6.3.21 Question Help * A survey found that women\'s heights are normally distrib
ID: 3309638 • Letter: 6
Question
6.3.21 Question Help * A survey found that women's heights are normally distributed with mean 63.5 in and standard deviation 2.2 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. 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? Click to view page 1 of the table. Click to view page 2 of the table.Explanation / Answer
Solution:- Given that mean = 63.5 and sd = 2.2
a. P(58 < X < 80) = P((58 - 63.5)/2.2 < Z < (80 - 63.5)/2.2 )
= P(-2.5 < Z < 7.5)
= 0.9938
probability of women that are too short or too tall = 1-0.9938 = 0.0062 = 0.62%
b. =>Let cutoff height for shortest 1% Lower: x = -2.3265*2.2 + 63.5 = 58.3817
=>Let cutoff height for tallest 2% higher: X = 2.05*2.2 + 63.5 = 68.01
new height requirements are 58.3817 and 68.01
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