http://www.youtube.com/watchv=xDIyAOBa_yU

TheStandardNormalDistributionNote:Thisfileisonmywebpageforthiscourse.
Seethesyllabusforthewebaddress.
Noticethatthexaxisisstandardscores,alsocalledzscores.
Thismeansthatthedistributionhasapopulationmeanofzero,andapopulationstandarddeviationof1.
Themostsignificantthingaboutthenormaldistributionisthatpredictableproportionsofcasesoccurinspecificregionsofthecurve.
50%50%Noticethat:(1)34.
13%ofthescoresliebetweenthemeanand1sdabovethemean,(2)13.
6%ofthescoresliebetween1sdabovethemeanand2sdsabovethemean,(3)2.
14%ofthescoresliebetween2sdsabovethemeanand3sdsabovethemean,and(4)0.
1%ofthescoresintheentireregionabove3sds.
Thecurveissymmetrical,sotheareabetw0and-1=34.
13%,-1to-2=13.
6%,etc.
Also,50%ofthecasesareabove0,50%below.
IQPercentileProblem1:IQ:m=100,s=15.
ConvertanIQscoreof115intoapercentile,usingthestandardnormaldistribution.
Step1:ConvertIQ=115intoazscore:z=xi-m/s=(115-100)/15=15/15;z=+1.
0(1sdabovethemean)IQ=115,z=1.
0Step2:Calculatetheareaunderthecurveforallscoresbelowz=1(percentile=%ofscoresfallingbelowascore).
Areaunderthecurvebelowz=1.
0:34.
13+50.
00=84.
13.
Wegetthisbyadding34.
13(theareabetweenthemeanandz=1)to50.
00(50%istheareaunderthecurveforvalueslessthanzero;i.
e.
,theentireleftsideofthebellcurve).
So,anIQscoreof115(z=1.
0)hasapercentilescoreof84.
13.
IQ=115,z=+1.
0Thepartwiththeslantylinesrepre-sentstheportionofthedistributionwe'relookingfor.
IQPercentileProblem2:IQ:m=100,s=15.
ConvertanIQscoreof85intoapercentile,usingthestandardnormaldistribution.
Step1:ConvertIQ=85intoazscore:z=xi-m/s=(85-100)/15=-15/15;z=-1.
0(1sdbelowthemean)IQ=85,z=-1.
0Step2:Calculatetheareaunderthecurveforallscoresbelowz=-1.
Areaunderthecurvevaluesbelowz=-1.
0:50.
00-34.
13=15.
87.
Wegetthisbysubtracting34.
13(theareabetweenthemeanandz=-1)from50.
00(50%isthetotalareaunderthecurveforvalueslessthanzero;i.
e.
,theentireleftsideofthebellcurve).
So,anIQscoreof85(z=-1.
0)hasapercentilescoreof15.
87.
IQ=85,z=-1.
0z=xi-m/s=85-100/15=-15/15z=-1.
0Areaunderthecurveforz=-1.
0:50.
00-34.
13=15.
87.
Wegetthisbysubtracting34.
13from50.
00(50.
00isthetotalareaunderthecurveforvalueslessthanzero;i.
e.
,theentireleftsideofthebellcurve.
)Wethereforewantthe50%minustheareabetweenzeroand-1.
0).
So,anIQscoreof85hasapercentilescoreof15.
87.
IQ=85,z=-1.
0