A survey found that women\'s heights are normally distributed with mean 63.4 in

Question

A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.4 in. A branch o f the military requires women's heights to be between 58 in and 80 in. Find the percentage o f women meeting the height requirement. Arc many women being denied the opportunity to join this branch of the military because they are too short or too tall? If this branch o f the military changes the height requirements so that all women arc eligible except the shortest 1% and the tallest 2%, what are the new height requirements? The percentage o f women who meet the height requirement is %. (Round to two decimal places as needed.)

Explanation / Answer

let X denotes the heights of the women.

by question X~N(63.4,2.42)

a branch of military requires the women's height to be between 58 in and 80 in

a) hence the percentage of women meeting the height requirement is

P[58<X<80]=P[(58-63.4)/2.4<(X-63.4)/2.4<(80-63.4)/2.4]=P[-2.25<Z<6.9167] where Z~N(0,1)

=P[Z<6.9167]-P[Z<-2.25]=1-0.0122245   [using TABLLE]

=0.9877755=98.78% [answer]

hence only 1.22% of women are denied for being too tall or too short

b) now all women are eligible except the shortest 1% and the tallest 2%

let the new lower bound be l and the new upper bound be u

so P[X<l]=1%=0.01 and P[X>u]=2%=0.02

now P[X<l]=0.01    or, P[(X-63.4)/2.4<(l-63.4)/2.4]=0.01

or, P[Z<(l-63.4)/2.4]=0.01 where Z~N(0,1)

or, P[Z<(l-63.4)/2.4]=0.01=P[Z<-2.32635]   [from table]

or, (l-63.4)/2.4=-2.32635

or, l=-2.32635*2.4+63.4=57.81676

again P[X>u]=0.02

or, P[X<u]=1-0.02=0.98

or, P[(X-63.4)/2.4<(u-63.4)/2.4]=0.98

or, P[Z<(u-63.4)/2.4]=0.98=P[Z<2.05375]      Z~N(0,1)    [from table]

so (u-63.4)/2.4=2.05375

or, u=2.05375*2.4+63.4=68.329

hence the new height requirements is that it should be between 57.81676 in. to 68.329 in.   [answer]

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