Chapter5ElementaryStatisticsLarsonFarberNNormalProbabilityDistributions1IntroductiontoNormalDistributionsThestandardNormalDistributionSection5.12PropertiesofaNormalDistributionThemean,median,andmodeareequalBellshapedandissymmetricaboutthemeanThetotalareathatliesunderthecurveisoneor100%x3Asthecurveextendsfartherandfartherawayfromthemean,itgetscloserandclosertothex-axisbutnevertouchesit.Thepointsatwhichthecurvaturechangesarecalledinflectionpoints.ThegraphcurvesdownwardbetweentheinflectionpointsandcurvesupwardpasttheinflectionpointstotheleftandtotherightxInflectionpointInflectionpointPropertiesofaNormalDistribution4MeansandStandardDeviations2012151810111314161719212291215181011131416171920Curveswithdifferentmeansdifferentstandarddeviations

Curveswithdifferentmeans,samestandarddeviation

5TheStandardScoreThestandardscore,orz-score,representsthenumberofstandarddeviationsarandomvariablexfallsfromthemean.Thetestscoresforacivilserviceexamarenormallydistributedwithameanof152andstandarddeviationof7.Findthestandardz-scoreforapersonwithascoreof:(a)161 (b)148 (c)152(c)(a)(b)6TheStandardNormalDistributionThestandardnormaldistributionhasameanof0andastandarddeviationof1.Usingz-scoresanynormaldistributioncanbetransformedintothestandardnormaldistribution.432101234z7CumulativeAreas

Thecumulativeareaiscloseto1forzscorescloseto3.49.Thecumulativeareaiscloseto0forz-scorescloseto-3.49.

Thecumulativeareaforz=0is0.5000Thetotalareaunderthecurveisone.0123-1-2-3z8Findthecumulativeareaforaz-scoreof-1.25.0123-1-2-3zCumulativeAreas0.1056Readdownthezcolumnonthelefttoz=-1.2andacrosstothecolumnunder.05.Thevalueinthecellis

0.1056,thecumulativearea.Theprobabilitythatzisatmost-1.25is0.1056.P(z

-1.25)=0.10569FindingProbabilitiesTofindtheprobabilitythatzislessthanagivenvalue,readthecumulativeareainthetablecorrespondingtothatz-score.0123-1-2-3zReaddownthez-columnto-1.4andacrossto.05.Thecumulativeareais0.0735.

FindP(z<-1.45)P(z<-1.45)=0.073510FindingProbabilitiesTofindtheprobabilitythatzisgreaterthanagivenvalue,subtractthecumulativeareainthetablefrom1.0123-1-2-3zP(z>-1.24)=0.8925RequiredareaFindP(z>-1.24)Thecumulativearea(areatotheleft)is0.1075.Sotheareatotherightis1-0.1075=

0.8925.0.10750.892511FindingProbabilitiesTofindtheprobabilityzisbetweentwogivenvalues,findthecumulativeareasforeachandsubtractthesmallerareafromthelarger.FindP(-1.25<z<1.17)1.P(z<1.17)=0.87902.P(z<-1.25)=0.10563.P(-1.25<z<1.17)=0.8790-0.1056=0.77340123-1-2-3z120123-1

-2-3zSummaryTofindtheprobabilitythatzislessthanagivenvalue,readthecorrespondingcumulativearea.0123-1-2-3zTofindtheprobabilityisgreaterthanagivenvalue,subtractthecumulativeareainthetablefrom1.0123-1-2-3zTofindtheprobabilityzisbetweentwogivenvalues,findthecumulativeareasforeachandsubtractthesmallerareafromthelarger.13NormalDistributionsFindingProbabilitiesSection5.214ProbabilitiesandNormalDistributions115100Ifarandomvariable,xisnormallydistributed,the

probability

thatxwillfallwithinanintervalisequaltotheareaunderthecurveintheinterval.IQscoresarenormallydistributedwithameanof100andstandarddeviationof15.FindtheprobabilitythatapersonselectedatrandomwillhaveanIQscorelessthan115.

