To find the mean and standard deviation of the sampling distribution of the sample proportion $\hat{p}$, we use the following formulas:

1. Mean of the sampling distribution of $\hat{p}$: $\mu_{\hat{p}} = p$ where $p$ is the probability of success in the population. In this case, $p = 0.12$.

2. Standard deviation of the sampling distribution of $\hat{p}$: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$ where $n$ is the sample size.

Let's calculate the mean and standard deviation for each case:

Part 1: $n = 5000$

• Mean: $\mu_{\hat{p}} = p = 0.12$

• Standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{0.12(1 - 0.12)}{5000}} = \sqrt{\frac{0.12(0.88)}{5000}} = \sqrt{\frac{0.1056}{5000}} \approx \sqrt{0.00002112} \approx 0.0046$

So for $n = 5000$:

• Mean = 0.12
• Standard deviation = 0.0046

Part 2: $n = 1000$

• Mean: $\mu_{\hat{p}} = p = 0.12$

• Standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{0.12(0.88)}{1000}} = \sqrt{\frac{0.1056}{1000}} \approx \sqrt{0.0001056} \approx 0.0103$

So for $n = 1000$:

• Mean = 0.12
• Standard deviation = 0.0103

Part 3: $n = 500$

• Mean: $\mu_{\hat{p}} = p = 0.12$

• Standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{0.12(0.88)}{500}} = \sqrt{\frac{0.1056}{500}} \approx \sqrt{0.0002112} \approx 0.0145$

So for $n = 500$:

• Mean = 0.12
• Standard deviation = 0.0145

Summary of Results:

1. For $n = 5000$:

   • Mean = 0.12
   • Standard deviation = 0.0046

2. For $n = 1000$:

   • Mean = 0.12
   • Standard deviation = 0.0103

3. For $n = 500$:

   • Mean = 0.12
   • Standard deviation = 0.0145

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Would you like any further clarification on the formulas or the calculations? Here's a set of related questions:

1. How would the standard deviation change if the sample size were doubled?
2. What does the standard deviation of the sample proportion tell us about the spread of possible sample proportions?
3. How would the standard deviation of the sample proportion behave if the probability of success were much higher (e.g., 0.9)?
4. What is the relationship between the sample size and the precision of an estimate of the population proportion?
5. How could you use the normal approximation to the binomial distribution with these sample sizes?

Tip: When calculating the standard deviation of the sample proportion, always ensure the sample size is large enough to use the normal approximation to the binomial distribution (i.e., both $np$ and $n(1 - p)$ should be at least 10).