Chapter 5: Probability Distributions: Discrete Probability Distributions

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University of Minnesota-Duluth, Econ-2030 (Dr. Tadesse)University of Minnesota-Duluth, Econ-2030 (Dr. Tadesse)

Chapter 5: Chapter 5: Probability Distributions: Probability Distributions:

Discrete Probability Discrete Probability DistributionsDistributions

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Learning ObjectivesLearning Objectives

Identifying Types of Discrete Probability Distribution Identifying Types of Discrete Probability Distribution and their Respective Functional Representations and their Respective Functional Representations

Calculating the Mean and Variance of a Discrete Calculating the Mean and Variance of a Discrete Random Variable with each of the different discrete Random Variable with each of the different discrete probability distribution identified.probability distribution identified.

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Random VariableRandom Variable

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Random VariablesRandom Variables

Any variable that is used to represent the outcomes any Any variable that is used to represent the outcomes any experiment of interest to us is called Random Variable. experiment of interest to us is called Random Variable.

A random variable can assume (take) any value (Positive, A random variable can assume (take) any value (Positive, negative, zero; finite, infinite; continuous and discrete negative, zero; finite, infinite; continuous and discrete values).values).

Depending upon the values it takes, we can identify two Depending upon the values it takes, we can identify two types of random variables:types of random variables:

1.1. Discrete VariablesDiscrete Variables

2.2. Continuous Variables.Continuous Variables.

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Note that both types of vNote that both types of variables canariables can assume either assume either a finite number of valuesa finite number of values or or an infinite sequence of valuesan infinite sequence of values..Note that both types of vNote that both types of variables canariables can assume either assume either a finite number of valuesa finite number of values or or an infinite sequence of valuesan infinite sequence of values..

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Let Let xx = number of TVs sold at a given store in one day. = number of TVs sold at a given store in one day.

The number of TV units that can be sold in a given day is The number of TV units that can be sold in a given day is

finite. It is also discrete: (0, 1, 2, 3, 4). finite. It is also discrete: (0, 1, 2, 3, 4).

XX can be considered as Discrete Random Variable can be considered as Discrete Random Variable

Let Let xx = number of TVs sold at a given store in one day. = number of TVs sold at a given store in one day.

The number of TV units that can be sold in a given day is The number of TV units that can be sold in a given day is

finite. It is also discrete: (0, 1, 2, 3, 4). finite. It is also discrete: (0, 1, 2, 3, 4).

XX can be considered as Discrete Random Variable can be considered as Discrete Random Variable

Example: JSL AppliancesExample: JSL Appliances

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

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Let Let YY = number of customers arriving in a store in one day. = number of customers arriving in a store in one day.

YY can take on the values 0, 1, 2, . . . can take on the values 0, 1, 2, . . .

Let Let YY = number of customers arriving in a store in one day. = number of customers arriving in a store in one day.

YY can take on the values 0, 1, 2, . . . can take on the values 0, 1, 2, . . .

Example: JSL AppliancesExample: JSL Appliances

Discrete random variable with an Discrete random variable with an infiniteinfinite sequence sequence of values of values

We can count the customers arriving. However, there is noWe can count the customers arriving. However, there is nofinite upper limit on the number that might arrive.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 a storehome to a store

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

ContinuousContinuous

Owning dog or Owning dog or catcat

ZZ = 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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Probability DistributionsProbability Distributions

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The The probability distributionprobability distribution for a random variable is a distribution for a random variable is a distributionthat describes the values that the random variable of interest takes.that describes the values that the random variable of interest takes. The The probability distributionprobability distribution for a random variable is a distribution for a random variable is a distributionthat describes the values that the random variable of interest takes.that describes the values that the random variable of interest takes.


It is defined by a It is defined by a probability functionprobability function, and is denoted by , and is denoted by ff((xx))---the probability of the values of the random variable.---the probability of the values of the random variable. It is defined by a It is defined by a probability functionprobability function, and is denoted by , and is denoted by ff((xx))---the probability of the values of the random variable.---the probability of the values of the random variable.

