33 Density Curves and Normal Distributions - People：33密度曲線與正態(tài)分布的人

3.3DensityCurvesandNormalDistributionsDensityCurvesMeasuringCenterandSpreadforDensityCurvesNormalDistributionsThe68-95-99.7RuleStandardizingObservationsUsingtheStandardNormalTableInverseNormalCalculationsNormalQuantilePlots1Wenowhaveakitofgraphicalandnumericaltoolsfordescribingdistributions.Wealsohaveastrategyforexploringdataonasinglequantitativevariable.Now,we’lladdonemoresteptothestrategy.ExploringQuantitativeDataAlwaysplotyourdata:Makeagraph.Lookfortheoverallpattern(shape,center,andspread)andforstrikingdeparturessuchasoutliers.Calculateanumericalsummarytobrieflydescribecenterandspread.Sometimes,theoverallpatternofalargenumberofobservationsissoregularthatwecandescribeitbyasmoothcurve.ExploringQuantitativeData 22DensitycurvesAdensitycurveisamathematicalmodelofadistribution…Thetotalareaunderthecurve,bydefinition,isequalto1,or100%.Theareaunderthecurveforarangeofvaluesistheproportionofallobservationsforthatrange.AreaunderDensityCurve~RelativeFrequencyofHistogramHistogramofasamplewiththesmoothed,densitycurvedescribingtheoreticallythepopulation.rel.freqoflefthistogram=287/947=.303area=.293underrt.curve4Adensitycurve

isacurvethat:IsalwaysonorabovethehorizontalaxisHasanareaofexactly1underneathitAdensitycurvedescribestheoverallpatternofadistribution.Theareaunderthecurveandaboveanyrangeofvaluesonthehorizontalaxisistheproportionofallobservationsthatfallinthatrange.DensityCurvesDensitycurvescomeinmanyshapes.Somearewellknownmathematicallyandothersaren’t–buttheyalllieabovethehorizontalaxisandhavetotalarea=1.6Ourmeasuresofcenterandspreadapplytodensitycurvesaswellastoactualsetsofobservations.Themedian

ofadensitycurveistheequal-areaspoint―thepointthatdividestheareaunderthecurveinhalf.Themean

ofadensitycurveisthebalancepoint,atwhichthecurvewouldbalanceifmadeofsolidmaterial.Themedianandthemeanarethesameforasymmetricdensitycurve.Theybothlieatthecenterofthecurve.Themeanofaskewedcurveispulledawayfromthemedianinthedirectionofthelongtail.DistinguishingtheMedianandMeanofaDensityCurve6DensityCurves7Themeanandstandarddeviationcomputedfromactualobservations(data)aredenotedby

ands,

respectively,andarecalledthesamplemeanandsamplestandarddeviation.Themeanandstandarddeviationoftheidealizeddistributionrepresentedbythedensitycurvearedenotedbyμ(“mu”)and(“sigma”),

respectively,andaresometimescalledthepopulationmeanandpopulationstandarddeviation.DensityCurves8OneparticularlyimportantclassofdensitycurvesaretheNormalcurves,whichdescribeNormaldistributions.AllNormalcurvesaresymmetric,single-peaked,andbell-shaped.AspecificNormalcurveisdescribedbygivingitsmeanμandstandarddeviationσ.NormalDistributions9ANormaldistribution

isdescribedbyaNormaldensitycurve.AnyparticularNormaldistributioniscompletelyspecifiedbytwonumbers:itsmeanμandstandarddeviationσ.ThemeanofaNormaldistributionisthecenterofthesymmetricNormalcurve.

Thestandarddeviationisthedistancefromthecentertothechange-of-curvaturepointsoneitherside,thepointsofinflectionofthedensity.WeabbreviatetheNormaldistributionwithmeanμandstandarddeviationσasN(μ,σ).NormalDistributionsNormaldistributionse=2.71828…Thebaseofthenaturallogarithmπ=pi=3.14159…Normal–orGaussian–distributionsareafamilyofsymmetrical,bell-shapeddensitycurvesdefinedbyameanm(mu)andastandarddeviations(sigma):N(m,s).xxAfamilyofdensitycurvesHere,meansaredifferent

