K
Suppose a simple random sample of size $n=200$ is obtained from a population whose size is $N=10,000$ and whose population proportion with a specified characteristic is $p=02$

Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
D. Approximately normal because $n \leq 0.05 \mathrm{~N}$ and $n p(1-p)< 10$

Determine the mean of the sampling distribution of $\hat{p}$
$\mu_{\hat{p}}=0.2$ (Round to one decimal place as needed)
Determine the standard deviation of the sampling distribution of $\hat{p}$
$\sigma_{\hat{p}}=0.028284^{\prime}$ (Round to six decimal places as needed.)
(b) What is the probability of obtaining $x=42$ or more individuals with the characteristic? That is, what is $P(\hat{p} \geq 0.21)$ ?
$P(\hat{p} \geq 0.21)=0.3618$ (Round to four decimal places as needed)
(c) What is the probability of obtaining $x=34$ or fewer individuals with the characteristic? That is, what is $P(\hat{p} \leq 0.17)$ ?
$P(\hat{p} \leq 0.17)=0.1444$ (Round to four decimal places as needed)
Calculator
Next
Date Worked
$11 / 19 / 23$
12:11pm
$11 / 17 / 23$
8:17am
$11 / 16 / 23$
5:07pm
$11 / 15 / 23$

Step 1 ：The mean of the sampling distribution of the proportion (p-hat) is equal to the population proportion (p). So, \(\mu_{\hat{p}} = p = 0.2\).

Step 2 ：The standard deviation of the sampling distribution of the proportion is given by the formula \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Substituting the given values, we get \(\sigma_{\hat{p}} = \sqrt{\frac{0.2(1-0.2)}{200}} = 0.028284\).

Step 3 ：To find the probability of obtaining 42 or more individuals with the characteristic, we first need to convert 42 to a proportion by dividing by the sample size. So, 42/200 = 0.21. We then need to find the z-score for this proportion, which is given by the formula \(z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}\). Substituting the given values, we get \(z = \frac{0.21 - 0.2}{0.028284} = 0.3536\). Looking up this z-score in the standard normal distribution table, we find that the area to the left of this z-score is 0.6382. Since we want the probability of obtaining 0.21 or more, we need to find the area to the right of the z-score, which is 1 - 0.6382 = 0.3618.

Step 4 ：Similarly, to find the probability of obtaining 34 or fewer individuals with the characteristic, we first convert 34 to a proportion, which is 34/200 = 0.17. The z-score for this proportion is \(z = \frac{0.17 - 0.2}{0.028284} = -1.0615\). Looking up this z-score in the standard normal distribution table, we find that the area to the left of this z-score is 0.1444, which is the probability we are looking for.

Step 5 ：So, the mean of the sampling distribution of the proportion is \(\boxed{0.2}\), the standard deviation is \(\boxed{0.028284}\), the probability of obtaining 0.21 or more is \(\boxed{0.3618}\), and the probability of obtaining 0.17 or less is \(\boxed{0.1444}\).