(1.24)-PowerPoint Ch 06商务统计课件

Chapter6

ContinuousProbabilityDistributionsUniformProbabilityDistributionf(x)

xUniformxf(x)Normalxf(x)ExponentialNormalProbabilityDistributionNormalApproximationofBinomialProbabilitiesExponentialProbabilityDistributionContinuousProbabilityDistributionsAcontinuousrandomvariablecanassumeanyvalueinanintervalonthereallineorinacollectionofintervals.Itisnotpossibletotalkabouttheprobabilityoftherandomvariableassumingaparticularvalue.Instead,wetalkabouttheprobabilityoftherandomvariableassumingavaluewithinagiveninterval.ContinuousProbabilityDistributionsTheprobabilityoftherandomvariableassumingavaluewithinsomegivenintervalfromx1tox2isdefinedtobetheareaunderthegraphoftheprobabilitydensityfunctionbetweenx1

andx2.f(x)

xUniform

x1

x2xf(x)Normal

x1

x2

x1

x2Exponentialxf(x)

x1

x2UniformProbabilityDistributionwhere:a=smallestvaluethevariablecanassume

b=largestvaluethevariablecanassumef(x)=1/(b–a)fora

<

x

<

b

=0elsewhereArandomvariableisuniformlydistributedwhenevertheprobabilityisproportionaltotheinterval’slength.Theuniformprobabilitydensityfunctionis:Var(x)=(b-a)2/12E(x)=(a+b)/2UniformProbabilityDistributionExpectedValueofxVarianceofxUniformProbabilityDistributionExample:Slater'sBuffetSlatercustomersarechargedfortheamountofsaladtheytake.Samplingsuggeststhattheamountofsaladtakenisuniformlydistributedbetween5ouncesand15ounces.UniformProbabilityDensityFunction

f(x)=1/10for5<

x

<15=0elsewherewhere:x=saladplatefillingweightUniformProbabilityDistributionExpectedValueofxE(x)=(a+b)/2=(5+15)/2=10Var(x)=(b-a)2/12 =(15–5)2/12 =8.33UniformProbabilityDistributionVarianceofxUniformProbabilityDistribution forSaladPlateFillingWeightf(x)x1/10SaladWeight(oz.)UniformProbabilityDistribution510150f(x)x1/10SaladWeight(oz.)510150P(12<

x

<15)=1/10(3)=.3 Whatistheprobabilitythatacustomerwilltakebetween12and15ouncesofsalad? UniformProbabilityDistribution12AreaasaMeasureofProbabilityTheareaunderthegraphoff(x)andprobabilityareidentical.Thisisvalidforallcontinuousrandomvariables.Theprobabilitythatxtakesonavaluebetweensomelowervaluex1andsomehighervaluex2canbefoundbycomputingtheareaunderthegraphoff(x)overtheintervalfromx1tox2.NormalProbabilityDistributionThenormalprobabilitydistributionisthemostimportantdistributionfordescribingacontinuousrandomvariable.Itiswidelyusedinstatisticalinference.Ithasbeenusedinawidevarietyofapplicationsincluding:HeightsofpeopleRainfallamountsTestscoresScientificmeasurementsAbrahamdeMoivre,aFrenchmathematician,publishedTheDoctrineofChancesin1733.Hederivedthenormaldistribution.NormalProbabilityDistributionNormalProbabilityDensityFunction

=mean

=standarddeviation

=3.14159e=2.71828where:Thedistributionissymmetric;itsskewnessmeasureiszero.NormalProbabilityDistributionCharacteristicsxTheentirefamilyofnormalprobabilitydistributionsisdefinedbyits

mean

mandits

standarddeviation

s.NormalProbabilityDistributionCharacteristicsStandardDeviationsMeanmxThehighestpointonthenormalcurveisatthe

mean,whichisalsothemedianandmode.NormalProbabilityDistributionCharacteristicsxNormalProbabilityDistributionCharacteristics-10025Themeancanbeanynumericalvalue:negative,zero,orpositive.xNormalProbabilityDistributionCharacteristicss=15s=25Thestandarddeviationdeterminesthewidthofthecurve:largervaluesresultinwider,flattercurves.xProbabilitiesforthenormalrandomvariablearegivenbyareasunderthecurve.Thetotalareaunderthecurveis1(.5totheleftofthemeanand.5totheright).NormalProbabilityDistributionCharacteristics.5.5xNormalProbabilityDistributionCharacteristics(basisfortheempiricalrule)ofvaluesofanormalrandomvariablearewithinofitsmean.68.26%+/-1standarddeviationofvaluesofanormalrandomvariablearewithinofitsmean.95.44%+/-2standarddeviationsofvaluesofanormalrandomvariablearewithinofitsmean.99.72%+/-3standarddeviationsNormalProbabilityDistributionCharacteristics(basisfortheempiricalrule)xm–3sm–1sm–2sm+1sm+2sm+3sm68.26%95.44%99.72%StandardNormalProbabilityDistributionArandomvariablehavinganormaldistributionwithameanof0andastandarddeviationof1issaidtohaveastandardnormalprobabilitydistribution.Characteristicss=10zTheletterzisusedtodesignatethestandardnormalrandomvariable.StandardNormalProbabilityDistributionCharacteristicsConvertingtotheStandardNormalDistribution

StandardNormalProbabilityDistributionWecanthinkofzasameasureofthenumberofstandarddeviationsxisfrom

.StandardNormalProbabilityDistributionExample:PepZonePepZonesellsautopartsandsuppliesincludingapopularmulti-grademotoroil.Whenthestockofthisoildropsto20gallons,areplenishmentorderisplaced.Thestoremanagerisconcernedthatsalesarebeinglostduetostockoutswhilewaitingforareplenishmentorder.Ithasbeendeterminedthatdemandduringreplenishmentlead-timeisnormallydistributedwithameanof15gallonsandastandarddeviationof6gallons.StandardNormalProbabilityDistributionExample:PepZoneThemanagerwouldliketoknowtheprobabilityofastockoutduringreplenishmentlead-time.Inotherwords,whatistheprobabilitythatdemandduringlead-timewillexceed20gallons?

