when females were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for males only. the ACES-II ejection seats were designed for males weighing between 139.8 lb and 210.7 lb. weights of females are now normally distributed with a mean of 174 lb and a standard deviation of 49 lb. answer the question : if 1 female is randomly selected, find the probability that their weight is between 139.8 lb and 210.7 lb. round to four decimal places as needed

To find the probability that the weight of a randomly selected female is between 139.8 lb and 210.7 lb, we need to calculate the area under the normal distribution curve between these two values. The weights are normally distributed with a mean \( \mu = 174 \) lb and a standard deviation \( \sigma = 49 \) lb.

We will use the standard normal distribution (Z-distribution) to solve this problem. First, we convert the given weights into Z-scores using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

### Step 1: Calculate the Z-scores

1. **For 139.8 lb:**
\[
Z_1 = \frac{139.8 - 174}{49} = \frac{-34.2}{49} \approx -0.698
\]

2. **For 210.7 lb:**
\[
Z_2 = \frac{210.7 - 174}{49} = \frac{36.7}{49} \approx 0.749
\]

### Step 2: Find the corresponding probabilities

Now, we look up the probabilities corresponding to the Z-scores from the standard normal distribution table:

- For \( Z_1 = -0.698 \), the cumulative probability is approximately 0.2432.
- For \( Z_2 = 0.749 \), the cumulative probability is approximately 0.7734.

### Step 3: Calculate the probability between the two Z-scores

The probability that the weight is between 139.8 lb and 210.7 lb is the difference between these two cumulative probabilities:

\[
P(139.8 \leq X \leq 210.7) = P(Z_2) - P(Z_1) = 0.7734 - 0.2432 = 0.5302
\]

### Final Answer:
The probability that the weight of a randomly selected female is between 139.8 lb and 210.7 lb is approximately **0.5302** (rounded to four decimal places).

---

Would you like more details, or do you have any questions about this? 

Here are 5 related questions you might find interesting:
1. What is the probability that a female weighs less than 139.8 lb?
2. How does the probability change if the mean female weight increases to 180 lb?
3. What if the standard deviation was larger, say 60 lb? How would the result change?
4. How would you calculate the probability for a male with the same weight range if male weights follow a different distribution?
5. How would you interpret a Z-score of 1.5 in this context?

**Tip:** Understanding Z-scores helps you compare values from different normal distributions and assess how extreme or typical a value is in relation to the mean.

When females were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for males only. The ACES-II ejection seats were designed for males weighing between 139.8 lb and 210.7 lb. Weights of females are now normally distributed with a mean of 174 lb and a standard deviation of 49 lb. Answer the question: if 1 female is randomly selected, find the probability that their weight is between 139.8 lb and 210.7 lb. Round to four decimal places as needed.

Learn how to calculate the probability that a randomly selected female weighs between 139.8 lb and 210.7 lb, given a normal distribution with a mean of 174 lb and a standard deviation of 49 lb. This problem involves the use of Z-scores and the standard normal distribution. Suitable for high school and college introductory levels.

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