypergeometric Probabilities

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

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End of Chapter 5End of Chapter 5

slides prepared by john s. loucks st. edward’s university

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