A survey found that women\'s heights are normally distributed with mean 63.5 in

Question

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 heightrequirements?

Explanation / Answer

Mean ( u ) =63.5
Standard Deviation ( sd )=2.2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 58) = (58-63.5)/2.2
= -5.5/2.2 = -2.5
= P ( Z <-2.5) From Standard Normal Table
= 0.00621
P(X < 80) = (80-63.5)/2.2
= 16.5/2.2 = 7.5
= P ( Z <7.5) From Standard Normal Table
= 1
P(58 < X < 80) = 1-0.00621 = 0.9938                  

Not many women denied

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 - 63.5/2.2 ) = 0.01
That is, ( x - 63.5/2.2 ) = -2.33
--> x = -2.33 * 2.2 + 63.5 = 58.3828                  
P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.05
P( x-u/ (s.d) > x - 63.5/2.2) = 0.02
That is, ( x - 63.5/2.2) = 2.05
--> x = 2.05 * 2.2+63.5 = 68.0188                  

58.3828 is the shortest, 68.0188 is the longest

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