Answer:

(a) Description of the Sampling Distribution of $\bar{x}$:

The sampling distribution of $\bar{x}$ is approximately normal due to the Central Limit Theorem (since $n = 64 \geq 30$). The mean of the sampling distribution ($\mu_{\bar{x}}$) is the same as the population mean, $\mu_{\bar{x}} = \mu = 76$. The standard deviation of the sampling distribution ($\sigma_{\bar{x}}$) is given by $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{24}{\sqrt{64}} = 3$.

(b) Comparison to the Population Distribution:

The sampling distribution of $\bar{x}$ is much less skewed than the population distribution, tending towards normality regardless of the population's shape. While the population distribution is skewed right, the sampling distribution of $\bar{x}$ is symmetric and normal due to the Central Limit Theorem. Additionally, the variability (standard deviation) of the sampling distribution is smaller than the population's variability.

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Would you like further clarification or details?

Here are 5 related questions to expand your understanding:

1. Why does the Central Limit Theorem apply in this scenario?
2. How would the standard deviation of the sampling distribution change if $n$ increased?
3. What happens to the shape of the sampling distribution if $n$ were smaller (e.g., $n < 30$)?
4. How does skewness in the population affect the sampling distribution for small sample sizes?
5. Can the mean of the sampling distribution ever differ from the population mean? Why or why not?

Tip: Always remember that the Central Limit Theorem ensures normality for large sample sizes, even if the population is not normal.