P(x>20)=?z=(x-

)/

=(20-15)/6=.83SolvingfortheStockoutProbability

Step1:Convertxtothestandardnormaldistribution.Step2:Findtheareaunderthestandardnormalcurvetotheleftofz=.83.

seenextslideStandardNormalProbabilityDistributionCumulativeProbabilityTablefor theStandardNormalDistributionP(z

<.83)StandardNormalProbabilityDistributionP(z>.83)=1–P(z

<.83) =1-.7967=.2033SolvingfortheStockoutProbability

Step3:Computetheareaunderthestandardnormalcurvetotherightofz=.83.ProbabilityofastockoutP(x>20)StandardNormalProbabilityDistributionSolvingfortheStockoutProbability

0.83Area=.7967Area=1-.7967=.2033zStandardNormalProbabilityDistributionStandardNormalProbabilityDistributionStandardNormalProbabilityDistributionIfthemanagerofPepZonewantstheprobabilityofastockoutduringreplenishmentlead-timetobenomorethan.05,whatshouldthereorderpointbe?---------------------------------------------------------------(Hint:Givenaprobability,wecanusethestandardnormaltableinaninversefashiontofindthecorrespondingzvalue.)SolvingfortheReorderPoint

0Area=.9500Area=.0500zz.05StandardNormalProbabilityDistributionSolvingfortheReorderPointStep1:Findthez-valuethatcutsoffanareaof.05 intherighttailofthestandardnormal distribution.Welookupthecomplementofthetailarea(1-.05=.95)StandardNormalProbabilityDistributionSolvingfortheReorderPointStep2:Convertz.05tothecorrespondingvalueofx.x=

+z.05

=15+1.645(6)=24.87or25Areorderpointof25gallonswillplacetheprobabilityofastockoutduringleadtimeat(slightlylessthan).05.StandardNormalProbabilityDistributionNormalProbabilityDistributionSolvingfortheReorderPoint15x24.87Probabilityofastockoutduringreplenishmentlead-time=.05Probabilityofnostockoutduringreplenishmentlead-time=.95SolvingfortheReorderPointByraisingthereorderpointfrom20gallonsto25gallonsonhand,theprobabilityofastockoutdecreasesfromabout.20to.05.ThisisasignificantdecreaseinthechancethatPepZonewillbeoutofstockandunabletomeetacustomer’sdesiretomakeapurchase.StandardNormalProbabilityDistributionNormalApproximationofBinomialProbabilitiesWhenthenumberoftrials,n,becomeslarge,evaluatingthebinomialprobabilityfunctionbyhandorwithacalculatorisdifficult.Thenormalprobabilitydistributionprovidesaneasy-to-useapproximationofbinomialprobabilitieswherenp>

5andn(1-p)>

5.Inthedefinitionofthenormalcurve,set

=npandAddandsubtractacontinuitycorrectionfactor

becauseacontinuousdistributionisbeingusedtoapproximateadiscretedistribution.Forexample,P(x=12)forthediscretebinomialprobabilitydistributionisapproximatedby

P(11.5<

x

<12.5)forthecontinuousnormaldistribution.NormalApproximationofBinomialProbabilitiesNormalApproximationofBinomialProbabilitiesExample

Supposethatacompanyhasahistoryofmakingerrorsin10%ofitsinvoices.Asampleof100invoiceshasbeentaken,andwewanttocomputetheprobabilitythat12invoicescontainerrors.

Inthiscase,wewanttofindthebinomialprobabilityof12successesin100trials.So,weset:

m=np=100(.1)=10=[100(.1)(.9)]½=3NormalApproximationofBinomialProbabilitiesNormalApproximationtoaBinomialProbabilityDistributionwithn=100andp=.1

m=10P(11.5<

x

<12.5)(Probabilityof12Errors)x11.512.5s=3NormalApproximationtoaBinomialProbabilityDistributionwithn=100andp=.1

10P(x

<12.5)=.7967x12.5NormalApproximationofBinomialProbabilitiesNormalApproximationofBinomialProbabilitiesNormalApproximationtoaBinomialProbabilityDistributionwithn=100andp=.1

10P(x

<11.5)=.6915x11.5NormalApproximationofBinomialProbabilities10P(x=12)=.7967-.6915=.1052x11.512.5TheNormalApproximationtotheProbabilityof12Successesin100Trialsis.1052

ExponentialProbabilityDistributionTheexponentialprobabilitydistributionisusefulindescribingthetimeittakestocompleteatask.TimebetweenvehiclearrivalsatatollboothTimerequiredtocompleteaquestionnaireDistancebetweenmajordefectsinahighwayTheexponentialrandomvariablescanbeusedtodescribe:Inwaitinglineapplications,theexponentialdistributionisoftenusedforservicetimes.ExponentialProbabilityDistributionApropertyoftheexponentialdistributionis