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 140 lb and 201 lb. Weights of women are now normally distributed with a mean of 175 lb and a standard deviation of 48 lb. Complete parts​ (a) through​ (c) below.
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Part 1
a. If 1 woman is randomly​ selected, find the probability that her weight is between 140 lb and 201 lb.
The probability is approximately
select: 0.4733
0.4733. ​(Round to four decimal places as​ needed.)
Part 2
b. If 28 different women are randomly​ selected, find the probability that their mean weight is between 140 lb and 201 lb.
The probability is approximately

enter your response here. ​(Round to four decimal places as​ needed.)

Let's solve this step by step.

### Part (a)
We need to find the probability that a randomly selected woman's weight is between 140 lb and 201 lb, given that women's weights are normally distributed with:
- Mean \( \mu = 175 \, \text{lb} \)
- Standard deviation \( \sigma = 48 \, \text{lb} \)

We are looking for:
\[
P(140 \leq X \leq 201)
\]
where \( X \) represents the weight of a randomly selected woman. We will use the **standard normal distribution (Z)** to solve this.

#### Step 1: Convert to Z-scores
The formula to convert a value \( x \) to its corresponding Z-score is:
\[
Z = \frac{x - \mu}{\sigma}
\]

For \( x = 140 \, \text{lb} \):
\[
Z_1 = \frac{140 - 175}{48} = \frac{-35}{48} \approx -0.7292
\]

For \( x = 201 \, \text{lb} \):
\[
Z_2 = \frac{201 - 175}{48} = \frac{26}{48} \approx 0.5417
\]

#### Step 2: Use Z-table or calculator to find probabilities
We need the area under the standard normal curve between these Z-scores.

- \( P(Z_1 \leq Z \leq Z_2) = P(-0.7292 \leq Z \leq 0.5417) \)

Using a Z-table or calculator, we get:
- \( P(Z \leq -0.7292) \approx 0.2335 \)
- \( P(Z \leq 0.5417) \approx 0.7052 \)

The probability is:
\[
P(140 \leq X \leq 201) = 0.7052 - 0.2335 = 0.4717
\]
Thus, the probability is approximately **0.4717**.

---

### Part (b)
Now, if 28 women are randomly selected, we need to find the probability that their **mean weight** is between 140 lb and 201 lb.

The central limit theorem tells us that the sampling distribution of the mean will also be normally distributed with:
- Mean \( \mu_{\bar{X}} = \mu = 175 \, \text{lb} \)
- Standard error \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{48}{\sqrt{28}} \approx 9.0706 \, \text{lb} \)

We now need to find:
\[
P(140 \leq \bar{X} \leq 201)
\]
where \( \bar{X} \) is the mean weight of the sample.

#### Step 1: Convert to Z-scores for the mean
For \( \bar{X} = 140 \, \text{lb} \):
\[
Z_1 = \frac{140 - 175}{9.0706} = \frac{-35}{9.0706} \approx -3.858
\]

For \( \bar{X} = 201 \, \text{lb} \):
\[
Z_2 = \frac{201 - 175}{9.0706} = \frac{26}{9.0706} \approx 2.865
\]

#### Step 2: Use Z-table or calculator to find probabilities
- \( P(Z_1 \leq Z \leq Z_2) = P(-3.858 \leq Z \leq 2.865) \)

Using a Z-table or calculator, we get:
- \( P(Z \leq -3.858) \approx 0.00006 \)
- \( P(Z \leq 2.865) \approx 0.9979 \)

The probability is:
\[
P(140 \leq \bar{X} \leq 201) = 0.9979 - 0.00006 = 0.9978
\]
Thus, the probability is approximately **0.9978**.

---

### Summary of Results:
- Part (a): The probability that a randomly selected woman's weight is between 140 lb and 201 lb is approximately **0.4717**.
- Part (b): The probability that the mean weight of 28 randomly selected women is between 140 lb and 201 lb is approximately **0.9978**.

Do you need further details or have any questions?

Here are 5 related questions to extend this topic:
1. How would the probability change if the sample size in part (b) were 50 women instead of 28?
2. What if the weight range changed to 130 lb to 190 lb? How would this affect the probabilities?
3. Can you explain how the Central Limit Theorem applies in part (b)?
4. How would increasing the standard deviation of women's weights affect the probabilities?
5. What is the probability of selecting a woman who weighs more than 201 lb?

**Tip**: Remember, larger sample sizes lead to a smaller standard error, meaning the sample mean is more likely to be closer to the population mean!

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 140 lb and 201 lb. Weights of women are now normally distributed with a mean of 175 lb and a standard deviation of 48 lb. Complete parts (a) through (c) below. Part 1: a. If 1 woman is randomly selected, find the probability that her weight is between 140 lb and 201 lb. Part 2: b. If 28 different women are randomly selected, find the probability that their mean weight is between 140 lb and 201 lb.

Solve a probability problem using normal distribution and Z-scores to determine the likelihood that women's weights fall between 140 lb and 201 lb. This problem also applies the Central Limit Theorem to calculate the probability for the mean weight of 28 women. Includes step-by-step calculations and insights into statistical concepts.

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