Men\'s heights are normally distributed with mean 68.7 in. and standard deviatio
ID: 2921450 • Letter: M
Question
Men's heights are normally distributed with mean 68.7 in. and standard deviation of 2.8 in.
Women's heaights are normally distributed with mean 63.2 in. and standard deviation of 2.5 in.
The standard doorway height is 80 in. 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? If a statistician designs a house so that all of the doorways have heights that are sufficient for all men exceot the tallest 5%, what doorway height would be used?
a. The percentage of men who are too tall to fit through a standard door without being is ____________ % ( Round to two decimal places as needed.)
The percentage of women who are too tall to fit through a standard door without bending is _______ % (Round to two decimal places as needed.)
b. The statistician would design a house with doorway height [ ] in. (Round to the nearesr tenth as needed.)
Explanation / Answer
We will use the normal distribution params to solve the problem:
For Men:Mu = 68.7, Stdev = 2.8
For Women:Mu = 63.2, Stdev = 2.5
Stair d height = 80
a. P(event) = P(X>80) = P(Z> 80-68.7/2.8) = P(Z>4.036) = 1-.99997 = 00.00%
P(event) = P(X>80) = P(Z> 80-63.2/2.5) = P(Z>6.72) = 1-1 = 00.00%
b.
To build height that are suffieicnt for all men except the tallest 5% = ?
P(X>=c) = .05
Z for top 5% is 1.645
So, (c-68.7)/2.8 = 1.645, c = 68.7+2.8*1.645 = 73.3. The height of door must be 73.3 inches to not let top 5% of men through the door.