Question

A survey found that​ women's heights are normally distributed with mean

63.5

in. and standard deviation

3.1

in. The survey also found that​ men's heights are normally distributed with mean

68.5

in. and standard deviation

3.4

in. Most of the live characters employed at an amusement park have height requirements of a minimum of

55

in. and a maximum of

64

in. Complete parts​ (a) and​ (b) below.

A) The percentage of men who meet the height requirements is?

B) If the height requirements are changed to exclude only  the tallest 50% of men and the shortest 5% of men what are the new height requirements?

MIN & MAX

Answer 1

Let X in. be the height of a randomly selected woman. X is normally distributed with mean in. and standard deviation in.

Let Y in be the height of a randomly selected man. Y is normally distributed with mean in. and standard deviation in

A) The percentage of men who meet the height requirements is?

The proportion of men who meet the height requirements is same as The probability that a randomly selected man has a height between 55 and 64 in.

ans: The percentage of men who meet the height requirements is 9.34%

B) If the height requirements are changed to exclude only  the tallest 50% of men and the shortest 5% of men what are the new height requirements?

Let 50% of the men be above q inches in height. This is same as the probability of a randomly selected man having a height greater than q is 0.50.

P(Y<q)=0.5

We know that Y (the height of men) is normally distributed and the area under the normal curve to the right (or to the left) of mean is 0.5. Hence q must be the mean of Y, which is 68.5 inches. That is P(Y<68.5)=0.5

Or 50% of the men are above 68.5 inches in height. That is, we have to exclude men above 68.5 in in height

Let the shortest 5% of men be below r inches in height. This is same as the probability of a randomly selected man having a height less than r is 0.05.

The z value corresponding to the probability 0.05 is

However, we know that P(Z<0)=0.5. Since we need a probability less than 0.5, the value of z must be negative (less than 0).

Hence

Using the standard normal tables, we get for z=1.645, P(Z<1.645)=0.95

This means,

We need

We can equate the z score of r to -1.645 and get

That is, the shortest 5% of men are below 62.907 in. in height

ans:

MIN: 62.91 inches & MAX=68.5 inches