survey found that women\'s heights are normally distributed with mean 63 9 in an
ID: 2908548 • Letter: S
Question
survey found that women's heights are normally distributed with mean 63 9 in and standard deviation 2.4 in. A branch of the military requires women's heights between 58 in and 80 in short or too tall? requirements? a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this br b. I this branch of the military changes the height women are e anch of the military because they requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new heigh a. The percentage of women who meet the height requirement isExplanation / Answer
1) P(X < A) = P(Z < (A - mean)/standard deviation)
Mean = 63.9 in
Standard deviation = 2.4 in
a) P(58 < X < 80) = P(X < 80) - P(X < 58)
= P(Z < (80 - 63.9)/2.4) - P(Z < (58 - 63.9)/2.4)
= P(Z < 6.71) - P(Z < -2.46)
= 1 - 0.0069
= 0.9931
b) Let the limit of shortest height be S and the limit of tallest height be T
P(X < S) = 0.01
P(Z < (S - 63.9)/2.4) = 0.01
(S - 63.9)/2.4 = -2.33
S = 58.31 in
P(X < T) = 0.98
P(Z < (T - 63.9)/2.4) = 0.98
(T - 63.9)/2.4 = 2.05
T = 68.82 in
2) Let the minimum table clearance for mean be M
P(X < M) = 0.95
P(Z < (M - 21.3)/1.2) = 0.95
(M - 21.3)/1.2 = 1.645
M = 23.3 in
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