The question in the image is asking to find $\mu_{\hat{p}}$ (the mean of the sampling distribution of the sample proportion) and $\sigma_{\hat{p}}$ (the standard deviation of the sampling distribution of the sample proportion), given $n = 152$ and $p = 0.5$.

Step 1: Find $\mu_{\hat{p}}$

The mean of the sampling distribution of the sample proportion is given by:

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

Since $p = 0.5$, we have:

$\mu_{\hat{p}} = 0.5$

Step 2: Find $\sigma_{\hat{p}}$

The standard deviation of the sampling distribution of the sample proportion is given by:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}$

Substitute $p = 0.5$ and $n = 152$:

$\sigma_{\hat{p}} = \sqrt{\frac{0.5(1 - 0.5)}{152}} = \sqrt{\frac{0.5 \times 0.5}{152}} = \sqrt{\frac{0.25}{152}} = \sqrt{0.001645}$

Now, let's calculate the value of $\sigma_{\hat{p}}$.

Calculation:The standard deviation of the sampling distribution, $\sigma_{\hat{p}}$, is approximately $0.0406$.

Final Answers:

1. $\mu_{\hat{p}} = 0.5$
2. $\sigma_{\hat{p}} \approx 0.0406$

Would you like more details or have any other questions?

Here are 5 related questions to expand your understanding:

1. How does the sample size $n$ affect the standard deviation $\sigma_{\hat{p}}$?
2. What happens to $\sigma_{\hat{p}}$ if the population proportion $p$ is closer to 0 or 1?
3. Can you explain the concept of sampling distributions?
4. How would you use the Central Limit Theorem to justify these calculations for large samples?
5. What is the relationship between the sample proportion $\hat{p}$ and the population proportion $p$?

Tip: In sampling distributions, larger sample sizes lead to a smaller standard deviation, making the sample proportion $\hat{p}$ more precise.