Men\'s heights are normally distributed with mean 70 in and standard deviation o
ID: 3176539 • 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 63 5 in and standard deviation of 2.5 in. the standard doorway height is 80 in. a. What percentage of men are too fall to fit through a standard doorway without bonding, 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? 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.
Mean ( u ) =70
Standard Deviation ( sd )=2.8
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X < 80) = (80-70)/2.8
= 10/2.8= 3.5714
= P ( Z <3.5714) From Standard Normal Table
= 0.9998
99.98% of men going without bending
b.
Mean ( u ) =63.5
Standard Deviation ( sd )=2.5
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X < 80) = (80-63.5)/2.5
= 16.5/2.5= 6.6
= P ( Z <6.6) From Standard Normal Table
= 1
all of women are going without bending
c.
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 70/2.8) = 0.05
That is, ( x - 70/2.8) = 1.6449
--> x = 1.6449 * 2.8+70 = 74.6056