Test- Chapter $7 \& 8$ Question 3 of 24 This test: 24 point(s) possible This question: 1 point(s) possible Submit test K Suppose a simple random sample of size $n=1000$ is obtained from a population whose size is $N=1,500,000$ and whose population proportion with a specified characteristic is $p=0.49$. Complete parts $(a)$ through $(c)$ below. (a) Describe the sampling distribution of $\hat{p}$. A. Approximately normal, $\mu_{\hat{p}}=0.49$ and $\sigma_{\hat{p}} \approx 0.0002$ B. Approximately normal, $\mu_{\hat{p}}=0.49$ and $\sigma_{\hat{p}} \approx 0.0158$ C. Approximately normal, $\mu_{\hat{p}}=0.49$ and $\sigma_{\hat{p}} \approx 0.0004$ (b) What is the probability of obtaining $x=530$ or more individuals with the characteristic? $P(x \geq 530)=\square$ (Round to four decimal places as needed.) (c) What is the probability of obtaining $x=470$ or fewer individuals with the characteristic? $P(x \leq 470)=\square$ (Round to four decimal places as needed.) (1) Time Remaining: 00:27:18 Next

Step 1 ：Given that the sample size (n) is 1000, the population proportion (p) is 0.49, the mean of the sampling distribution of the proportion (mu) is 0.49, and the standard deviation of the sampling distribution of the proportion (sigma) is approximately 0.0158.

Step 2 ：For part (a), the sampling distribution of the proportion is approximately normal because the sample size is large enough (n > 30). The mean of the sampling distribution of the proportion is equal to the population proportion (p), and the standard deviation of the sampling distribution of the proportion is \(\sqrt{\frac{p(1-p)}{n}}\). Therefore, the sampling distribution of \(\hat{p}\) is approximately normal with \(\mu_{\hat{p}}=0.49\) and \(\sigma_{\hat{p}} \approx 0.0158\).

Step 3 ：For part (b), we can use the z-score formula to find the probability. The z-score is calculated as \(\frac{x - \mu}{\sigma}\), where x is the number of successes, \(\mu\) is the expected number of successes (n*p), and \(\sigma\) is the standard deviation of the distribution \(\sqrt{n*p*(1-p)}\). The z-score for obtaining 530 or more individuals with the characteristic is approximately 2.53. Using the standard normal distribution, the probability is approximately 0.0057. Therefore, the probability of obtaining \(x=530\) or more individuals with the characteristic is \(P(x \geq 530)=\boxed{0.0057}\).

Step 4 ：For part (c), using the same z-score formula, the z-score for obtaining 470 or fewer individuals with the characteristic is approximately -1.27. Using the standard normal distribution, the probability is approximately 0.1029. Therefore, the probability of obtaining \(x=470\) or fewer individuals with the characteristic is \(P(x \leq 470)=\boxed{0.1029}\).