Question 1011627
A survey found that​ women's heights are normally distributed with mean 63.6 in and standard deviation 2.3 in. A branch of the military requires​ women's heights to be between 58 in and 80 in. 
a. Find the percentage of women meeting the height requirement. 
Using the z-scores you would get normalcdf(58,80,63.6,2.3) = 99.26%
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Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall?:: No.
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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?
invNorm(0.01) = -2.3265 ; invNorm(0.98) = 2.0538
Lower: x = -2.3265*2.3 + 63.6 
Upper: x = +2.0538*2.3 + 63.6 

a. The percentage of women who meet the height requirement is 
​Comment: You just eliminated 1% of the shortes and 2% of the tallest.
 
Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? 
A.
​No, because the percentage of women who meet the height requirement is fairly small.
B.
​Yes, because the percentage of women who meet the height requirement is fairly large.
C.
​Yes, because a large percentage of women are not allowed to join this branch of the military because of their height.
D.
​No, because only a small percentage of women are not allowed to join this branch of the military because of their height. 
b. For the new height​ requirements, this branch of the military requires​ women's heights to be at least given above in and at most given above in.
​(Round to one decimal place as​ needed.)
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Cheers,
Stan H.
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