Mens heights are normally distributed with mean 687 in and s

Men\'s heights are normally distributed with mean 68.7 in and standard deviation of 2.8 in. Women\'s heights are norammly distributed with mean 63.3 in and standard deviation of 2.5 in. The standard doorway hieght is 80 in. a) what percentage of men are too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b) If a statistician designs a house so that all of the doorways have heights that are sufficient for all the men except the tallest 5%, what doorway height would be used?

Solution

a)
Mean ( u ) =    68.7
Standard Deviation ( sd )=2.8
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 80) = (80-68.7)/2.8
= 11.3/2.8= 4.0357
= P ( Z <4.0357) From Standard Normal Table
= 1  

100% are MEN too tall to to fit through a standard doorway
without bending          
b)
P(X < 80) = (80-63.3)/2.5
= 16.7/2.5= 6.68
= P ( Z <6.68) From Standard Normal Table
= 1                  

100% are WOMEN too tall to to fit through a standard doorway
without bending
c)              
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.64
P( x-u/ (s.d) > x - 68.7/2.8) = 0.05
That is, ( x - 68.7/2.8) = 1.64
--> x = 1.64 * 2.8+68.7 = 73.306