Let's address these questions one at a time.

Question 4

1. Mean of the Distribution of Sample Means $(\mu_{\bar{X}})$: The mean of the distribution of sample means (also called the sampling distribution mean) is the same as the mean of the population. Given: $\mu = 24.2$ Therefore: $\mu_{\bar{X}} = 24.2$

2. Standard Deviation of the Distribution of Sample Means $(\sigma_{\bar{X}})$: The standard deviation of the sampling distribution (also called the standard error) is calculated as: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ where $\sigma = 96.7$ and $n = 201$. Plugging in the values: $\sigma_{\bar{X}} = \frac{96.7}{\sqrt{201}} \approx \frac{96.7}{14.177} \approx 6.82$ So, the standard deviation of the distribution of sample means is approximately $6.82$ (to two decimal places).

Question 5

The Central Limit Theorem states that if the population distribution is normal, then the sampling distribution of the sample mean will also be normal regardless of the sample size. Here, since the population distribution is given as normal, the sampling distribution will be normal even though $n = 10$.

Thus, the correct answer is:

• Yes, the sampling distribution is normally distributed because the population is normally distributed.

Would you like further details or have any other questions on these concepts?

Additional Related Questions:

1. How would the standard error change if the sample size increased?
2. If the population distribution was not normal, what sample size would make the sampling distribution approximately normal?
3. Why is the Central Limit Theorem important for inferential statistics?
4. What happens to the sampling distribution's mean if the population mean changes?
5. How would changing the population standard deviation affect the sampling distribution?

Tip:

When calculating the standard error, remember that increasing the sample size decreases the standard error, leading to more precise estimates of the population mean.