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For any probability function the following conditions must beFor any probability function the following conditions must besatisfied:satisfied:For any probability function the following conditions must beFor any probability function the following conditions must besatisfied:satisfied:


1.1. ff((xx) ) >> 0 01.1. ff((xx) ) >> 0 0

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

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Discrete Probability DistributionsDiscrete Probability Distributions

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A A Discrete ProbabilityDiscrete Probability Distribution is a tabular, graphic or Distribution is a tabular, graphic orFunctional representation of a Random Variable with discrete Functional representation of a Random Variable with discrete outcomes that follows the principle of probability distribution outcomes that follows the principle of probability distribution

A A Discrete ProbabilityDiscrete Probability Distribution is a tabular, graphic or Distribution is a tabular, graphic orFunctional representation of a Random Variable with discrete Functional representation of a Random Variable with discrete outcomes that follows the principle of probability distribution outcomes that follows the principle of probability distribution

Discrete Probability DistributionsDiscrete Probability Distributions

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

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

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• Using past data on sales, Using past data on sales, a tabular representation of a tabular representation of the probability distribution for TV sales was developed.the probability distribution for TV sales was developed.

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

Discrete Probability Distributions--Discrete Probability Distributions--ExampleExample

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1. Uniform Distribution1. Uniform Distribution

2. Binomial Distribution2. Binomial Distribution

3. Poisson Distribution3. Poisson Distribution

There are four basic types of discrete probability There are four basic types of discrete probability distributions:distributions:

4. Hyper-Geometric Distribution4. Hyper-Geometric Distribution

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5.1) Discrete Uniform Probability 5.1) Discrete Uniform Probability DistributionDistribution

The The probability distributionprobability distribution of a discrete probability of a discrete probability distribution is given by the following formula.distribution is given by the following formula. The The probability distributionprobability distribution of a discrete probability of a discrete probability distribution is given by the following formula.distribution is given by the following formula.

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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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

5.2) Binomial Distribution5.2) Binomial Distribution

The probability distribution of a Binomial The probability distribution of a Binomial Distribution is given by the following functionDistribution is given by the following function

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The probability of a Poisson Distribution is The probability of a Poisson Distribution is given by the following functiongiven by the following function

5.3) Poisson Distribution5.3) Poisson Distribution

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 = 2.71828

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The probability of a Hyper-geometric The probability of a Hyper-geometric Distribution is given by the following FunctionDistribution is given by the following Function

5.4) Hypergeometric Distribution5.4) Hypergeometric Distribution

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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5.1) Discrete Uniform Probability 5.1) Discrete Uniform Probability DistributionDistribution

The The probability distributionprobability distribution of a discrete probability of a discrete probability distribution is given by the following formula.distribution is given by the following formula. The The probability distributionprobability distribution of a discrete probability of a discrete probability distribution is given by the following formula.distribution is given by the following formula.

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 (Mean) and Variance of Expected Value (Mean) and Variance of Discrete Probability Distributions Discrete Probability Distributions

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Expected Value (Mean) for …Expected Value (Mean) for ……..Discrete Uniform Distribution…..Discrete Uniform Distribution

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.

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

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Variance and Standard Deviation of Discrete Variance and Standard Deviation of Discrete Uniform Probability DistributionUniform Probability Distribution

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

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

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

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• The likelihood of selling TV sets in any given day is considered The likelihood of selling TV sets in any given day is considered equally likely (Uniform). The following table summarizes sales equally likely (Uniform). The following table summarizes sales data on the past 200 days.data on the past 200 days.

