数据、模型与决策（第12版）课件 第11章 概率与统计

ProbabilityandStatisticsChapter11Copyright©2016PearsonEducation,Inc.

TypesofProbabilityFundamentalsofProbabilityStatisticalIndependenceandDependenceExpectedValueChapterTopicsCopyright©2016PearsonEducation,Inc.

Deterministictechniquesassumethatnouncertaintyexistsinmodelparameters.Chapters2-10introducedtopicsthatarenotsubjecttouncertaintyorvariation.Probabilistictechniquesincludeuncertaintyandassumethattherecanbemorethanonemodelsolution.Thereissomedoubtaboutwhichoutcomewilloccur.Solutionsmaybeintheformofaverages.OverviewCopyright©2016PearsonEducation,Inc.

Classical,orapriori(priortotheoccurrence),

probabilityisanobjectiveprobabilitythatcanbestatedpriortotheoccurrenceoftheevent.Itisbasedonthelogicoftheprocessproducingtheoutcomes.Objectiveprobabilitiesthatarestatedaftertheoutcomesofaneventhavebeenobservedarerelativefrequencies,basedonobservationofpastoccurrences.Relativefrequencyisthemorewidelyuseddefinitionofobjectiveprobability.TypesofProbabilityObjectiveProbabilityCopyright©2016PearsonEducation,Inc.

Subjectiveprobabilityisanestimatebasedonpersonalbelief,experience,orknowledgeofasituation.Itisoftentheonlymeansavailableformakingprobabilisticestimates.Frequentlyusedinmakingbusinessdecisions.Differentpeopleoftenarriveatdifferentsubjectiveprobabilities.Objectiveprobabilities

areusedinthistextunlessotherwiseindicated.TypesofProbabilitySubjectiveProbabilityCopyright©2016PearsonEducation,Inc.

Anexperimentisanactivitythatresultsinoneofseveralpossibleoutcomeswhicharetermedevents.Theprobabilityofaneventisalwaysgreaterthanorequaltozeroandlessthanorequaltoone.Theprobabilities

ofalltheeventsincludedinanexperimentmustsumtoone.Theeventsinanexperimentaremutuallyexclusiveifonlyonecanoccuratatime.Theprobabilitiesofmutuallyexclusiveeventssumtoone.FundamentalsofProbabilityOutcomesandEventsCopyright©2016PearsonEducation,Inc.

Afrequencydistribution

isanorganizationofnumericaldataabouttheeventsinanexperiment.Alistofcorrespondingprobabilitiesforeacheventisreferredtoasaprobabilitydistribution.Asetofeventsiscollectivelyexhaustivewhenitincludesalltheevents

thatcanoccurinanexperiment.FundamentalsofProbabilityDistributionsCopyright©2016PearsonEducation,Inc.

StateUniversity,3000students,managementsciencegradesforpastfouryears.FundamentalsofProbabilityAFrequencyDistributionExampleCopyright©2016PearsonEducation,Inc.

A

marginalprobability

istheprobabilityofasingleeventoccurring,denotedbyP(A).Formutuallyexclusiveevents,theprobabilitythatoneortheotherofseveraleventswilloccurisfoundbysummingtheindividualprobabilitiesoftheevents:P(AorB)=P(A)+P(B)A

Venndiagram

isusedtoshowmutuallyexclusiveevents.FundamentalsofProbabilityMutuallyExclusiveEvents&MarginalProbabilityCopyright©2016PearsonEducation,Inc.

Figure11.1VennDiagramforMutuallyExclusiveEventsFundamentalsofProbabilityMutuallyExclusiveEvents&MarginalProbabilityCopyright©2016PearsonEducation,Inc.

Probabilitythatnon-mutuallyexclusiveeventsAandBorbothwilloccurexpressedas:P(AorB)=P(A)+P(B)-P(AB)A

jointprobability,P(AB),istheprobabilitythattwoormoreeventsthatarenotmutuallyexclusivecanoccursimultaneously.FundamentalsofProbabilityNon-MutuallyExclusiveEvents&JointProbabilityCopyright©2016PearsonEducation,Inc.

Figure11.2Venndiagramfornon–mutuallyexclusiveeventsandthejointeventFundamentalsofProbabilityNon-MutuallyExclusiveEvents&JointProbabilityM=studentstakingmanagementscienceF=studentstakingfinanceCopyright©2016PearsonEducation,Inc.

Canbedevelopedbyaddingtheprobabilityofaneventtothesumofallpreviouslylistedprobabilitiesinaprobabilitydistribution.ProbabilitythatastudentwillgetagradeofCorhigher:P(AorBorC)=P(A)+P(B)+P(C)=.10+.20+.50=.80FundamentalsofProbabilityCumulativeProbabilityDistributionCopyright©2016PearsonEducation,Inc.

