A survey found that women\'s heights are normally distributed with mean 63.8in a

Question

A survey found that women's heights are normally distributed with mean 63.8in and standard deviation 2.3in. A branch of the military requires women's heights to be between 58in and 80in.

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

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?

a). The percentage of women who meet the height requirement is _____ %. (Round to two decimal places as needed.)

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 only a small percentage of women are not allowed to join this branch of the military because of their height.

B.- Yes, because the percentage of women who meet the height requirement is fairly large.

C.- No, because the percentage of women who meet the height requirement is fairly small.

D.- Yes, because a large 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 _____ in and at most ____ in. (Round to one decimal place as needed.)

Explanation / Answer

Mean = 63.8 in and standerd deviation = 2.3 in

Military requirements arein between 58 in and 80 in

so we will find how many percentage of women come under this requirement

P ( 58<= x<= 80; 63..8 ; 2.3) = P( X<= 80; 63.8 ; 2.3) - P ( x <= 58; 63.8 ; 2.3) where Probabilitywill be calculated by z - table

z value for 80 in. = (80 - 63.8)/2.3 = 7.04 so p - value for this = 1, because it is vey insignificant.

Z value for 58 in. = ( 58 - 63.8) /2.3 = - 2.52 so p - value = 0.0059

P ( 58<= x<= 80; 63..8 ; 2.3) = 1 - 0.0059 = 0.9941

Percentage of women meet the height requirement = 99.41%

option c is good as the percentage of women who meet the height requirement is fairly small.

for b,the criteria is that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements

so lets say women above x height are excluded, which is 2% of the tallest. so,

P ( x >= X ; 63..8 ; 2.3) = 0.02

from Z - table, Z value for p- value 0.98 ( 1- 0.02) is Z = + 2.055

so height of 2% tallest women = 63.8 + 2.055 * 2.3 = 68.5265 in

same in 1% shortest women case

women below y height will excluded, where heights of women below y height are 1%

P ( y<= Y ; 63..8 ; 2.3) = 0.01

from Z - table, Z value for p- value 0.01 is Z = - 2.325

so height of 2% tallest women = 63.8 - 2.325 * 2.3 = 58.4525 in

so filling in the blanks will be 58.45 in at least and 68.52 at most.

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