(1.23)-PowerPoint Ch 05商务统计课件

Chapter5

DiscreteProbabilityDistributions.10.20.30.400

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234RandomVariablesDiscreteProbabilityDistributionsExpectedValueandVarianceBinomialProbabilityDistributionPoissonProbabilityDistributionHypergeometricProbabilityDistributionArandomvariableisanumericaldescriptionoftheoutcomeofanexperiment.RandomVariablesAdiscreterandomvariablemayassumeeitherafinitenumberofvaluesoraninfinitesequenceofvalues.Acontinuousrandomvariablemayassumeanynumericalvalueinanintervalorcollectionofintervals.Letx=numberofTVssoldatthestoreinoneday, wherexcantakeon5values(0,1,2,3,4)Example:JSLAppliancesDiscreteRandomVariablewithaFiniteNumberofValuesWecancounttheTVssold,andthereisafiniteupperlimitonthenumberthatmightbesold(whichisthenumberofTVsinstock).Letx=numberofcustomersarrivinginoneday, wherexcantakeonthevalues0,1,2,...DiscreteRandomVariablewithanInfiniteSequenceofValuesWecancountthecustomersarriving,butthereisnofiniteupperlimitonthenumberthatmightarrive.Example:JSLAppliancesRandomVariablesQuestionRandomVariablexTypeFamilysizex=NumberofdependentsreportedontaxreturnDiscreteDistancefromhometostorex=DistanceinmilesfromhometothestoresiteContinuousOwndogorcatx=1ifownnopet;=2ifowndog(s)only;=3ifowncat(s)only;=4ifowndog(s)andcat(s)DiscreteTheprobabilitydistributionforarandomvariabledescribeshowprobabilitiesaredistributedoverthevaluesoftherandomvariable.Wecandescribeadiscreteprobabilitydistributionwithatable,graph,orformula.DiscreteProbabilityDistributionsTheprobabilitydistributionisdefinedbya

probabilityfunction,denotedbyf(x),whichprovidestheprobabilityforeachvalueoftherandomvariable.Therequiredconditionsforadiscreteprobabilityfunctionare:DiscreteProbabilityDistributionsf(x)>0

f(x)=1atabularrepresentationoftheprobabilitydistributionforTVsaleswasdeveloped.UsingpastdataonTVsales,… Number

UnitsSold

ofDays 0 80 1 50 2 40 3 10 4 20 200

x

f(x)0 .401 .252 .203 .054 .101.0080/200DiscreteProbabilityDistributionsExample:JSLAppliances.10.20.30.40.500

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4ValuesofRandomVariablex(TVsales)ProbabilityDiscreteProbabilityDistributionsExample:JSLAppliancesGraphicalrepresentationofprobabilitydistributionDiscreteUniformProbabilityDistributionThediscreteuniformprobabilitydistributionisthesimplestexampleofadiscreteprobabilitydistributiongivenbyaformula.Thediscreteuniformprobabilityfunctionisf(x)=1/nwhere:

n=thenumberofvaluestherandom variablemayassumethevaluesoftherandomvariableareequallylikelyExpectedValueTheexpectedvalue,ormean,ofarandomvariableisameasureofitscentrallocation.Theexpectedvalueisaweightedaverageofthevaluestherandomvariablemayassume.Theweightsaretheprobabilities.Theexpectedvaluedoesnothavetobeavaluetherandomvariablecanassume.E(x)=

=

xf(x)VarianceandStandardDeviationThevariancesummarizesthevariabilityinthevaluesofarandomvariable.Thevarianceisaweightedaverageofthesquareddeviationsofarandomvariablefromitsmean.Theweightsaretheprobabilities.Var(x)=

2=

(x-

)2f(x)Thestandarddeviation,

,isdefinedasthepositivesquarerootofthevariance.expectednumberofTVssoldinaday

x

f(x)

xf(x)0 .40 .001 .25 .252 .20 .403 .05 .154 .10 .40

E(x)=1.20ExpectedValueExample:JSLAppliances01234-1.2-0.20.81.82.81.440.040.643.247.84.40.25.20.05.10.576.010.128.162.784x-

