A survey found that women\'s heights are normally distributed with mean 63.7 in
ID: 3179708 • Letter: A
Question
A survey found that women's heights are normally distributed with mean 63.7 in and standard deviation23in A branch of the military requires women's heights to be between 58 in and 8o in Find the percentage of women meeting the height requirement Are many women boing derive the opportunity to join this branch of the military because they are too short or two if this branch of the military changes the height requirements so that all women are eight concept the shortest 1% and the tallest 21% what area the new height requirement? The percentage of women who meet the height requirements is (Round to two decimal places as needed) Are many women being derived the opportunity to join this branch of the military because they are too short or too tall? Yes, because a large percentage of women are not allowed to join this branch of the military because of her height No, because the percentage of women who meet the height requirement is fairy small. No because only a small percentage of women are not allowed to join this branch of military because of their height. Yes, because only a small percentage of women who meet the height requirement is large. For the new height requirements this branch of the military requires women's heights too be at in and a at most in (Round to one decimal place as needed)Explanation / Answer
a. From information given, mu=63.7, and sigma=2.3. Compute Z scores for x1=58 and x2=80 by substituting values in following Z score formula.
Z=(x-mu)/sigma, where, x is raw score, mu is population mean, and sigma is population standard deviation.
Z1=(58-63.7)/2.3=-2.48 and Z2=7.09
The two Z scores are opposite in signs, therefore, find areas between mean and respective Z scores and add them
P(58<X<80)=0.4934+0.4999=0.9933~99.33%
From the ans, explanation C is correct.
b. Compute the raw score corresponding to Z=-2.33 (representing shortest 1%) and Z=2.05 (representing tallest 2%).
X1=-2.33*2.3+63.7=58.34 (ans)
X2=2.05*2.3+63.7=68.41 (ans)
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