[ppt]statistics for business and economics - … · web viewtitle statistics for business and...

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© 2003 Thomson/South-Western© 2003 Thomson/South-Western

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 valuesvaluesLet Let xx = number of TV sets sold at the store in = number of TV sets sold at the store in one dayone day 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 Discrete random variable with an infinite sequence of valuessequence of valuesLet 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 We can count the customers arriving, but there is no finite upper limit on the number that might is no finite upper limit on the number that might arrive.arrive.

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Random VariablesRandom Variables

QuestionQuestion Random Variable Random Variable xx Type Type

Family Family xx = Number of dependents in Discrete = Number of dependents in Discretesize family reported on tax return size family reported on tax return

Distance from Distance from xx = Distance in miles from = Distance in miles from Continuous Continuoushome to store home to the store site home to store home to the store site

Own dog Own dog xx = 1 if own no pet; = 1 if own no pet; Discrete Discreteor cat or cat = 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)

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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

.10.20.30.40.50

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

Prob

abilit

yPr

obab

ility

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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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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.

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. 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.

Expected Value and VarianceExpected Value and 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) = 1.20The expected number of TV sets sold in a day is The expected number of TV sets sold in a day is 1.21.2

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Variance and Standard DeviationVariance and Standard Deviation of a Discrete Random Variableof a Discrete Random Variablexx 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 sets.The standard deviation of sales is 1.2884 TV sets.


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Binomial Probability DistributionBinomial Probability Distribution

Properties of a Binomial ExperimentProperties of a Binomial Experiment1.1. The experiment consists of a sequence of The experiment consists of a sequence of nn

identical trials.identical trials.2.2. Two outcomes, Two outcomes, successsuccess and and failurefailure, are , are

possible on each trial. possible on each trial. 3.3. 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.4.4. The trials are independent.The trials are independent.

StationarityStationarityAssumptionAssumption

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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.

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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Number of Experimental Outcomes Number of Experimental Outcomes Providing Exactly Providing Exactly xx Successes in Successes in nn Trials Trials

where:where: nn! = ! = nn((nn – 1)( – 1)(nn – 2) . . . (2)(1) – 2) . . . (2)(1) 0! = 10! = 1

!!( )!

n nx x n x

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Probability of a Particular Sequence of Probability of a Particular Sequence of Trial Outcomes with x Successes in Trial Outcomes with x Successes in nn Trials Trials

( )(1 )x n xp p

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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 x nx n x

p px n x( ) !!( )!

( )( )

1

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Example: Evans ElectronicsExample: Evans Electronics

Binomial Probability DistributionBinomial Probability Distribution 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 at random, Choosing 3 hourly employees at 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?

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Using the Binomial Probability FunctionUsing the Binomial Probability FunctionLetLet: p: p = .10, = .10, nn = 3, = 3, xx = 1 = 1

= (3)(0.1)(0.81)= (3)(0.1)(0.81) = .243 = .243

f x nx n x

p px n x( ) !!( )!

( )( )

1

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 p

n 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


1st Worker 2nd Worker 3rd Worker x Probab.

Leaves (.1)

Stays (.9)

3

2

0

22

Leaves (.1)

Leaves (.1)S (.9)

Stays (.9)

Stays (.9)

S (.9)

S (.9)

S (.9)

L (.1)

L (.1)

L (.1)

L (.1) .0010

.0090

.0090

.7290

.0090

1

11

.0810

.0810

.0810

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SD( ) ( )x np p 1

Expected ValueExpected Value

EE((xx) = ) = = = npnp VarianceVariance

Var(Var(xx) = ) = 22 = = npnp(1 - (1 - pp)) Standard DeviationStandard Deviation

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Binomial Probability DistributionBinomial Probability Distribution• 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

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Poisson Probability DistributionPoisson Probability Distribution

A discrete random variable following this A discrete random variable following this distribution is often useful in estimating the distribution is often useful in estimating the number of occurrences over a number of occurrences over a specified interval specified interval of time or spaceof time or space..

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

ExamplesExamples: : • the number of knotholes in 14 linear feet of the number of knotholes in 14 linear feet of

pine boardpine board• the number of vehicles arriving at a toll booth the number of vehicles arriving at a toll booth

in one hourin one hour

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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 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 x ex

x( )

!

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Example: Mercy HospitalExample: Mercy Hospital

Using the Poisson Probability FunctionUsing the Poisson Probability FunctionPatients 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 = 44 33 (2.71828)(4) .16804!f

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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

Example: Mercy HospitalExample: Mercy Hospital

Using the Tables of Poisson ProbabilitiesUsing the Tables of Poisson Probabilities

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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.

The key differences are:The key differences are:• the trials are not independentthe trials are not independent• probability of success changes from trial to probability of success changes from trial to

trialtrial

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nN

xnrN

xr

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 trialstrials 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

• is the number of ways a sample of size is the number of ways a sample of size nn can be selected from a population of size can be selected from a population of size NN..

• is the number of ways is the number of ways xx successes can be successes can be selected from a total of selected from a total of rr successes in the successes in the population.population.

• is the number of ways is the number of ways nn – – xx failures can failures can be selected from a total of be selected from a total of NN – – rr failures in the failures in the population.population.

Nn

rx

N rn x

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Example: NevereadyExample: Neveready

Hypergeometric Probability DistributionHypergeometric Probability DistributionBob 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 Hypergeometric Probability DistributionHypergeometric Probability Distribution

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

167.61

!2!2!4

!2!0!2

!0!2!2

24

02

22

)(

nN

xnrN

xr

xf

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End of Chapter 5End of Chapter 5

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