Let's analyze the problem step-by-step.

Part (a): Describe the Sampling Distribution

The question states:

• 37% of adults do not own a credit card, so $p = 0.37$.
• A sample of 500 adults is taken, so $n = 500$.

We need to determine whether the sampling distribution of $\hat{p}$ (the sample proportion) is approximately normal.

To verify if the sampling distribution of $\hat{p}$ is approximately normal, we check two conditions:

1. Large sample condition: $n \geq 0.05N$ (where $N$ is the population size, typically large).
2. Success-failure condition: Both $np \geq 10$ and $n(1 - p) \geq 10$.

Calculating:

• $np = 500 \times 0.37 = 185$ (which is greater than 10).
• $n(1 - p) = 500 \times (1 - 0.37) = 500 \times 0.63 = 315$ (also greater than 10).

Both conditions are satisfied, so the sampling distribution of $\hat{p}$ is approximately normal.

Thus, Option A is correct: "Approximately normal because $n \geq 0.05N$ and $np(1 - p) \geq 10$."

Part (b): Determine the Mean of the Sampling Distribution of $\hat{p}$

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

Since $p = 0.37$, we have: $\mu_{\hat{p}} = 0.37$

So, the mean of the sampling distribution of $\hat{p}$ is 0.37.

Would you like further explanation on these calculations or details on how sampling distributions work?

Related Questions

1. How is the standard deviation of the sampling distribution of $\hat{p}$ calculated?
2. What is the significance of the success-failure condition in determining normality?
3. How does sample size affect the shape of the sampling distribution?
4. Can the sampling distribution of $\hat{p}$ ever be exactly normal?
5. How would the results change if $p$ were closer to 0 or 1?

Tip

When verifying the normality of a sampling distribution of a proportion, always check both the sample size and the success-failure conditions to ensure accurate results.