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3) The grade point averages of a large population of college studenta is approximately normally distributed with mean 2.4 and







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Answer #1

3.

X follows a Normal distribution with mean = 2.4 and S.D = 0.8 Then z=(x-mean)/SD = (x-2.4)/0.8 follows standard Normal.

When x=3, z=(3-2.4)/0.8 = 0.75 P(Z<0.75)=0.7734 [using Normal table or Excel : NORM.S.DIST(0.75, TRUE)] P(X> 3) = 1-PCX < 3)

Ans : 0.2266 or 22.66 %

4.

When x=1.9, z=(1.9-2.4)/0.8 = -0.63 P(Z<-0.63) =0.2643 [using Normal table or Excel : NORM.S.DIST(-0.63, TRUE)] P(X < 1.9) =

Ans : 26.43 %

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