(x-

)2f(x)(x-

)2f(x)Varianceofdailysales=s2=1.660xTVssquaredStandarddeviationofdailysales=1.2884TVsVarianceExample:JSLAppliancesBinomialProbabilityDistributionFourPropertiesofaBinomialExperiment3.Theprobabilityofasuccess,denotedbyp,doesnotchangefromtrialtotrial.4.Thetrialsareindependent.2.Twooutcomes,successandfailure,arepossibleoneachtrial.1.Theexperimentconsistsofasequenceofnidenticaltrials.stationarityassumptionBinomialProbabilityDistributionOurinterestisinthenumberofsuccessesoccurringinthentrials.Weletxdenotethenumberofsuccessesoccurringinthentrials.where:

x=thenumberofsuccessesp=theprobabilityofasuccessononetrialn=thenumberoftrials

f(x)=theprobabilityofxsuccessesinn

trialsn!=n(n–1)(n–2)…..(2)(1)

BinomialProbabilityDistributionBinomialProbabilityFunctionBinomialProbabilityDistributionBinomialProbabilityFunctionProbabilityofaparticularsequenceoftrialoutcomeswithxsuccessesinntrialsNumberofexperimentaloutcomesprovidingexactlyxsuccessesinntrialsBinomialProbabilityDistributionExample:EvansElectronicsEvansElectronicsisconcernedaboutalowretentionrateforitsemployees.Inrecentyears,managementhasseenaturnoverof10%ofthehourlyemployeesannually.Choosing3hourlyemployeesatrandom,whatistheprobabilitythat1ofthemwillleavethecompanythisyear? Thus,foranyhourlyemployeechosenatrandom,managementestimatesaprobabilityof0.1thatthepersonwillnotbewiththecompanynextyear.BinomialProbabilityDistributionExample:EvansElectronicsTheprobabilityofthefirstemployeeleavingandthesecondandthirdemployeesstaying,denoted(S,F,F),isgivenby

p(1–p)(1–p)Witha.10probabilityofanemployeeleavingonanyonetrial,theprobabilityofanemployeeleavingonthefirsttrialandnotonthesecondandthirdtrialsisgivenby

(.10)(.90)(.90)=(.10)(.90)2=.081BinomialProbabilityDistributionExample:EvansElectronicsTwootherexperimentaloutcomesalsoresultinonesuccessandtwofailures.Theprobabilitiesforallthreeexperimentaloutcomesinvolvingonesuccessfollow.ExperimentalOutcome(S,F,F)(F,S,F)(F,F,S)ProbabilityofExperimentalOutcomep(1–p)(1–p)=(.1)(.9)(.9)=.081(1–p)p(1–p)=(.9)(.1)(.9)=.081(1–p)(1–p)p=(.9)(.9)(.1)=.081Total=.243BinomialProbabilityDistributionLet:p=.10,n=3,x=1Example:EvansElectronicsUsingtheprobabilityfunctionBinomialProbabilityDistribution

1stWorker2ndWorker3rdWorkerxProb.Leaves(.1)Stays(.9)32022Leaves(.1)Leaves(.1)S(.9)Stays(.9)Stays(.9)S(.9)S(.9)S(.9)L(.1)L(.1)L(.1)L(.1).0010.0090.0090.7290.009011.0810.0810.08101Example:EvansElectronicsUsingatreediagramBinomialProbabilitiesandCumulativeProbabilitiesWithmoderncalculatorsandthecapabilityofstatisticalsoftwarepackages,suchtablesarealmostunnecessary.Thesetablescanbefoundinsomestatisticstextbooks.Statisticianshavedevelopedtablesthatgiveprobabilitiesandcumulativeprobabilitiesforabinomialrandomvariable.BinomialProbabilityDistributionE(x)=

