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.

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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.

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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!