https://file.aiquickdraw.com/mathbot/file/1729213321136hm93bopt.jpg

Let's work through the problems shown in your image.

We are dealing with a normally distributed variable \( X \), which represents the height of adult women in the U.S., with:
- Mean \( \mu = 65 \) inches
- Standard deviation \( \sigma = 3.5 \) inches.

### Part (a)
We are asked to find the probability of selecting a female shorter than 58 inches tall:
\[
P(X < 58)
\]

To standardize this to the standard normal distribution, we calculate the z-score using the formula:
\[
z = \frac{X - \mu}{\sigma} = \frac{58 - 65}{3.5} = \frac{-7}{3.5} = -2
\]

Now, using a standard normal distribution table or calculator, the probability corresponding to \( z = -2 \) is approximately 0.02275. 

So the correct probability should be:
\[
P(X < 58) \approx 0.02275
\]

It looks like the value 0.02280 you entered is very close but slightly off.

### Part (b)
This asks for the 90th percentile of the distribution.

We are looking for the value \( x \) such that \( P(X < x) = 0.90 \). Using a normal distribution percentile calculator, the z-score corresponding to 0.90 is approximately 1.28.

Now, use the z-score formula to solve for \( x \):
\[
x = \mu + z\sigma = 65 + 1.28 \times 3.5 \approx 65 + 4.48 = 69.48
\]

So the correct answer is:
\[
x \approx 69.5 \text{ inches}
\]

This answer seems to be correct based on the image.

### Part (c)
This asks for the probability that the sample mean height of 100 randomly chosen women is less than 64.5 inches.

For this, we use the **Central Limit Theorem**. The mean of the sampling distribution of the sample mean is \( \mu_{\bar{X}} = 65 \), and the standard deviation is \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3.5}{\sqrt{100}} = 0.35 \).

Now, calculate the z-score:
\[
z = \frac{64.5 - 65}{0.35} = \frac{-0.5}{0.35} \approx -1.43
\]

Using a standard normal distribution table or calculator, the probability corresponding to \( z = -1.43 \) is approximately 0.0764.

So the correct answer is:
\[
P\left(\bar{X} < 64.5\right) \approx 0.0764
\]

This matches the value you entered, so this answer is correct.

---

Let me know if you'd like more details on any part of the solution. Here are 5 related questions you might find helpful:

1. How do you calculate probabilities for different ranges of heights, such as \( P(60 < X < 70) \)?
2. How does increasing the sample size affect the probability in part (c)?
3. What is the 95th percentile for this distribution?
4. How do you interpret the z-score in a real-world context?
5. What happens to the probability in part (c) if the sample size is halved?

**Tip**: Whenever you're working with sampling distributions, always remember to adjust the standard deviation by dividing it by the square root of the sample size!

a) What is the probability of randomly selecting a female shorter than 58 inches tall? b) Determine the 90th percentile for the distribution. c) If 100 women were randomly chosen, what is the probability that the sample mean of this group would be less than 64.5 inches?

Solve problems involving normal distribution and probability for the heights of American women. Learn how to calculate probabilities, find percentiles, and apply the Central Limit Theorem to determine the probability that a sample mean is below a certain value. Suitable for high school students.

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