Given that the population mean $\mu = 100$ and the standard deviation $\sigma = 21$, we will solve the problem as follows:

Part (a)

1. Mean of the sampling distribution of the sample mean: The mean of the sampling distribution of the sample mean ($\overline{X}$) is the same as the population mean, so: $\text{Mean of } \overline{X} = \mu = 100$

2. Standard deviation of the sampling distribution of the sample mean: The standard deviation of the sampling distribution, also known as the standard error (SE), is calculated as: $\text{Standard error} = \frac{\sigma}{\sqrt{n}} = \frac{21}{\sqrt{49}} = \frac{21}{7} = 3$

Thus:

• The mean of the sampling distribution of the sample mean is 100.
• The standard deviation of the sampling distribution of the sample mean is 3.

Part (b)

For the sampling distribution of the sample mean to be approximately normally distributed with the values calculated above, we rely on the Central Limit Theorem (CLT). The CLT states that the sampling distribution of the sample mean will be approximately normal if the sample size is sufficiently large, regardless of the population's distribution.

The answer for part (b) is C: No assumption was made.

Let me know if you’d like more explanation or have further questions.

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Follow-up questions for practice:

1. How would the standard error change if the sample size increased to 100?
2. What is the probability of selecting a sample with a mean less than 97?
3. How would a smaller population standard deviation affect the sampling distribution?
4. What sample size would yield a standard error of 2?
5. How does the Central Limit Theorem apply when the sample size is smaller?

Tip: Remember, a larger sample size decreases the standard error, making the sample mean more precise.