A survey found that​ women's heights are normally distributed with mean 63.8 in. and standard deviation 2.9 in.

Question

The survey also found that​ men's heights are normally distributed with mean 67.5 in. and standard deviation 3.2 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 63 in.

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?

Answer 1

You need to use the information given to find how to set height requirements that will exclude the percent of the population. You know the Z score is a measure of how many standard deviations a measurement is above the mean (positive score) or below the mean (negative score). On the high end, you want to add a number of inches that will exclude the tallest 50% of the men. That's easy: the mean is the dividing line between the tallest 50% and the shortest 50%, so if you set the top height requirement (upper bound) at the mean (67.5 in.), it will exclude the tallest 50% of the men. Now you need to set the lower bound to exclude the shortest 5% of the men. Looking in the body of a Z table, we find the closest value to 5% (.05) corresponds to a Z score halfway between -1.65 (.0495) and -1.64 (.0505), so we call it -1.645. We multiply that by the standard deviation for men (3.2 in.) and we get 5.264 in. Subtracting this value from the mean (67.5 in.), we get 62.236, which we round up to 63.3 for the lower bound.