统计学入门——正态分布 36页

Copyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-1Chap6-1Chapter6TheNormalDistributionBasicBusinessStatistics12thEdition\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-2Chap6-2LearningObjectivesInthischapter,youlearn:TocomputeprobabilitiesfromthenormaldistributionHowtousethenormaldistributiontosolvebusinessproblemsTousethenormalprobabilityplottodeterminewhetherasetofdataisapproximatelynormallydistributed\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-3Chap6-3ContinuousProbabilityDistributionsAcontinuousrandomvariableisavariablethatcanassumeanyvalueonacontinuum(canassumeanuncountablenumberofvalues)thicknessofanitemtimerequiredtocompleteatasktemperatureofasolutionheight,ininchesThesecanpotentiallytakeonanyvaluedependingonlyontheabilitytopreciselyandaccuratelymeasure\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-4Chap6-4TheNormalDistribution‘BellShaped’SymmetricalMean,MedianandModeareEqualLocationisdeterminedbythemean,μSpreadisdeterminedbythestandarddeviation,σTherandomvariablehasaninfinitetheoreticalrange:+toMean=Median=ModeXf(X)μσ\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-5Chap6-5TheNormalDistributionDensityFunctionTheformulaforthenormalprobabilitydensityfunctionisWheree=themathematicalconstantapproximatedby2.71828π=themathematicalconstantapproximatedby3.14159μ=thepopulationmeanσ=thepopulationstandarddeviationX=anyvalueofthecontinuousvariable\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-6Chap6-6Byvaryingtheparametersμandσ,weobtaindifferentnormaldistributionsManyNormalDistributions\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-7Chap6-7TheNormalDistributionShapeXf(X)μσChangingμshiftsthedistributionleftorright.Changingσincreasesordecreasesthespread.\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-8Chap6-8TheStandardizedNormalAnynormaldistribution(withanymeanandstandarddeviationcombination)canbetransformedintothestandardizednormaldistribution(Z)NeedtotransformXunitsintoZunitsThestandardizednormaldistribution(Z)hasameanof0andastandarddeviationof1\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-9Chap6-9TranslationtotheStandardizedNormalDistributionTranslatefromXtothestandardizednormal(the“Z”distribution)bysubtractingthemeanofXanddividingbyitsstandarddeviation:TheZdistributionalwayshasmean=0andstandarddeviation=1\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-10Chap6-10TheStandardizedNormalDistributionAlsoknownasthe“Z”distributionMeanis0StandardDeviationis1Zf(Z)01ValuesabovethemeanhavepositiveZ-values,valuesbelowthemeanhavenegativeZ-values\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-11Chap6-11ExampleIfXisdistributednormallywithmeanof$100andstandarddeviationof$50,theZvalueforX=$200isThissaysthatX=$200istwostandarddeviations(2incrementsof$50units)abovethemeanof$100.\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-12Chap6-12ComparingXandZunitsZ$1002.00$200$XNotethattheshapeofthedistributionisthesame,onlythescalehaschanged.Wecanexpresstheproblemintheoriginalunits(Xindollars)orinstandardizedunits(Z)(μ=$100,σ=$50)(μ=0,σ=1)\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-13Chap6-13FindingNormalProbabilitiesabXf(X)PaXb()≤Probabilityismeasuredbytheareaunderthecurve≤PaXb()<<=(Notethattheprobabilityofanyindividualvalueiszero)\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-14Chap6-14f(X)XμProbabilityasAreaUndertheCurve0.50.5Thetotalareaunderthecurveis1.0,andthecurveissymmetric,sohalfisabovethemean,halfisbelow\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-15Chap6-15TheStandardizedNormalTableTheCumulativeStandardizedNormaltableinthetextbook(AppendixtableE.2)givestheprobabilitylessthanadesiredvalueofZ(i.e.,fromnegativeinfinitytoZ)Z02.000.9772Example:P(Z<2.00)=0.9772\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-16Chap6-16TheStandardizedNormalTable(TableE2)ThevaluewithinthetablegivestheprobabilityfromZ=uptothedesiredZ-value.97722.0P(Z<2.00)=0.9772TherowshowsthevalueofZtothefirstdecimalpointThecolumngivesthevalueofZtotheseconddecimalpoint2.0...(continued)Z0.000.010.02…0.00.1\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-17Chap6-17GeneralProcedureforFindingNormalProbabilitiesDrawthenormalcurvefortheproblemintermsofXTranslateX-valuestoZ-valuesUsetheStandardizedNormalTable(TableE2)TofindProb(a18.6)X18.618.0\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-22Chap6-22NowFindP(X>18.6)…(continued)Z0.120Z0.120.547801.0001.0-0.5478=0.4522P(X>18.6)===P(Z>0.12)=1.0-P(Z≤.12)=1-.5478=.4522FindingNormalUpperTailProbabilities(TableE2)\nCopyright©2012PearsonEducation,Inc.publishingasPrenticeHallChap6-23Chap6-23FindingaNormalProbabilityBetweenTwoValuesSupposeXisnormalwithmean18.0andstandarddeviation5.0.FindP(18