A survey found that women\'s heights are normally distributed with moan 63 4 in

Question

A survey found that women's heights are normally distributed with moan 63 4 in and standard deviation 2.5 in. A branch of the military requires women's heights to be between 58 in and 00 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? 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? the percentage of women who meet the height requirement is %. (Round to two decimal places as needed)

Explanation / Answer

Mean ( u ) =63.4
Standard Deviation ( sd )=2.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 58) = (58-63.4)/2.5
= -5.4/2.5 = -2.16
= P ( Z <-2.16) From Standard Normal Table
= 0.01539
P(X < 80) = (80-63.4)/2.5
= 16.6/2.5 = 6.64
= P ( Z <6.64) From Standard Normal Table
= 1
P(58 < X < 80) = 1-0.01539 = 0.9846                  
percentage of women that qualified are 98.46%
no, we have majority of women that are qualified
b.
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.326
P( x-u/s.d < x - 63.4/2.5 ) = 0.01
That is, ( x - 63.4/2.5 ) = -2.33
--> x = -2.33 * 2.5 + 63.4 = 57.5841 ~ 57.59                  
P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.0537
P( x-u/ (s.d) > x - 63.4/2.5) = 0.02
That is, ( x - 63.4/2.5) = 2.0537
--> x = 2.0537 * 2.5+63.4 = 68.5344 ~ 68.544              

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