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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 ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 "Body Data" in Appendix B). a. If I woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb. b. If 25 different women are randomly selected, find the probability that their mean weight is between 140 lb and 211 lb. c. When redesigning the fighter jet ejection seats to better accommodate women, which probability is more relevant: the result from part (a) or the result from part (b)? Why?

Step by step solution

01

- Identify Key Parameters for Part (a)

For part (a), we need to find the probability that a woman's weight is between 140 lb and 211 lb. The key parameters are the mean (\u03bc) and standard deviation (). Given: \(\) = 171 lb and \(\) = 46 lb.

02

- Convert Weights to Z-Scores for Part (a)

We convert the weights of 140 lb and 211 lb to Z-scores using the formula: \ \( Z = \frac{X - \mu}{s} \). For 140 lb: \ \( Z_{140} = \frac{140 - 171}{46} = -0.6748 \). For 211 lb: \ \( Z_{211} = \frac{211 - 171}{46} = 0.8696 \).

03

- Find Corresponding Probabilities for Z-Scores (Part (a))

Use Z-tables to find probabilities for the calculated Z-scores. For \ \( Z_{140} = -0.6748 \): P(Z < -0.6748) 鈮� 0.25. For \ \( Z_{211} = 0.8696 \): P(Z < 0.8696) 鈮� 0.80.

04

- Calculate Probability for Part (a)

The probability that a randomly selected woman's weight is between 140 lb and 211 lb is \ \( P(140 < X < 211) \): P(Z < 0.8696) - P(Z < -0.6748) = 0.80 - 0.25 = 0.55.

05

- Identify Key Parameters for Part (b)

For part (b), we need to find the probability that the mean weight (\bar{X}) of 25 women is between 140 lb and 211 lb. Use the same mean and standard deviation, but adjust the standard deviation using the formula: \ \( \bar{\sigma} = \frac{\sigma}{\sqrt{n}} \). \(\) = 171 lb, \(\) = 46 lb, and \(\) = 25.

06

- Calculate Standard Error for Part (b)

Calculate the standard error: \ \( \bar{\sigma} = \frac{46}{\sqrt{25}} = 9.2 \).

07

- Convert Mean Weights to Z-Scores for Part (b)

Convert the weights of 140 lb and 211 lb to Z-scores using the standard error: For 140 lb: \ \( Z_{140} = \frac{140 - 171}{9.2} = -3.37 \). For 211 lb: \ \( Z_{211} = \frac{211 - 171}{9.2} = 4.35 \).

08

- Find Corresponding Probabilities for Z-Scores (Part (b))

Use Z-tables to find probabilities for the Z-scores. For \ \( Z_{140} = -3.37 \): P(Z < -3.37) 鈮� 0.0004. For \ \( Z_{211} = 4.35 \): P(Z < 4.35) 鈮� 1.

09

- Calculate Probability for Part (b)

The probability that the mean weight of 25 randomly selected women is between 140 lb and 211 lb is \ \( P(140 < \bar{X} < 211) \): P(Z < 4.35) - P(Z < -3.37) = 1 - 0.0004 = 0.9996.

10

- Determine Relevance for Part (c)

When redesigning the ejection seats, the probability related to the mean weight of a sample (part (b)) is more relevant since the seat design must accommodate a range of weights, not just individual cases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

probability

Probability helps us determine the likelihood of an event occurring. In our case, we want to find out the chances of a randomly selected woman's weight falling within a certain range. This is done by first understanding the distribution of weights and then converting the problem into a probability question. Probabilities range from 0 to 1, where 0 means the event cannot happen and 1 means it will certainly happen. Probability is foundational in statistics and engineering for making predictions and informed decisions.

normal distribution

The normal distribution, often called the 'bell curve,' is a way to describe how data points are spread out around a mean. It's symmetric, with most values clustering around the center. In our problem, women's weights are normally distributed with a mean weight of 171 lb and a standard deviation of 46 lb. The shape of this distribution helps us understand the likelihood of certain weights occurring. Engineers use normal distribution to model real-world variables like weights, temperatures, or measurement errors.

z-scores

A Z-score tells us how many standard deviations a data point is from the mean. For our weight distribution, calculating the Z-score helps convert actual weights into a standardized form, making it easier to find probabilities. The formula is: \(Z = \frac{X - \bar{\text{X}}}{\text{S}}\). For instance, a weight of 140 lb corresponds to a Z-score of -0.6748. It means that 140 lb is 0.6748 standard deviations below the mean. A Z-table then helps us look up the probability that a weight is below this Z-score.

sample mean

The sample mean is the average of a sample of data points. In part (b) of our exercise, we calculate the mean weight of 25 randomly selected women. Instead of focusing on individual weights, we're interested in the average weight of these women. We use the standard error, which adjusts the standard deviation for the sample size (=25), to calculate Z-scores for the sample mean. Finding these probabilities helps when designing equipment that will be used by groups of people, not just individuals.