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

Example-TV Sales in a Given Example-TV Sales in a Given StoreStore

Given this data what is the Average Number of TVs sold Given this data what is the Average Number of TVs sold in a day?in a day?

xx FF((xx)) 00 .40 .40 11 .65 .65 22 .85 .85 33 .90 .90 44 1.00 1.00

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Given the data, we can find the Expected Value (Mean Number) of Given the data, we can find the Expected Value (Mean Number) of TVs sold in a day as followsTVs sold in a day as follows

expected number expected number of TVs sold in a dayof TVs sold in a day

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 Value (MEAN) and VarianceExpected Value (MEAN) and Variance

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

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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

.40.40

.25.25

.20.20

.05.05

.10.10

.576.576

.010.010

.128.128

.162.162

.784.784

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

TVs squaredTVs squaredStandard deviation of daily sales = 1.2884 TVsStandard deviation of daily sales = 1.2884 TVs

Find the Variance and Standard Deviation of the Number Find the Variance and Standard Deviation of the Number of TVs Sold in a given day.of TVs Sold in a given day.

Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx)= 1.66)= 1.66Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx)= 1.66)= 1.66

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

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5.2) The Binomial Distribution5.2) The Binomial Distribution

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5.2) Binomial Distribution5.2) Binomial Distribution

A discrete probability distribution with binomial A discrete probability distribution with binomial distribution has the following propertiedistribution has the following propertie

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

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Typical Examples of a Binomial Experiment:Typical Examples of a Binomial Experiment:

•Lottery: Win or LoseLottery: Win or Lose

•Election: A Candidate Wins or LosesElection: A Candidate Wins or Loses

•Gender of an Employee: is Male or FemaleGender of an Employee: is Male or Female

•Flipping a coin: Heads or TailsFlipping a coin: Heads or Tails

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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.

let let XX denote the number of successes occurring in the denote the number of successes occurring in the nn trials. trials.let let XX denote the number of successes occurring in the denote the number of successes occurring in the nn trials. trials.

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where:where: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 ff((xx) = the probability of ) = the probability of xx successes in successes in nn trials trials

( )!( ) (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


!!( )!

nx n x

!!( )!

nx n x

( )(1 )x n xp p ( )(1 )x n xp p

Binomial Probability FunctionBinomial Probability Function

Probability of a particularProbability of a particular sequence of outcomessequence of outcomes

with x successeswith x successes

Probability of a particularProbability of a particular sequence of outcomessequence of outcomes

with x successeswith x successes

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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Binomial DistributionBinomial Distribution

Example: Evans ElectronicsExample: Evans Electronics

A local Electronics company is concerned A local Electronics company is concerned about its low retention of employees. In about its low retention of employees. In recent years, management has seen an annual recent years, management has seen an annual turnover of 10% in its hourly employees. turnover of 10% in its hourly employees. Thus, for any hourly employee chosen at Thus, for any hourly employee chosen at random, the company estimates that there is random, the company estimates that there is 0.1 probability that the person will leave the 0.1 probability that the person will leave the company in a year time.company in a year time.

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Given the above information, if we randomly Given the above information, if we randomly select 3 hourly employees, what is the select 3 hourly employees, what is the probability that 1 of them will leave the probability that 1 of them will leave the company in one year ?company in one year ?

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 = 1Solution:Solution:

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Using Tables of Binomial ProbabilitiesUsing Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .12501 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

pn x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .12501 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

p


Page….592-600Page….592-600

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Mean and Variance of A Binomial Mean and Variance of A Binomial DistributionDistribution

(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

Given that p=0.1, for the 3 randomly selected hourly Given that p=0.1, for the 3 randomly selected hourly employees, what is the employees, what is the expected numberexpected number and and variancevariance of workers who might leave the company this of workers who might leave the company this year?year?

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5.3) Poisson Distribution5.3) Poisson Distribution

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Poisson distribution refers to the probability Poisson distribution refers to the probability distribution of a trial that involves cases of rare distribution of a trial that involves cases of rare events that occur over events that occur over a fixed timea fixed time interval interval or or within awithin a specified regionspecified region

6.4. Poisson Distribution6.4. Poisson Distribution

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Examples….Examples….