Tofindtheareainthisinterval,firstfindthestandardscoreequivalenttox=115.1501ProbabilitiesandNormalDistributionsFindP(z<1)115100StandardNormalDistributionFindP(x<115)NormalDistributionP(z<1)=0.8413,soP(x<115)=0.8413SAMESAME16Monthlyutilitybillsinacertaincityarenormallydistributedwithameanof$100andastandarddeviationof$12.Autilitybillisrandomlyselected.Findtheprobabilityitisbetween$80and$115.P(80<x<115)NormalDistributionP(-1.67<z<1.25)0.8944-0.0475=0.8469Theprobabilityautilitybillisbetween$80and$115is0.8469.Application17NormalDistributionsFindingValuesSection5.318z0.9803FromAreastoz-scoresLocate0.9803intheareaportionofthetable.Readthevaluesatthebeginningofthecorrespondingrowandatthetopofthecolumn.Thez-scoreis2.06.Findthez-scorecorrespondingtoacumulativeareaof0.9803.z=2.06correspondsroughlytothe98thpercentile.4321012340.980319Findingz-scoresFromAreasFindthez-scorecorrespondingtothe90thpercentile.z0.90Theclosesttableareais.8997.Therowheadingis1.2andcolumnheading.08.Thiscorrespondstoz=1.28.Az-scoreof1.28correspondstothe90thpercentile.20Findingz-scoresFromAreasFindthez-scorewithanareaof.60fallingtoitsright..60.400zzWith.60totheright,cumulativeareais.40.Theclosestareais.4013.Therowheadingis–0.2andcolumnheadingis.05.Thez-scoreis–0.25.Az-scoreof–0.25hasanareaof.60toitsright.Italsocorrespondstothe40thpercentile21Findingz-scoresFromAreasFindthez-scoresuchthat45%oftheareaunderthecurvefallsbetween–zandz.0z-zThearearemaininginthetailsis.55.Halfthisareaisineachtail,sosince.55/2=.275isthecumulativeareaforthenegativezvalueand.275+.45=.725isthecumulativeareaforthepositivez.Theclosesttableareais.2743andthez-scoreis–0.60.Thepositivezscoreis0.60..45.275.27522Fromz-ScorestoRawScoresThetestscoresforacivilserviceexamarenormallydistributedwithameanof152andstandarddeviationof7.Findthetestscoreforapersonwithastandardscoreof(a)2.33(b)-1.75(c)0(a)x=152+(2.33)(7)=168.31(b)x=152+(-1.75)(7)=139.75(c)x=152+(0)(7)=152Tofindthedatavalue,xwhengivenastandardscore,z: 23FindingPercentilesorCut-offvaluesMonthlyutilitybillsinacertaincityarenormallydistributedwithameanof$100andastandarddeviationof$12.Whatisthesmallestutilitybillthatcanbeinthetop10%ofthebills?10%90%Findthecumulativeareainthetablethatisclosestto0.9000(the90thpercentile.)Thearea0.8997correspondstoaz-scoreof1.28.x=100+1.28(12)=115.36.$115.36isthesmallestvalueforthetop10%.zTofindthecorrespondingx-value,use24TheCentralLimitTheoremSection5.425SamplingDistributionsAsamplingdistributionistheprobabilitydistributionofasamplestatisticthatisformedwhensamplesofsizenarerepeatedlytakenfromapopulation.Ifthesamplestatisticisthesamplemean,thenthedistributionisthesamplingdistributionofsamplemeans.SampleSampleSampleSampleSampleSampleThesamplingdistributionconsistsofthevaluesofthesamplemeans,

