a. Let x be the height of men and y be the height of women.
X follows Normal distribution with mean 68.7 in and standard deviation of 2.8 in.
Y follows Normal distribution with 64 in and standard deviation of 2.5 in.
The standard doorway height is 80 in.
The percentage of men who are too tall to fit through a standard doorway without bending is
P(X > 80) = [tex] P(\frac{x-mean}{standard deviation} > \frac{80 - 68.7}{2.8} ) [/tex]
= P( Z > 4.0357)
= 1 - P(Z < 4.04)
= 1 - 1
The z score table contains probability below 3. The area below 3 is 1 so probability below 4 is 1
P(X > 80) = 0
The percentage of men who are too tall to fit through a standard doorway without bending is 0%
The percentage of women who are too tall to fit through a standard doorway without bending is
P(Y > 80) = [tex] P(\frac{x-mean}{standard deviation} > \frac{80 - 64}{2.5} ) [/tex]
= P(Z > 6.4)
We know that z table contains values from -3 to +3 only. It is because the probability below -3 is 0 and probability above z=+3 is 0.
P(Y > 80) = 0
The percentage of women who are too tall to fit through a standard doorway without bending is 0%
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 doorway height that separates the tallest 5% men is
P(Z > z) =0.05
It means first we have to find z score such that probability above it is 0.05 and below is 1-0.05 =0 .95
Using z score table to find probability close to 0.95
We do not accurate 0.95 probability value, there are two probability values close to 0.95 as 0.9495 and 0.9505.
The z score corresponding to them is 1.64 and 1.65. So the required z score is
(1.64 +1.65) /2 = 1.645
So we have P(Z < 1.645)= 0.95
Now find x using z=1.645, μ= 68.7 and σ=2.8
x = (1.645 * 2.8) + 68.7
x = 73.3
So the height of doorway needs to be 73.3 inches so that it will be sufficient for all men except the tallest 5%
