A survey found that women\'s heights are normally distributed with mean 63.6 in

Question

A survey found that women's heights are normally distributed with mean 63.6 in and standard deviation 2.3 in. A branch of the military requires women's heights etc be between 58 in and 80 in. Find the percentage of women meeting the height requirement. Are many women being derived 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

Explanation / Answer

Normal Distribution
Mean ( u ) =63.6
Standard Deviation ( sd )=2.3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 58) = (58-63.6)/2.3
= -5.6/2.3 = -2.4348
= P ( Z <-2.4348) From Standard Normal Table
= 0.00745
P(X < 80) = (80-63.6)/2.3
= 16.4/2.3 = 7.1304
= P ( Z <7.1304) From Standard Normal Table
= 1
P(58 < X < 80) = 1-0.00745 = 0.9925                  
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 - 63.6/2.3 ) = 0.01
That is, ( x - 63.6/2.3 ) = -2.33
--> x = -2.33 * 2.3 + 63.6 = 58.2502                  
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 - 63.6/2.3) = 0.02
That is, ( x - 63.6/2.3) = 2.05
--> x = 2.05 * 2.3+63.6 = 68.3242  

Shortest 58.2502 , tallest 68.3242   

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