This question is asking you to analyze the sampling distribution for a proportion $\hat{p}$, given certain conditions.

Part 1: Sampling Distribution of $\hat{p}$

The problem states that 37% of adults do not own a credit card. You have a simple random sample of 600 adults ($n = 600$), and you are asked to determine whether the sampling distribution is approximately normal.

The general rule of thumb for the normal approximation of a proportion $\hat{p}$ is that:

• The sample size $n$ must be large enough so that both $np \geq 10$ and $n(1 - p) \geq 10$.
• Also, the sample size $n$ should be small enough that it doesn’t exceed 5% of the population size $N$, i.e., $n \leq 0.05N$.

Let’s check the conditions:

1. $np = 600 \times 0.37 = 222 \geq 10$
2. $n(1 - p) = 600 \times (1 - 0.37) = 600 \times 0.63 = 378 \geq 10$

Both conditions for normal approximation are satisfied, so the correct answer would be:

• Answer: Option B - Approximately normal because $n \leq 0.05N$ and $np(1 - p) \geq 10$.

Part 2: Mean of the Sampling Distribution of $\hat{p}$

The mean of the sampling distribution of the sample proportion $\hat{p}$ is given by:

$\mu_{\hat{p}} = p$

Since $p = 0.37$, the mean of the sampling distribution is:

$\mu_{\hat{p}} = 0.37$

Summary:

1. Sampling Distribution: Option B is correct.
2. Mean of Sampling Distribution: $\mu_{\hat{p}} = 0.37$ (rounded to two decimal places).

Would you like further clarification or detailed explanations on any part of this?

Relative Questions:

1. What is the variance of the sampling distribution of $\hat{p}$?
2. How does increasing the sample size affect the normal approximation?
3. How do you calculate the standard deviation of the sampling distribution?
4. What is the Central Limit Theorem, and how does it apply here?
5. Can the approximation still be used if $np < 10$?

Tip: Always check the conditions for normal approximation when dealing with sample proportions!