Assume that heights of men are normally distributed with a m

Assume that heights of men are normally distributed with a mean of 69.0 in. and a standard deviation of 2.8 in. Also assume that heights of women are normally distributed with a mean of 63.6 in. and a standard deviation of 2.5 in. (based on data from the National Health Survey). The Mark VI monorail used at Disney World and the Boeing 757-200 ER airline have doors with a height of 72 inches. A. What percentage of adult men can fit through the doors without bending? What percentage of a women can fit through the door without bending? What doorway height would allow 98% of adult men to fit without bending?

Solution

a)
Men avaerage ( u ) =69
Standard Deviation ( sd )=2.8
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 72) = (72-69)/2.8
= 3/2.8= 1.0714
= P ( Z <1.0714) From Standard Normal Table
= 0.858                  

b)
Female average( u ) =63.6
Standard Deviation ( sd )=2.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 72) = (72-63.6)/2.5
= 8.4/2.5= 3.36
= P ( Z <3.36) From Standard Normal Table
= 0.9996                  

c)
P ( Z < x ) = 0.98
Value of z to the cumulative probability of 0.98 from normal table is 2.054
P( x-u/s.d < x - 69/2.8 ) = 0.98
That is, ( x - 69/2.8 ) = 2.05
--> x = 2.05 * 2.8 + 69 = 74.7512