We are dealing with a normally distributed population of women’s weights with a mean $\mu = 143 \, \text{lb}$ and a standard deviation $\sigma = 29 \, \text{lb}$. To solve the given questions, we will use Z-scores and the properties of the standard normal distribution.

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Part (a):

If 1 woman is randomly selected, find the probability that her weight is between 108 lb and 179 lb.

We convert both bounds (108 lb and 179 lb) into Z-scores using:

$Z = \frac{X - \mu}{\sigma}$

For 108 lb:

$Z_1 = \frac{108 - 143}{29} = \frac{-35}{29} \approx -1.21$

For 179 lb:

$Z_2 = \frac{179 - 143}{29} = \frac{36}{29} \approx 1.24$

Using standard normal distribution tables or a calculator:

• $P(Z < -1.21) \approx 0.1131$
• $P(Z < 1.24) \approx 0.8925$

Now, the probability that the woman's weight is between 108 and 179 lb:

$P(108 < X < 179) = P(Z < 1.24) - P(Z < -1.21) = 0.8925 - 0.1131 = 0.7794$

So, the probability is:

$\boxed{0.7794}$

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Part (b):

If 6 women are randomly selected, find the probability that they have a mean weight between 108 lb and 179 lb.

When dealing with the sample mean, the standard deviation becomes:

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$

Here, $n = 6$, so:

$\sigma_{\bar{X}} = \frac{29}{\sqrt{6}} \approx 11.83$

Now, we recalculate the Z-scores with the new standard deviation:

For 108 lb:

$Z_1 = \frac{108 - 143}{11.83} \approx -2.96$

For 179 lb:

$Z_2 = \frac{179 - 143}{11.83} \approx 3.04$

Using standard normal distribution tables:

• $P(Z < -2.96) \approx 0.0015$
• $P(Z < 3.04) \approx 0.9988$

Now, the probability that the mean weight is between 108 and 179 lb:

$P(108 < \bar{X} < 179) = P(Z < 3.04) - P(Z < -2.96) = 0.9988 - 0.0015 = 0.9973$

So, the probability is:

$\boxed{0.9973}$

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Part (c):

If 95 women are randomly selected, find the probability that they have a mean weight between 108 lb and 179 lb.

For this larger sample size $n = 95$, the standard deviation of the sample mean becomes:

$\sigma_{\bar{X}} = \frac{29}{\sqrt{95}} \approx 2.97$

Now, we recalculate the Z-scores:

For 108 lb:

$Z_1 = \frac{108 - 143}{2.97} \approx -11.78$

For 179 lb:

$Z_2 = \frac{179 - 143}{2.97} \approx 12.12$

Using the standard normal table:

• $P(Z < -11.78)$ is extremely close to 0.
• $P(Z < 12.12)$ is extremely close to 1.

Thus, the probability that the mean weight is between 108 and 179 lb:

$P(108 < \bar{X} < 179) = P(Z < 12.12) - P(Z < -11.78) \approx 1 - 0 = 1$

So, the probability is:

$\boxed{1.0000}$

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Summary of Results:

• (a) Probability for 1 woman: $0.7794$
• (b) Probability for 6 women: $0.9973$
• (c) Probability for 95 women: $1.0000$

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Do you need further clarification or more details?

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Here are 5 related questions:

1. What is the probability that a randomly selected woman weighs more than 200 lb?
2. How does increasing the sample size affect the standard deviation of the sample mean?
3. What is the Z-score corresponding to a weight of 143 lb?
4. How would the probabilities change if the standard deviation were smaller?
5. What is the meaning of the central limit theorem in the context of these problems?

Tip: As the sample size increases, the variability of the sample mean decreases, leading to probabilities closer to 1 for values near the population mean.