Asuccessionofeventsthatdonotaffecteachotherare

independentevents.Theprobabilityofindependenteventsoccurringinasuccessioniscomputedbymultiplyingtheprobabilitiesofeachevent.Aconditionalprobabilityistheprobabilitythataneventwilloccurgiventhatanothereventhasalreadyoccurred,denotedasP(A

B).IfeventsAandBareindependent,then:P(AB)=P(A)

P(B)andP(A

B)=P(A)StatisticalIndependenceandDependenceIndependentEventsCopyright©2016PearsonEducation,Inc.

ForcointossedthreeconsecutivetimesFigure11.3StatisticalIndependenceandDependenceIndependentEvents–ProbabilityTreesProbabilityofgettingheadon1sttoss,tailon2nd,tailon3rdis:P(HTT)=P(H)

P(T)

P(T)=(.5)(.5)(.5)=.125Copyright©2016PearsonEducation,Inc.

PropertiesofaBernoulliProcess:Therearetwopossibleoutcomesforeachtrial.Theprobabilityoftheoutcomeremainsconstantovertime.Theoutcomesofthetrialsareindependent.Thenumberoftrialsisdiscreteandinteger.StatisticalIndependenceandDependenceIndependentEvents–BernoulliProcessDefinitionCopyright©2016PearsonEducation,Inc.

Abinomialprobabilitydistributionfunction

isusedtodeterminetheprobabilityofanumberofsuccessesinntrials.Itisadiscreteprobabilitydistribution

sincethenumberofsuccessesandtrialsisdiscrete.

where: p=probabilityofasuccess q=1-p=probabilityofafailure n=numberoftrials r=numberofsuccessesinntrialsStatisticalIndependenceandDependenceIndependentEvents–BinomialDistributionCopyright©2016PearsonEducation,Inc.

Determineprobabilityofgettingexactlytwotailsinthreetossesofacoin.StatisticalIndependenceandDependenceBinomialDistributionExample–TossedCoinsCopyright©2016PearsonEducation,Inc.

Microchipproduction;sampleoffouritemsperbatch,20%ofallmicrochipsaredefective.Whatistheprobabilitythateachbatchwillcontainexactlytwodefectives?StatisticalIndependenceandDependenceBinomialDistributionExample–QualityControlCopyright©2016PearsonEducation,Inc.

Fourmicrochipstestedperbatch;iftwoormorefounddefective,batchisrejected.Whatisprobabilityofrejectingentirebatchifbatchinfacthas20%defective?Probabilityoflessthantwodefectives:P(r<2)=P(r=0)+P(r=1)=1.0-[P(r=2)+P(r=3)+P(r=4)] =1.0-.1808=.8192StatisticalIndependenceandDependenceBinomialDistributionExample–QualityControlCopyright©2016PearsonEducation,Inc.

Figure11.4DependenteventsStatisticalIndependenceandDependenceDependentEvents(1of2)Copyright©2016PearsonEducation,Inc.

Iftheoccurrenceofoneeventaffectstheprobabilityoftheoccurrenceofanotherevent,theeventsaredependent.Cointosstoselectbucket,drawforblueball.Iftailoccurs,1/6chanceofdrawingblueballfrombucket2;ifheadresults,nopossibilityofdrawingblueballfrombucket1.Probabilityofevent“drawingablueball”dependentonevent“flippingacoin.”StatisticalIndependenceandDependenceDependentEvents(2of2)Copyright©2016PearsonEducation,Inc.

Unconditional:P(H)=.5;P(T)=.5,mustsumtoone.Figure11.5AnothersetofdependenteventsStatisticalIndependenceandDependenceDependentEvents–UnconditionalProbabilitiesCopyright©2016PearsonEducation,Inc.

Conditional:P(R

H)=.33,P(W

H)=.67,P(R

T)=.83,P(W

T)=.17StatisticalIndependenceandDependenceDependentEvents–ConditionalProbabilitiesFigure11.6ProbabilitytreefordependenteventsCopyright©2016PearsonEducation,Inc.

GiventwodependenteventsAandB:P(A

B)=P(A

B)/P(B)Withdatafrompreviousexample:P(RH)=P(R

H)

P(H)=(.33)(.5)=.165P(WH)=P(W

H)

P(H)=(.67)(.5)=.335P(RT)=P(R

T)

P(T)=(.83)(.5)=.415P(WT)=P(W

T)

P(T)=(.17)(.5)=.085StatisticalIndependenceandDependenceMathFormulationofConditionalProbabilitiesCopyright©2016PearsonEducation,Inc.

Figure11.7Probabilitytreewithmarginal,conditionalandjointprobabilitiesStatisticalIndependenceandDependenceSummaryofExampleProblemProbabilitiesCopyright©2016PearsonEducation,Inc.

