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The problem in the image involves a normal distribution of women's heights with a mean of 63.6 inches and a standard deviation of 2.3 inches. Two parts of the question are being asked:

1. **Part (a)** asks for the percentage of women meeting a height requirement of between 58 and 80 inches.
2. **Part (b)** asks for new height requirements that exclude the shortest 1% and the tallest 2% of women.

### Part (a):
To find the percentage of women meeting the height requirement of between 58 inches and 80 inches, we use the properties of the normal distribution.

- **Mean (μ) = 63.6 inches**
- **Standard deviation (σ) = 2.3 inches**

We want to find the percentage of the population that falls between 58 and 80 inches.

First, convert these heights to their corresponding z-scores using the formula:
\[
z = \frac{x - \mu}{\sigma}
\]
- For 58 inches:
\[
z = \frac{58 - 63.6}{2.3} = \frac{-5.6}{2.3} \approx -2.43
\]
- For 80 inches:
\[
z = \frac{80 - 63.6}{2.3} = \frac{16.4}{2.3} \approx 7.13
\]

Now, using z-tables or a standard normal distribution calculator:
- The probability corresponding to \( z = -2.43 \) is approximately 0.0075.
- For \( z = 7.13 \), the probability is nearly 1 (since it's well beyond the usual range).

Therefore, the percentage of women between these heights is approximately:
\[
(1 - 0.0075) \times 100 \approx 99.24\%
\]

So, **99.24% of women meet the height requirement**.

### Part (b):
The new requirement asks to find heights such that all women are eligible except for the shortest 1% and the tallest 2%.

1. For the shortest 1%, find the z-score corresponding to 1% in the lower tail. From z-tables, the z-score for 1% is approximately **-2.33**.
2. For the tallest 2%, the z-score corresponding to 98% (since we exclude the top 2%) is about **2.05**.

Now, convert these z-scores back to heights using the formula:
\[
x = z \cdot \sigma + \mu
\]
- For \( z = -2.33 \):
\[
x = -2.33 \cdot 2.3 + 63.6 \approx 58.3 \text{ inches}
\]
- For \( z = 2.05 \):
\[
x = 2.05 \cdot 2.3 + 63.6 \approx 68.3 \text{ inches}
\]

Thus, the new height requirements are **at least 58.3 inches** and **at most 68.3 inches**.

Would you like more details on any part of this solution? Here are some related questions for deeper exploration:

1. How would the percentage change if the height requirements shifted further?
2. Can you explain how the z-scores relate to probabilities in a normal distribution?
3. What effect does changing the standard deviation have on the percentage of women meeting the requirements?
4. How would the calculation differ if the mean height changed to 65 inches?
5. What other factors might be considered when setting military height requirements?

**Tip**: Z-scores are helpful in comparing data points to the overall distribution, especially when converting between raw scores and probabilities.

A survey found that women’s heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.3 inches. A branch of the military requires women's heights to be between 58 inches and 80 inches. What percentage of women meet the height requirement, and what would the new requirements be if they exclude the shortest 1% and tallest 2%?

Solve a problem involving the percentage of women meeting military height requirements using the normal distribution. Learn how to find the z-scores for specific heights and calculate new height requirements that exclude the shortest 1% and tallest 2%.

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