Question

When women were finally allowed to become pilots of fighter​ jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ejection seats were designed for men weighing between 140lb and 201lb. Weights of women are now normally distributed with a mean of 173 and a standard deviation of 37.

Complete parts​ (a) through​ (c) below.

a. If 1 woman is randomly​ selected, find the probability that her weight is between 140lb and 201lb.

The probability is approximately _____. ​(Round to four decimal places as​ needed.)

b. If 27 different women are randomly​ selected, find the probability that their mean weight is between 140lb and 201lb.

The probability is approximately _____. ​(Round to four decimal places as​ needed.)

c. When redesigning the ejection​ seat, which probability is more​ relevant?

A. The part​ (a) probability is more relevant because the seat performance for a single pilot is more important.

B. The part​ (a) probability is more relevant because the seat performance for a sample of pilots is more important.

C. The part​ (b) probability is more relevant because the seat performance for a sample of pilots is more important.

D. The part​ (b) probability is more relevant because the seat performance for a single pilot is more important.

Answer 1

a. 0.73

b. 1.00

c. A

Explanation:

a. Mean weight of women = 173 lb

Standard deviation of women's weight = 37 lb

Let's find the z-scores for the lower and upper limits:

For 140 lb:

z1 = (140 - 173) / 37 ≈ -0.8919

For 201 lb:

z2 = (201 - 173) / 37 ≈ 0.7568

P(140 lb < weight < 201 lb) = P(-0.8919 < Z < 0.7568)

P(-0.8919 < Z < 0.7568) ≈ 0.7264

So, the probability that a randomly selected woman's weight is between 140 lb and 201 lb is 0.7264

b.

n = 27 (sample size)

Standard deviation = 37 / √27 ≈ 7.124

Finding z-score

For 140 lb:

z1 = (140 - 173) / 7.124 ≈ -4.628

For 201 lb:

z2 = (201 - 173) / 7.124 ≈ 3.932

= P(-4.628 < Z < 3.932)

P(-4.628 < Z < 3.932) ≈ 1.0000

So, the probability is 1.0000.

c.

A. the probability in part (a) is ....