Suppose a simple random sample of size $n=75$ is obtained from a population whose size is $N=15,000$ and whose population proportion with a specified characteristic is $p=0.4$. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). A. Not normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$. B. Not normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p)<10$. C. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$. D. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p)<10$. Determine the mean of the sampling distribution of $\hat{p}$. $\mu_{p}=0.4$ (Round to one decimal place as needed.) Determine the standard deviation of the sampling distribution of $\hat{p}$. $\sigma_{\hat{p}}=0.056569$ (Round to six decimal places as needed.) (b) What is the probability of obtaining $x=36$ or more individuals with the characteristic? That is, what is $P(\hat{p} \geq 0.48)$ ? $P(\hat{p} \geq 0.48)=\square$ (Round to four decimal places as needed.)

Step 1 ：Given that the sample size \(n=75\), the population size \(N=15000\), and the population proportion \(p=0.4\).

Step 2 ：First, we need to check if the distribution is normal or not. The conditions for a distribution to be approximately normal are that \(n \leq 0.05 N\) and \(n p(1-p) \geq 10\).

Step 3 ：Calculating these values using the given values of \(n\), \(N\), and \(p\), we find that both conditions are satisfied. Therefore, the distribution is approximately normal.

Step 4 ：Next, we calculate the mean and standard deviation of the sampling distribution of \(\hat{p}\). The mean of the sampling distribution of \(\hat{p}\) is equal to the population proportion \(p\), which is \(0.4\).

Step 5 ：The standard deviation of the sampling distribution of \(\hat{p}\) can be calculated using the formula \(\sqrt{\frac{p(1-p)}{n}}\), which gives us approximately \(0.056569\).

Step 6 ：Finally, we need to find the probability of obtaining \(x=36\) or more individuals with the characteristic. This is equivalent to finding the probability that \(\hat{p} \geq 0.48\).

Step 7 ：We can find this probability using the standard normal distribution, by first converting \(\hat{p}\) to a z-score using the formula \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}\), and then finding the area to the right of this z-score in the standard normal distribution.

Step 8 ：Doing these calculations, we find that the probability is approximately \(0.0786\).

Step 9 ：Final Answer: The distribution is approximately normal because \(n \leq 0.05 N\) and \(n p(1-p) \geq 10\). The mean of the sampling distribution of \(\hat{p}\) is \(\boxed{0.4}\). The standard deviation of the sampling distribution of \(\hat{p}\) is \(\boxed{0.056569}\). The probability of obtaining \(x=36\) or more individuals with the characteristic is \(\boxed{0.0786}\).