(m=10,15,and20)whilestandarddeviationsarethe

same(s=3).Here,meansarethesame(m=15)whilestandarddeviationsaredifferent(s=2,4,and6).12The68-95-99.7RuleIntheNormaldistributionwithmeanμandstandarddeviationσ:Approximately68%oftheobservationsfallwithinσofμ.Approximately95%oftheobservationsfallwithin2σofμ.Approximately99.7%oftheobservationsfallwithin3σofμ.The68-95-99.7RuleHere’saN(64.5”,2.5”)distributionofheightsofcollege-agedfemales.13ThedistributionofIowaTestofBasicSkills(ITBS)vocabularyscoresfor7th-gradestudentsinGary,Indiana,isclosetoNormal.SupposethedistributionisN(6.84,1.55).SketchtheNormaldensitycurveforthisdistribution.WhatpercentofITBSvocabularyscoresarelessthan3.74?Whatpercentofthescoresarebetween5.29and9.94?The68-95-99.7Rule14AllNormaldistributionsarethesameifwemeasureinunitsofsizeσfromthemeanμascenter.Ifavariablexhasadistributionwithmeanμandstandarddeviationσ,thenthestandardizedvalueofx,oritsz-score,isNotez=thenumberofs.d.’sawayfrommuthatxis…(sorryaboutthegrammer!)ThestandardNormaldistribution

istheNormaldistributionwithmean0andstandarddeviation1.Thatis,thestandardNormaldistributionisN(0,1)–itisrepresentedbyZandwewriteZ~N(0,1)StandardizingObservations15BecauseallNormaldistributionsarethesamewhenwestandardize,wecanfindareasunderanyNormalcurvefromasingletable.TheStandardNormalTableTableAisatableofareasunderthestandardNormalcurve.Thetableentryforeachvaluezistheareaunderthecurvetotheleftofz.TheStandardNormalTable16Z0.000.010.020.70.75800.76110.76420.80.78810.79100.79390.90.81590.81860.8212P(z<0.81)=0.7910SupposewewanttofindtheproportionofobservationsfromthestandardNormaldistributionthatarelessthan0.81.WecanuseTableA:TheStandardNormalTableTipsonusingTableATocalculatetheareabetween2z-values,firstgettheareaunderN(0,1)totheleftforeachz-valuefromTableA.areabetweenz1andz2=arealeftofz1–arealeftofz2Acommonmistakemadebystudentsistosubtractbothzvalues-itistheareasthataresubtracted,notthez-scores!Thensubtractthesmallerareafromthelargerarea.arearightofz=1-arealeftofz18ExpresstheproblemintermsoftheobservedvariableX.DrawapictureofthedistributionofXandshadetheareaofinterestunderthecurve.Performcalculations.StandardizeXtorestatetheproblemintermsofastandardNormalvariableZ.UseTableAandthefactthatthetotalareaunderthecurveis1tofindtherequiredareaunderthestandardNormalcurve.Writeyourconclusioninthecontextoftheproblem.HowtoSolveProblemsInvolvingNormalDistributionsNormalCalculations19AccordingtotheHealthandNutritionExaminationStudyof1976–1980,theheightsX(ininches)ofadultmenaged18–24areN(70,2.8).Howtallmustamanbe(?below)tobeinthelower10%formenaged18–24?0.10?70

N(70,2.8)InverseNormalCalculations20z0.070.080.09–1.30.08530.08380.0823–1.2.10200.10030.0985–1.10.12100.11900.1170Howtallmustamanbeinthelower10%formenaged18–24?0.10?70

N(70,2.8)Lookuptheclosestprobability(closestto0.10)inthetable.Findthecorrespondingzthestandardizedscore.Thevalueyouseekisthatmanystandarddeviationsfromthemean.Z=–1.28InverseNormalCalculations21Howtallmustamanbeinthelower10%formenaged18–24?0.10?70

N(70,2.8)Weneedto“unstandardize”thez-scoretofindtheobservedvalue(x):Z=–1.28x=70+z(2.8)=70+[(-1.28)

(2.8)]=70+(–3.58)=66.42Amanwouldhavetobeapproximately66.42inchestallorlesstoplaceinthelower10%ofallmeninthepopulation.NormalCalculationsOnewaytoassessifadistributionisindeedapproximatelyNormalistoplotthedataonaNormalquantileplot.Thedatapointsareorderedfromsmallesttolargestandtheirpercentileranksareconvertedtoz-scoreswithTableA.Thesez-scoresarethenplottedagainstthedatatocreateaNormalquantileplot.IfthedistributionisindeedNormal,theplotwillshowastraightline,indicatingagoodmatchbetweenthedataandaNormaldistribution–inJMPthepointsfallwithinthedottedlines.Systematicdeviationsfromastraightlineindicateanon-Normaldistribution.Outliersappearas