• The number of errors a typist makes per pageThe number of errors a typist makes per page

• The number of cars entering a service station per hourThe number of cars entering a service station per hour

• The number of telephone calls received by a switchboard per The number of telephone calls received by a switchboard per hour.hour.

• The number of bank failures during a given economic recession. The number of bank failures during a given economic recession.

• The number of housing foreclosures in a given city during a The number of housing foreclosures in a given city during a given year.given year.

• The number of car accidents in one day on I-35 stretch from the The number of car accidents in one day on I-35 stretch from the Twin Cities to Duluth Twin Cities to Duluth

6.4. Poisson Distribution6.4. Poisson Distribution

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We use a Poisson distribution to estimate We use a Poisson distribution to estimate the number of the number of occurrences of these discrete events that often occur overoccurrences of these discrete events that often occur over a specified interval of time or spacea specified interval of time or space

We use a Poisson distribution to estimate We use a Poisson distribution to estimate the number of the number of occurrences of these discrete events that often occur overoccurrences of these discrete events that often occur over a specified interval of time or spacea specified interval of time or space

That is, a Poisson distributed random variable is discrete; That is, a Poisson distributed random variable is discrete; Often times it assumes an Often times it assumes an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ).(x = 0, 1, 2, . . . ).

That is, a Poisson distributed random variable is discrete; That is, a Poisson distributed random variable is discrete; Often times it assumes an Often times it assumes an infinite sequence of valuesinfinite sequence of values (x = 0, 1, 2, . . . ).(x = 0, 1, 2, . . . ).

Poisson DistributionPoisson Distribution

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The number of successes (events) that occur in a certain time The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in interval is independent of the number of successes that occur in another time interval.another time interval.

The probability of a success in a certain time interval isThe probability of a success in a certain time interval is

• the same for all time intervals of the same size; proportional to the same for all time intervals of the same size; proportional to the length of the interval.the length of the interval.

– The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

Properties of a Poisson ExperimentProperties of a Poisson Experiment

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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 = 2.71828

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On average 6 patients arrive On average 6 patients arrive

per hour at Mercy Hospital per hour at Mercy Hospital

emergency room on weekend emergency room on weekend

evenings. evenings.

What is the probability of What is the probability of

4 arrivals in 30 minutes on a 4 arrivals in 30 minutes on a

weekend evening?weekend evening?

Poisson Distribution--ExamplePoisson Distribution--Example

MERCYMERCY

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Poisson Distribution-ExamplePoisson Distribution-Example

Using the Poisson Probability FunctionUsing the Poisson Probability Function

4 33 (2.71828)(4) .1680

4!f

4 33 (2.71828)

(4) .16804!

f

MERCYMERCY

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

f xex

x( )

!

f x

ex

x( )

!

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Using Poisson Probability TablesUsing Poisson Probability Tables

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

MERCYMERCY

Pages: 602-607Pages: 602-607

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Mean and Variance of Poisson DistributionMean and Variance of Poisson Distribution

Another special property of the Poisson distribution Another special property of the Poisson distribution is that the mean and variance are equal.is that the mean and variance are equal.

Another special property of the Poisson distribution Another special property of the Poisson distribution is that the mean and variance are equal.is that the mean and variance are equal.

= = 22

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Poisson DistributionPoisson DistributionMERCYMERCY

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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5.4) Hyper-Geometric Distribution5.4) Hyper-Geometric Distribution

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Hyper-geometric DistributionHyper-geometric Distribution

The The hyper-geometric distributionhyper-geometric distribution is closely related is closely related to a binomial distribution. to a binomial distribution. The The hyper-geometric distributionhyper-geometric distribution is closely related is closely related to a binomial distribution. to a binomial distribution.