=npVar(x)=

2=np(1-

p)ExpectedValueVarianceStandardDeviationBinomialProbabilityDistributionE(x)=np

=3(.1)=.3employeesoutof3Var(x)=np(1–p)=3(.1)(.9)=.27ExpectedValueVarianceStandardDeviationExample:EvansElectronicsAPoissondistributedrandomvariableisoftenusefulinestimatingthenumberofoccurrencesoveraspecifiedintervaloftimeorspaceItisadiscreterandomvariablethatmayassumeaninfinitesequenceofvalues(x=0,1,2,...).PoissonProbabilityDistributionExamplesofaPoissondistributedrandomvariable:thenumberofknotholesin14linearfeetofpineboardthenumberofvehiclesarrivingatatollboothinonehourPoissonProbabilityDistributionBellLabsusedthePoissondistributiontomodelthearrivalofphonecalls.PoissonProbabilityDistributionTwoPropertiesofaPoissonExperimentTheoccurrenceornonoccurrenceinanyintervalisindependentoftheoccurrenceornonoccurrenceinanyotherinterval.Theprobabilityofanoccurrenceisthesameforanytwointervalsofequallength.PoissonProbabilityFunctionPoissonProbabilityDistributionwhere:

x=thenumberofoccurrencesinanintervalf(x)=theprobabilityofxoccurrencesinaninterval

=meannumberofoccurrencesinaninterval

e

=2.71828x!=x(x–1)(x–2)...(2)(1)PoissonProbabilityDistributionPoissonProbabilityFunctionInpracticalapplications,xwilleventuallybecomelargeenoughsothatf(x)isapproximatelyzeroandtheprobabilityofanylargervaluesofxbecomesnegligible.Sincethereisnostatedupperlimitforthenumberofoccurrences,theprobabilityfunctionf(x)isapplicableforvaluesx=0,1,2,…withoutlimit.PoissonProbabilityDistributionExample:MercyHospitalPatientsarriveattheemergencyroomofMercyHospitalattheaveragerateof6perhouronweekendevenings.Whatistheprobabilityof4arrivalsin30minutesonaweekendevening?PoissonProbabilityDistribution

=6/hour=3/half-hour,x=4Example:MercyHospitalUsingtheprobabilityfunctionPoissonProbabilityDistributionPoissonProbabilities0.000.050.100.150.200.25012345678910NumberofArrivalsin30MinutesProbabilityactually,thesequencecontinues:11,12,…Example:MercyHospitalPoissonProbabilityDistributionApropertyofthePoissondistributionisthatthemeanandvarianceareequal.

m=s2PoissonProbabilityDistributionVarianceforNumberofArrivals During30-MinutePeriods m=s

2=3Example:MercyHospitalHypergeometricProbabilityDistributionThehypergeometricdistributioniscloselyrelatedtothebinomialdistribution.However,forthehypergeometricdistribution:thetrialsarenotindependent,andtheprobabilityofsuccesschangesfromtrialtotrial.HypergeometricProbabilityFunctionHypergeometricProbabilityDistributionwhere:

x

=numberofsuccesses

n

=numberoftrialsf(x)=probabilityofxsuccessesinntrials

N=numberofelementsinthepopulation

r=numberofelementsinthepopulation labeledsuccessHypergeometricProbabilityFunctionHypergeometricProbabilityDistributionfor0<

x

<

rnumberofwaysxsuccessescanbeselectedfromatotalofrsuccessesinthepopulationnumberofwaysn–xfailurescanbeselectedfromatotalofN–rfailuresinthepopulationnumberofwaysnelementscanbeselectedfromapopulationofsizeNHypergeometricProbabilityDistributionHypergeometricProbabilityFunctionIfthesetwoconditionsdonotholdforavalueofx,thecorrespondingf(x)equals0.However,onlyvaluesofxwhere:1)x

<

rand2)n–x

<

N–rarevalid.Theprobabilityfunctionf(x)onthepreviousslideisusuallyapplicableforvaluesofx=0,1,2,…n.HypergeometricProbabilityDistributionBobNevereadyhasremovedtwodeadbatteriesfromaflashlightandinadvertentlymingledthemwiththetwogoodbatteriesheintendedasreplacem