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The problem provided involves normal distribution with the following given parameters:

- Mean weight \( \mu = 143 \, \text{lb} \)
- Standard deviation \( \sigma = 29 \, \text{lb} \)

Let's solve each part of the problem:

### (a) Probability that a randomly selected woman has a weight below 108 lb:

To find this probability, we can standardize the value using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \( X = 108 \) (the value we're interested in),
- \( \mu = 143 \),
- \( \sigma = 29 \).

Substitute the values:
\[
Z = \frac{108 - 143}{29} = \frac{-35}{29} \approx -1.21
\]

Now, using the standard normal distribution table, the probability corresponding to \( Z = -1.21 \) is approximately **0.1131**.

So, the probability that a randomly selected woman has a weight below 108 lb is **0.1131** or **11.31%**.

### (b) Probability that the mean weight of 4 randomly selected women is below 108 lb:

When calculating the mean for a sample of \( n \) individuals, we use the standard error of the mean \( \sigma_{\bar{X}} \), which is given by:
\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}
\]
For \( n = 4 \), we have:
\[
\sigma_{\bar{X}} = \frac{29}{\sqrt{4}} = \frac{29}{2} = 14.5
\]

Now, we can calculate the Z-score for the sample mean:
\[
Z = \frac{108 - 143}{14.5} = \frac{-35}{14.5} \approx -2.41
\]

Using the standard normal distribution table, the probability corresponding to \( Z = -2.41 \) is approximately **0.0080**.

So, the probability that the mean weight of 4 women is below 108 lb is **0.0080** or **0.80%**.

### (c) Probability that the mean weight of 50 randomly selected women is below 108 lb:

For \( n = 50 \), we calculate the standard error of the mean:
\[
\sigma_{\bar{X}} = \frac{29}{\sqrt{50}} \approx 4.10
\]

Now, calculate the Z-score for the sample mean:
\[
Z = \frac{108 - 143}{4.10} = \frac{-35}{4.10} \approx -8.54
\]

The probability for \( Z = -8.54 \) is extremely small, practically **0**.

Thus, the probability that the mean weight of 50 women is below 108 lb is effectively **0**.

---

Would you like more details or have any questions?

Here are some related questions for further understanding:
1. What is the significance of the Z-score in normal distribution problems?
2. How does the sample size affect the standard error of the mean?
3. Can you explain why the probability decreases as the sample size increases?
4. How would the probabilities change if the standard deviation were larger?
5. What is the importance of using a normal distribution table in these types of problems?

**Tip:** When dealing with sample means, remember that the distribution of the sample means approaches normality as the sample size increases, according to the Central Limit Theorem.

Assume that women's weights are normally distributed with a mean given by μ = 143 lb and a standard deviation given by σ = 29 lb. (a) If 1 woman is randomly selected, find the probability that her weight is below 108 lb. (b) If 4 women are randomly selected, find the probability that they have a mean weight below 108 lb. (c) If 50 women are randomly selected, find the probability that they have a mean weight below 108 lb.

Learn how to calculate the probability of women’s weights using normal distribution. This detailed solution involves calculating probabilities for individual women and groups of women using the Z-score formula and sampling distribution concepts. Ideal for students in Grades 11-12 or college.

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