Part 2 of 3 HW score: $66.67 \%, 4.67$ of 7 points Points: 0 of 1 Save Suppose a simple random sample of size $n=1000$ is obtained from a population whose size is $N=2,000,000$ and whose population proportion with a specified characteristic is $p=0.76$. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Describe the sampling distribution of $\hat{p}$ A. Approximately normal, $\mu_{\hat{p}}=0.76$ and $\sigma_{\hat{p}} \approx 0.0135$ B. Approximately normal, $\mu_{\hat{p}}=0.76$ and $\sigma_{\hat{p}} \approx 0.0003$ C. Approximately normal, $\mu_{\hat{p}}=076$ and $\sigma_{\hat{p}} \approx 0.0002$ (b) What is the probability of obtaining $x=790$ or more individuals with the characteristic? $P(x \geq 790)=\square$ (Round to four decimal places as needed) n example Get more help. Clear all Check answer

Step 1 ：The sampling distribution of \(\hat{p}\) is approximately normal because the sample size is large enough (n=1000).

Step 2 ：The mean of the sampling distribution, \(\mu_{\hat{p}}\), is equal to the population proportion, p=0.76.

Step 3 ：The standard deviation of the sampling distribution, \(\sigma_{\hat{p}}\), can be calculated using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where p is the population proportion and n is the sample size.

Step 4 ：Calculate \(\sigma_{\hat{p}}\) as follows: \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.76(1-0.76)}{1000}} \approx 0.0135\).

Step 5 ：To find the probability of obtaining x=790 or more individuals with the characteristic, we need to standardize the value and look it up in the standard normal distribution table.

Step 6 ：First, calculate the z-score using the formula \(z = \frac{x - np}{\sqrt{np(1-p)}}\), where x is the number of successes, n is the sample size, and p is the population proportion.

Step 7 ：Calculate z as follows: \(z = \frac{790 - 1000*0.76}{\sqrt{1000*0.76*(1-0.76)}} \approx 2.35\).

Step 8 ：The probability of obtaining a z-score of 2.35 or more is found by looking up the z-score in the standard normal distribution table and subtracting the value from 1 (because we want the probability of obtaining a z-score greater than 2.35).

Step 9 ：Calculate the probability as follows: \(P(x \geq 790) = 1 - P(z \leq 2.35) = 1 - 0.9906 = 0.0094\).

Step 10 ：\(\boxed{P(x \geq 790) = 0.0094}\)