26xthesamplemeanswillhavea

normaldistributionTheCentralLimitTheoremwithameanandstandarddeviationIfasamplen30istakenfromapopulationwith

anytypedistributionthathasamean= andstandarddeviation=27thedistributionofmeansofsamplesizen,willbenormal

withameanstandarddeviationTheCentralLimitTheoremxIfasampleofanysizeistakenfromapopulationwith

anormaldistributionwithmean=andstandarddeviation=

28Application69.2Distributionofmeansofsamplesize60,willbenormal.ThemeanheightofAmericanmen(ages20-29)isinches.Randomsamplesof60suchmenareselected.Findthemeanandstandarddeviation(standarderror)ofthesamplingdistribution.meanStandarddeviation29InterpretingtheCentralLimitTheoremThemeanheightofAmericanmen(ages20-29)is

=69.2”.Ifarandomsampleof60meninthisagegroupisselected,whatistheprobabilitythemeanheightforthesampleisgreaterthan70”?Assumethestandarddeviationis2.9”.Findthez-scoreforasamplemeanof70:standarddeviationmeanSincen>30thesamplingdistributionofwillbenormal30InterpretingtheCentralLimitTheorem2.14P(>70)zThereisa0.0162probabilitythatasampleof60menwillhaveameanheightgreaterthan70”.=P(z>2.14)=1-0.9838

=0.0162InterpretingtheCentralLimitTheorem

31ApplicationCentralLimitTheoremDuringacertainweekthemeanpriceofgasolineinCaliforniawas$1.164pergallon.Whatistheprobabilitythatthemeanpriceforthesampleof38gasstationsinCaliforniaisbetween$1.169and$1.179?Assumethestandarddeviation=$0.049.standarddeviationmeanCalculatethestandardz-scoreforsamplevaluesof$1.169and$1.179.Sincen>30thesamplingdistributionofwillbenormal32.631.90zApplicationCentralLimitTheoremP(0.63<z<1.90)=0.9713-0.7357=0.2356Theprobabilityis0.2356thatthemeanforthesampleisbetween$1.169and$1.179.33NormalApproximationtotheBinomialSection5.534BinomialDistributionCharacteristicsThereareafixednumberofindependenttrials.(n)Eachtrialhas2outcomes,SuccessorFailure.Theprobabilityofsuccessonasingletrialispandtheprobabilityoffailureisq.p+q=1Wecanfindtheprobabilityofexactlyxsuccessesoutofntrials.Wherex=0or1or2…n.

xisadiscreterandomvariablerepresentingacountofthenumberofsuccessesinntrials.35Application34%ofAmericanshavetypeA+blood.If500Americansaresampledatrandom,whatistheprobabilityatleast300havetypeA+blood?Usingtechniquesofchapter4youcouldcalculatetheprobabilitythatexactly300,exactly301…exactly500AmericanshaveA+bloodtypeandaddtheprobabilities.Or…youcouldusethenormalcurveprobabilitiestoapproximatethebinomialprobabilities.Ifnp

5andnq5,thebinomialrandomvariablexisapproximatelynormallydistributedwithmeanand36Whydowerequirenp5andnq5?

01234544n=5p=0.25,q=.75np=1.25nq=3.75n=20p=0.25np=5nq=15n=50p=0.25np=12.5nq=37.50102030405037BinomialProbabilitiesThebinomialdistributionisdiscretewithaprobabilityhistogramgraph.The

probabilitythataspecificvalueof

x

willoccurisequaltothearea

oftherectanglewithmidpointatx.

Ifn=50andp=0.25findP(15

x17)Addtheareasoftherectangleswithmidpointsatx=15,x=16,x=17.1516170.0890.0660.0430.089+0.066+.043=0.198=.20P(15

x17)=0.197Approximately20%38151617CorrectionforContinuityCheckthatnp=12.5

5andnq=37.5

5.

UsethenormalapproximationtothebinomialtofindP(14

x16)ifn=50andp=0.25Valuesforthebinomialrandomvariablexare15,16and17.39151617CorrectionforContinuityCheckthatnp=12.5

5andnq=37.5

5.

UsethenormalapproximationtothebinomialtofindP(15

x17)ifn=50andp=0.25Theintervalofvaluesunderthenormalcurveis14.5

x17.5.Toensuretheboundariesofeachrectangleareincludedintheinterval,subtract0.5fromaleft-handboundaryandadd0.5toa