Table11.1JointprobabilitytableStatisticalIndependenceandDependenceSummaryofExampleProblemProbabilitiesCopyright©2016PearsonEducation,Inc.

InBayesiananalysis,additionalinformationisusedtoalterthemarginalprobabilityoftheoccurrenceofanevent.Aposteriorprobabilityisthealteredmarginalprobabilityofaneventbasedonadditionalinformation.Bayes’Rulefortwoevents,AandB,andthirdevent,C,conditionallydependentonAandB:StatisticalIndependenceandDependenceBayesianAnalysisCopyright©2016PearsonEducation,Inc.

Machinesetup;ifcorrectthereisa10%chanceofadefectivepart;ifincorrect,a40%chanceofadefectivepart.50%chancesetupwillbecorrectorincorrect.Whatisprobabilitythatmachinesetupisincorrectifasamplepartisdefective?Solution:P(C)=.50,P(IC)=.50,P(D|C)=.10,P(D|IC)=.40 whereC=correct,IC=incorrect,D=defectiveStatisticalIndependenceandDependenceBayesianAnalysis–Example(1of2)Copyright©2016PearsonEducation,Inc.

Posteriorprobabilities:StatisticalIndependenceandDependenceBayesianAnalysis–Example(2of2)Copyright©2016PearsonEducation,Inc.

Whenthevaluesofvariablesoccurinnoparticularorderorsequence,thevariablesarereferredtoas

randomvariables.Randomvariablesarerepresentedsymbolicallybyaletterx,y,z,etc.Althoughexactvaluesofrandomvariablesarenotknownpriortoevents,itispossibletoassignaprobabilitytotheoccurrenceofpossiblevalues.ExpectedValueRandomVariablesCopyright©2016PearsonEducation,Inc.

Machinesbreakdown0,1,2,3,or4timespermonth.Relativefrequencyofbreakdowns,oraprobabilitydistribution:ExpectedValueExample(1of4)Copyright©2016PearsonEducation,Inc.

Theexpectedvalue

ofarandomvariableiscomputedbymultiplyingeachpossiblevalueofthevariablebyitsprobabilityandsummingtheseproducts.Theexpectedvalueistheweightedaverage,ormean,oftheprobabilitydistributionoftherandomvariable.Expectedvalueofnumberofbreakdownspermonth: E(x)=(0)(.10)+(1)(.20)+(2)(.30)+(3)(.25)+(4)(.15) =0+.20+.60+.75+.60 =2.15breakdownsExpectedValueExample(2of4)Copyright©2016PearsonEducation,Inc.

Varianceisameasureofthedispersionofarandomvariable’svaluesaboutthemean.Varianceiscomputedasfollows:Squarethedifferencebetweeneachvalueandtheexpectedvalue.Multiplytheresultingamountsbytheprobabilityofeachvalue.Sumthevaluescompiledinstep2.Generalformula:

2=

[xi-E(x)]2P(xi)ExpectedValueExample(3of4)i=1nCopyright©2016PearsonEducation,Inc.

Standarddeviationiscomputedbytakingthesquarerootofthevariance.Forexampledata[E(x)=2.15]:

2=1.425(breakdownspermonth)2 standarddeviation=

=sqrt(1.425) =1.19breakdownspermonthExpectedValueExample(4of4)Copyright©2016PearsonEducation,Inc.

Acontinuousrandomvariable

cantakeonaninfinitenumberofvalueswithinsomeinterval.Continuousrandomvariableshavevaluesthatarenotspecificallycountableandareoftenfractional.Cannotassignauniqueprobabilitytoeachvalueofacontinuousrandomvariable.Inacontinuousprobabilitydistributiontheprobabilityreferstoavalueoftherandomvariablebeingwithinsomerange.TheNormalDistributionContinuousRandomVariablesThenormaldistribution

isacontinuousprobabilitydistributionthatissymmetricalonbothsidesofthemean.Thecenterofanormaldistributionisitsmean

.Theareaunderthenormalcurverepresentsprobability,andthetotalareaunderthecurvesumstoone.TheNormalDistributionDefinitionFigure11.8ThenormalcurveCopyright©2016PearsonEducation,Inc.

Meanweeklycarpetsalesof4,200yards,withastandarddeviationof1,400yards.Whatistheprobabilityofsalesexceeding6,000yards?

=4,200yd;

=1,400yd;probabilitythatnumberofyardsofcarpetwillbeequaltoorgreaterthan6,000expressedas:P(x

6,000).TheNormalDistributionExample(1of5)Copyright©2016PearsonEducation,Inc.

--Figure11.9ThenormaldistributionforcarpetdemandTheNormalDistributionExample(2of5)P(x≥6,000)Copyright©2016PearsonEducation,Inc.

