Suppose a simple random sample of size $n=75$ is obtained from a population whose size is $\mathrm{N}=25,000$ and whose population proportion with a specified characteristic is $p=0.6$. Complete parts (a) through (c) below.
(a) Describe the sampling distribution of $\hat{p}$. Choose the phrase that best describes the shape of the sampling distribution below.
A. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$.

B. Approximately normal because $\mathrm{n} \leq 0.05 \mathrm{~N}$ and $n p(1-p)< 10$.
C. Not nombl because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p)< 10$.
D. Not normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p) \geq 10$.

Step 1 ：Given that the sample size \(n = 75\), the population size \(N = 25000\), and the population proportion \(p = 0.6\).

Step 2 ：The shape of the sampling distribution of \(\hat{p}\) can be approximated as normal if two conditions are met: \n1. The sample size is less than or equal to 5% of the population size (\(n \leq 0.05N\)). \n2. The expected number of successes (\(np\)) and failures (\(n(1-p)\)) are both greater than or equal to 10.

Step 3 ：Check the first condition: \(n = 75\) and \(N = 25000\), so \(n \leq 0.05N\) is true.

Step 4 ：Check the second condition: Calculate the expected number of successes and failures. The expected number of successes is \(np = 75 * 0.6 = 45.0\) and the expected number of failures is \(n(1-p) = 75 * (1 - 0.6) = 30.0\). Both are greater than or equal to 10, so the second condition is also true.

Step 5 ：Since both conditions for the sampling distribution of \(\hat{p}\) to be approximately normal are met, the shape of the sampling distribution is approximately normal.

Step 6 ：Final Answer: \(\boxed{\text{A. Approximately normal because } n \leq 0.05 \mathrm{~N} \text{ and } n p(1-p) \geq 10}\)