In: Statistics and Probability
A survey found that women's heights are normally distributed with mean 62.9 in. and standard deviation 3.8 in. The survey also found that men's heights are normally distributed with mean 67.8 in. and standard deviation 3.5 in. Consider an executive jet that seats six with a doorway height of 56.4 in. Complete parts (a) through (c) below. a. What percentage of adult men can fit through the door without bending? The percentage of men who can fit without bending is?
Given in a survey it is found that women's heights are normally
distributed with mean
= 62.9 in. and a standard deviation of
= 3.8 in. The survey also found that men's heights are normally
distributed with mean
= 67.8 in. and a standard deviation of
= 3.5 in. Considering an executive jet that seats six with a
doorway height of X = 56.4 inches, the percentage of adult men can
fit through the door without bending is calculated using the
Z-score since the distribution is normal.
Here the percentage of adult men can fit through the door without bending are the men whose height is below X = 56.4, thus the Z score at X is calculated as:
Thus P( X <= 56.4) = P( Z < -3.257) proportion of adult men can fit through the door without bending is calculated using the excel formula for distribution which is =NORM.S.DIST(-3.257, TRUE), thus the proportion is computed as 0.000563, and converting into percentage we get 0.000563* 100 = 0.0563%.
Thus the percentage of adult men can fit through the door without bending is 0.0563%