A survey found that women\'s heights are normally distributed with mean 63.8 in
ID: 3130272 • Letter: A
Question
A survey found that women's heights are normally distributed with mean 63.8 in and standard deviation 2.4in. A branch of the military requires women's heights to be between 58 in 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.Yes, because the percentage of women who meet the height requirement is fairly large.
B.Yes, because a large percentage of women are not allowed to join this branch of the military because of their height.
C.No, because only a small percentage of women are not allowed to join this branch of the military because of their height.
D.No, because the percentage of women who meet the height requirement is fairly small.
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
A survey found that women's heights are normally distributed and we can drow a normal bell curve for the data set. As we know that as per Emperical Rule, 68% of data will fall within the first standard deviation, 95% will fall within the first two standard deviations, and 99.7% will fall within the first three standard deviations of the mean.
Given that :- X Bar= 63.8 and S = 2.4, Lower specification limit = 58, Upper specification limit = 80
Formula:- Z = ( x - ) /
Z= (58-63.8)/2.4= -2.41, P(Z=2.41) = 0.008
Z= (80-63.8)/2.4= 6.75, P(Z=6.75)= .999
Hence total data fall between 58 to 80in is =.999-.008=.991, So 99.10%
a. The percentage of women who meet the height requirement is 99.10%.
(C) No, because only a small percentage of women are not allowed to join this branch of the military because of their height.
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%
1% of lower specification limit (Left Tailed test)= 0.01, Hence Z value = -2.23
-2.23=(X-63.8)/2.4 =) -5.592=X-63.8
Hence X= 58.20 In
2% of Upper specification limit = 0.02, Hence Z value = 2.06
2.06=(X-63.8)/2.4 =) 4.95=X-63.8
Hence X= 68.74 In
For the new height requirements, this branch of the military requires women's heights to be at least 58.20 In in and at most 68.74 In.
Thank you.
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