Theareaorprobabilityunderanormalcurveismeasuredbydeterminingthenumberofstandarddeviationsthevalueofarandomvariablexisfromthemean.NumberofstandarddeviationsavalueisfromthemeandesignatedasZ.TheNormalDistributionStandardNormalCurve(1of2)Copyright©2016PearsonEducation,Inc.

TheNormalDistributionStandardNormalCurve(2of2)Figure11.10ThestandardnormaldistributionCopyright©2016PearsonEducation,Inc.

Figure11.11DeterminationoftheZvalueTheNormalDistributionExample(3of5) Z=(x-

)/=(6,000-4,200)/1,400 =1.29standarddeviations P(x

6,000)=.5000-.4015=.0985P(x≥6,000)Copyright©2016PearsonEducation,Inc.

Determinetheprobabilitythatdemandwillbe5,000yardsorless.Z=(x-

)/=(5,000-4,200)/1,400=.57standarddeviationsP(x

5,000)=.5000+.2157=.7157TheNormalDistributionExample(4of5)Figure11.12NormaldistributionforP(x

5,000yards)Copyright©2016PearsonEducation,Inc.

TheNormalDistributionExample(5of5)Figure11.13NormaldistributionwithP(3000yards

x

5000yards) Determinetheprobabilitythatdemandwillbebetween3,000yardsand5,000yards.Z=(3,000-4,200)/1,400=-1,200/1,400=-.86P(3,000

x

5,000)=.2157+.3051=.5208Copyright©2016PearsonEducation,Inc.

Thepopulation

meanandvariancearefortheentiresetofdatabeinganalyzed.Thesamplemeanandvariancearederivedfromasubsetofthepopulationdataandareusedtomakeinferencesaboutthepopulation.TheNormalDistributionSampleMeanandVarianceCopyright©2016PearsonEducation,Inc.

TheNormalDistributionComputingtheSampleMeanandVarianceSamplemeanSamplevarianceSamplevarianceshortcutformCopyright©2016PearsonEducation,Inc.

Samplemean=42,000/10=4,200ydSamplevariance=[(190,060,000)-(1,764,000,000/10)]/9 =1,517,777Samplestd.dev.=sqrt(1,517,777)=1,232ydTheNormalDistributionExampleProblemRevisitedCopyright©2016PearsonEducation,Inc.

Itcanneverbesimplyassumedthatdataarenormallydistributed.Thechi-squaretestisusedtodetermineifasetofdatafitaparticulardistribution.Thechi-squaretestcomparesanobservedfrequencydistributionwithatheoreticalfrequencydistribution(testingthegoodness-of-fit).TheNormalDistributionChi-SquareTestforNormality(1of2)Copyright©2016PearsonEducation,Inc.

Inthetest,theactualnumberoffrequenciesineachrangeoffrequencydistributioniscomparedtothetheoreticalfrequenciesthatshouldoccurineachrangeifthedatafollowaparticulardistribution.Achi-squarestatisticisthencalculatedandcomparedtoanumber,calledacriticalvalue,fromachi-squaretable.Iftheteststatisticisgreaterthanthecriticalvalue,thedistributiondoesnotfollowthedistributionbeingtested;ifitisless,thedistributionfits.Thechi-squaretestisaformofhypothesistesting.TheNormalDistributionChi-SquareTestforNormality(2of2)Copyright©2016PearsonEducation,Inc.

ArmorCarpetStoreexample-assumesamplemean=4,200yards,andsamplestandarddeviation=1,232yards.TheNormalDistributionExampleofChi-SquareTest(1of6)Copyright©2016PearsonEducation,Inc.

Figure11.14ThetheoreticalnormaldistributionTheNormalDistributionExampleofChi-SquareTest(2of6)Copyright©2016PearsonEducation,Inc.

Table11.2ThedeterminationofthetheoreticalrangefrequenciesTheNormalDistributionExampleofChi-SquareTest(3of6)Copyright©2016PearsonEducation,Inc.

TheNormalDistributionExampleofChi-SquareTest(4of6)Comparingtheoreticalfrequencieswithactualfrequencies:where:fo=observedfrequency ft=theoreticalfrequencyk=thenumberofclasses,p=thenumberofestimatedparameters k-p-1=degreesoffreedom.Copyright©2016PearsonEducation,Inc.

Table11.3Computationof

2teststatistic

TheNormalDistributionExampleofChi-SquareTest(5of6)Copyright©2016PearsonEducation,Inc.

2k-p-1=

(fo-ft)2/10=2.588k-p-1=6-2–1=3degreesoffreedom,withlevelofsignificance(degofconfidence)of.05(

=.05).fromTableA.2,

2.05,3=7.815;because7.815>2.588,weacceptthehypothesisthatthedistributionisnormal.TheNormalDistributionExampleofChi-SquareTest(6of6)Copyright©2016PearsonEducation,Inc.

Exhibit11.1StatisticalAnalysiswithExcel(1of2)Clickon“Data