A survey found that women\'s heights are normally distributed with mean 63.7in a

Question

A survey found that women's heights are normally distributed with mean 63.7in and standard deviation 2.3in. A branch of the military requires women's heights to be between 58 in and 80 in.

a) what percent of women meet the height requirement? ______ % (round to TWO decimal places as needed)

b) If this 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 height requirements? Women's heights to be at least _____inches and at most _____inches. (round to ONE decimal place as needed)

please circle, highlight, or put in bold the correct answer for a and b. please make it clear what the exact answer is that I should put in the blanks.

Explanation / Answer

Mean is 63.7 and SD is 2.3

a) P(58<x<80) =P((58-63.7)/2.3<z<(80-63.7)/2.3)=P(-2.48<z<7.09)=P(z<7.09)-(1-P(z<2.48))

from normal distribution table it gives 1-(1-0.9934)=0.9934

b) For the bottom 1% we need to look at the z at 0.99 which is actually 2.33 so for bottom 1% it is -2.33

thus x=63.7-2.33*2.3 =58.34

For tallest 2% we need to find the z at 0.98 i,e 2.06

thus x=63.7+2.3*2.06 =68.438

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