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