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The test scores for the analytical writing section of a particular standardized test can be approximated by a normal distribu
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Let X be the test score for the analytical writing. From the given graph we can write, X~N(3.6, 0.832).

a) The maximum score that can be in the bottom 20%, i. e. , the value of X=x for which the CDF of X is 0.20 is,

P(X<x) =0.20

=P(Z<(x-3.6) /0.83)=0.20 [where Z=(X-3.6) /0.83~N(0, 1) ]

=P(Z<(x-3.6) 0.83) =P(Z<-0.85) [from the standard normal table]

i.e., (x-3.6) /0.83=-0.85

Or, x=2.89(from the standard normal table)

The required score is=2.89.

b) Let the interval in which 60% of scores lie be (-x,x)

Thus, P(-X<X<X) =0.60

Or, P(X<x)-P(X<-x) =0.60

Or, 2*P(X<x) -1=0.60[ since P(X<-x) =1-P(X<x)]

Or, P(X<x) =0.80

Or, P(Z<(x-3.6) /0.83) =0.80

Or, P(Z< ( x-3. 6) /0.83) =P(Z<0.85) [from the standard normal table)

Or, (x-3.6) /0.83=0.85

Or, x =0.41(taking upto 2 decimal places)

Thus the interval in which the 60% of the scores lie is, (-0.41, 0.41)

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