Men's heights are normally distributed with mean 68.7 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?

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) =

= 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) =

= 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%

6e14 is the answer.