However, for the hyper-geometric distribution:However, for the hyper-geometric distribution: However, for the hyper-geometric distribution:However, for the hyper-geometric 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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Hyper-geometric Probability FunctionHyper-geometric 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 labeled as success = number of elements labeled as success

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Hyper-geometric Probability FunctionHyper-geometric 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 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 populationnumber of waysnumber of ways

xx successes can be selected successes can be selectedfrom a total of from a total of rr successes successes

in the populationin the populationnumber of waysnumber of ways

a sample of size a sample of size n n can be selectedcan be selectedfrom a population of size from a population of size NN

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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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Hyper-geometric Distribution: Hyper-geometric Distribution: InspectionInspection

Electric fuses produced by a given company are Electric fuses produced by a given company are packed in boxes. Each box carries 12 units of packed in boxes. Each box carries 12 units of electric fuses. The role of an inspector in electric fuses. The role of an inspector in the company is to make sure that all fuses in the company is to make sure that all fuses in each box are in good condition. each box are in good condition.

Consider the following scenario: A worker Consider the following scenario: A worker inadvertently places 5 defective items in a inadvertently places 5 defective items in a box. box. As part of her job, the inspector randomly As part of her job, the inspector randomly selects 3 fuses from the box that contains the selects 3 fuses from the box that contains the defective fuses.defective fuses.

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1.1. What is the probability that none of the three What is the probability that none of the three randomly selected fuses are defective?randomly selected fuses are defective?

2.2. What is the probability that the inspector What is the probability that the inspector finds only one of the three randomly selected finds only one of the three randomly selected fuses to be defective?fuses to be defective?

3.3. What is the probability that the inspector What is the probability that the inspector finds at least one of the three fuses finds at least one of the three fuses defective?defective?

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1.1. What is the probability that none of the What is the probability that none of the three randomly selected fuses are three randomly selected fuses are defective?defective?

Solution:Solution:N=12;N=12; r=5; n=3; x=0; r=5; n=3; x=0;

P(x=0)?P(x=0)?

1591.0220

35

!9!3!12

!4!3!7

!5!0!5

)(

))(()0(

123

51203

50

xf

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2.2. What is the probability that the inspector What is the probability that the inspector finds only one of the three randomly finds only one of the three randomly selected fuses to be defective?selected fuses to be defective?

Solution:Solution:N=12;N=12; r=5; n=3; x=1; r=5; n=3; x=1;

P(x=1)?P(x=1)?

4773.0220

215

!9!3!12

!5!2!7

!4!1!5

)(

))(()1(

123

51213

51

Xxf

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2.2. What is the probability that the inspector What is the probability that the inspector finds at least one of the three fuses finds at least one of the three fuses defective?defective?

Solution:Solution:

N=12;N=12; r=5; n=3; x 1 r=5; n=3; x 1

8409.0

1591.01

)0(1

?)1(

Xp

xP

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Example: NevereadyExample: NevereadyBob has removed two dead batteries from a Bob has removed two dead batteries from a flashlight and inadvertently mingled them flashlight and inadvertently mingled them with the two good batteries that he intended with the two good batteries that he intended to use as replacements. The four batteries to use as replacements. The four batteries look identical.look identical.

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

ZAPZAP ZA

PZ

AP

ZAP ZAPZAPZAP

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Using the Hyper-geometric FunctionUsing the Hyper-geometric Function

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

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

xx = 2 =number of = 2 =number of goodgood batteries to be selected batteries to be selected nn = 2 = number of batteries selected = 2 = number of batteries selected

167.06

1

!2!2

!4!2!0

!2

!0!2

!2

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When the population size is large, a hyper-geometricWhen the population size is large, a hyper-geometric 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 hyper-geometricWhen the population size is large, a hyper-geometric 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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Consider a hyper-geometric distribution with Consider a hyper-geometric 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.

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.

Chapter 5: Probability Distributions: Discrete Probability Distributions

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Transcript of Chapter 5: Probability Distributions: Discrete Probability Distributions