A survey found that women\'s heights are normally distributed with mean 63.6 in.

Question

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

63.6 in. and standard deviation 2.8 in. The survey also found that men's heights are normally distributed with a mean 67.4 in. and standard deviation 2.6

Complete parts a through c below.

a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the height requirement.The percentage of women who meet the height requirement is

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

The percentage of men who meet the height requirement is

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

Solution

Theory – concept Base

1. Height follows Normal Distribution.

2. If X follows Normal Distribution with mean, m, and standard deviation, s, then

P(X or t) = P[Z or {(t - m)/s}], where Z is Standard Normal Variate, i.e., Z has Normal Distribution with mean 0 and standard deviation 1

3. Percentage = 100 x probability

Now, to work out the solutions,[Note: all normal probabilities are taken directly from Standard Normal Table]

Part (a)

Let W = height of a woman. Then, we are given m = 63.6636 and s = 2.828.

P(a woman conforms to height requirement) = P(57.7 W 75.3)

= P[{(57.7 – 63.6636)/2.828} Z {(75.3 – 63.6636)/2.828}]

= P(- 2.109 Z 4.115) = P(Z 4.115) - P(- 2.109 Z) = 0.99997 – 0.0174 = 0.98257.

Hence, percentage of women meeting height requirement = 100 x 0.98257 = 98.257

= 98.26% ANSWER

Part (b)

Let M = height of a man. Then, we are given m = 67.4674 and s = 2.626.

P(a man conforms to height requirement) = P(57.7 M 75.3)

= P[{(57.7 – 67.4674)/2.626} Z {(75.3 – 67.4674)/2.626}]

= P(- 3.719 Z 2.983) = P(Z 2.983) - P(- 3.719 Z) = 0.9986 – 0.00013 = 0.99847.

Hence, percentage of men meeting height requirement = 100 x 0.99847 = 99.847

= 99.85% ANSWER

Part (c)

If H is the height above which top 5% of men have height, then we have P(M > H) = 0.05 or

P{Z > {(H – 67.4674)/2.626}] = 0.05 => from Standard Normal Table,

{(H – 67.4674)/2.626} = 1.645 or H = 71.787.

Similarly, if h is the height below which 5% of women have height, then we have

P(W < h) = 0.05 or

P{Z < {(h – 63.6626)/2.828}] = 0.05 => from Standard Normal Table,

{(h – 63.6626)/2.828} = - 1.645 or h = 59.011.

So, the revised height requirement is at least 59.0 in and at most 71.8 in ANSWER

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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