Men\'s heights are normally distributed with mean 70 in and standard deviation o
ID: 3257229 • Letter: M
Question
Men's heights are normally distributed with mean 70 in and standard deviation of 2.8 in. Women's heights are normally distributed with mean 64 in and standard deviation of 2.5 in. The standard doorway height 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 men except the tallest 5%, what doorway height would be used? Click to view page 1 of the table. Click to view page 2 of the table. a. The percentage of men who are too tall to fit through a standard door without bending is _ %. (Round to two decimal places as needed.)Explanation / Answer
a)
men's height
mean = 70 , s = 2.8 , x = 80
z = ( x - mean)/s
= ( 80 - 70)/2.8
= 3.571
P(x > 80) = P( z > 3.571) = 0.0002 = 0.02%
Women's height
mean = 64 , s = 2.5 , x = 80
z = ( x - mean)/s
= ( 80 - 64)/2.8
= 5.714
P(x > 80) = P( z > 5.714) = 0%
b)
z value at tall 5% = 1.645
mean = 70 , s = 2.8
z = ( x - mean)/ s
1.645 = ( x - 70) / 2.8
x = 74.606