*A survey found that women\'s heights are normally distributed with a mean 63.5

Question

*A survey found that women's heights are normally distributed with a mean 63.5 in and standard deviation 2.2 in. A branch of the militay requires women's heghts to be between 58 in and 80 in.

a) Find the percentage of women meeting the height requirement. And are many women being denied the opportunity to join this branch of the military because they are too short or tall?

b) If thi branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new least and most height requiements?

Explanation / Answer

given that mu =63.5 and sigma =2.2

58<x<80

Convert to z score as

58-63.5/2.2 <z<80-63.5/2.2

-2.5<z<7.5

P(-2.5<z<7.5) = 0.9938

Nearly 99.38% of women meet this height requirement.

No . Women are not denied the opportunity.

b) Shortest 1% and tallest 2% means

z<-0.255   or z>2.33

x<63.5-2.2(0.255) or X > 63.5+2.5(0.255)

x<62.99,or x>64.1375

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