A survey found that women\'s heights are normally distributed with mean 63.5 in.

Question

A survey found that women's heights are normally distributed with mean 63.5 in. and standard deviation 2.2 in. The survey also found that men's heights are normally distributed with a mean 68.3 in. and standard deviation 2.8. Complete parts a through c below.

Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 99 in. and a maximum of 6 ft 33 in. Find the percentage of women meeting the height requirement.

The percentage of women who meet the height requirement is %.(Round to two decimal places as needed.)

b. Find the percentage of men meeting the height requirement.

The percentage of men who meet the height requirement is %(Round to two decimal places as needed.)

c. If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?

The new height requirements are at least in. and at most   in.

Explanation / Answer

I think you meant 4 feet 9 inches and 6 ft 3 inches. because 33 inches and 99 inches will give absolutely different values.

This involves finding z values which tell you percentage from 0 to x.

z=(x-)/

(a) In this case our minimum sample point is 4 ft 9 inches or 4*12 + 9 = 57 inches.

The maximum sample point is 6 ft 3 inches or 6*12 + 3 = 75.

For the minimum pointP(X <= 57); z= (57-63.5)/2.2 = -2.95 which equates to 0.00159.

The maximum P(X<=75) ; z=(75-63.5)/2.2 = 5.23 which is =1

Therefore the required probability = P(X<=75) - P( X<=57) = 1-0.00159 = 0.99841

So there is a 99.84% chance of meeting the height requirement for women.

(b) For men the minimum P(X<=57); z = (57-68.3)/2.8 = -4.035 which is essentially 0 (0.0001).

The maximum P(X<=75); z = (75-68.3)/2.8 = 2.39 which is 0.99158

Therefore the required probability = P(X<=75) - P( X<=57) = 0.99158 - 0.00001 = 0.99157

So there is a 99.16% chance of meeting the height requirement for men.

c)   For part c, we must work backward from the table. Look for the Z-scores that give you 45% above the mean and 45% below the mean. These turn out to be: +1.64 and -1.64

Now we convert these back to inches away from the mean and add or subtract:

     men --      68.3 + 1.64(2.8) = 72.89 inches

     women -- 63.5 - 1.64(2.2) = 59.89 inches

    New range: At least 59.89 in and at most 72.89 in

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.