Suppose a simple random sample of size n = 200 is obtained from a population whose size is N = 15,000 and whose population proportion with a specified characteristic is p=0.8. Complete parts (a) through (c) below. (...) OC. Approximately normal because n ≤0.05N and np(1-p) < 10. OD. Not normal because n ≤0.05N and np(1-p) ≥ 10. Determine the mean of the sampling distribution of p. HA=0.8 (Round to one decimal place as needed.) al Determine the standard deviation of the sampling distribution of p OA 0.028284 (Round to six decimal places as needed.) P (b) What is the probability of obtaining x= 168 or more individuals with the characteristic? That is, what is P(p ≥ 0.84)? of io P02084)- (Round to four decimal places as needed.) (c) What is the probability of obtaining x-146 or fewer individuals with the characteristic? That is what is P(p ≤0.73)? mc P(p ≤0.73)= (Round to four decimal places as needed)

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### Example Problem: Sampling Distribution of Sample Proportion

Suppose a simple random sample of size \(n = 200\) is obtained from a population whose size is \(N = 15,000\) and whose population proportion with a specified characteristic is \(p = 0.8\). Complete parts (a) through (c) below.
 
(a) **Determine if the sampling distribution of \(\hat{p}\) is approximately normal:**

- \[C. \text{Approximately normal because } n \leq 0.05N \text{ and } np(1 - p) < 10.\]
- \[D. \text{Not normal because } n \leq 0.05N \text{ and } np(1 - p) \geq 10.\]

(b) **Determine the mean of the sampling distribution of \(\hat{p}\):**

- \[\mu_{\hat{p}} = 0.8 \quad \text{(Round to one decimal place as needed.)}\]

(c) **Determine the standard deviation of the sampling distribution of \(\hat{p}\):**

- \[\sigma_{\hat{p}} = 0.028284 \quad \text{(Round to six decimal places as needed.)}\]

### Example Calculations:

1. **Determine the Mean:**
   \[\mu_{\hat{p}} = p\]

2. **Determine the Standard Deviation:**
   \[\sigma_{\hat{p}} = \sqrt{\frac{p (1 - p)}{n}} \times \sqrt{\frac{N - n}{N - 1}}\]

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### Example Probability Calculation:

(d) **What is the probability of obtaining \(x = 168\) or more individuals with the characteristic? That is, what is \(P(\hat{p} \geq 0.84)\)?**

- \[P(\hat{p} \geq 0.84) = \quad \text{(Round to four decimal places as needed.)}\]

(e) **What is the probability of obtaining \(x = 146\) or fewer individuals with the characteristic? That is, what is \(P(\hat{p} \leq 0.73)\)?**

- \[P(\hat{p} \leq 0.73) = \quad \text
Transcribed Image Text:--- ### Example Problem: Sampling Distribution of Sample Proportion Suppose a simple random sample of size \(n = 200\) is obtained from a population whose size is \(N = 15,000\) and whose population proportion with a specified characteristic is \(p = 0.8\). Complete parts (a) through (c) below. (a) **Determine if the sampling distribution of \(\hat{p}\) is approximately normal:** - \[C. \text{Approximately normal because } n \leq 0.05N \text{ and } np(1 - p) < 10.\] - \[D. \text{Not normal because } n \leq 0.05N \text{ and } np(1 - p) \geq 10.\] (b) **Determine the mean of the sampling distribution of \(\hat{p}\):** - \[\mu_{\hat{p}} = 0.8 \quad \text{(Round to one decimal place as needed.)}\] (c) **Determine the standard deviation of the sampling distribution of \(\hat{p}\):** - \[\sigma_{\hat{p}} = 0.028284 \quad \text{(Round to six decimal places as needed.)}\] ### Example Calculations: 1. **Determine the Mean:** \[\mu_{\hat{p}} = p\] 2. **Determine the Standard Deviation:** \[\sigma_{\hat{p}} = \sqrt{\frac{p (1 - p)}{n}} \times \sqrt{\frac{N - n}{N - 1}}\] --- ### Example Probability Calculation: (d) **What is the probability of obtaining \(x = 168\) or more individuals with the characteristic? That is, what is \(P(\hat{p} \geq 0.84)\)?** - \[P(\hat{p} \geq 0.84) = \quad \text{(Round to four decimal places as needed.)}\] (e) **What is the probability of obtaining \(x = 146\) or fewer individuals with the characteristic? That is, what is \(P(\hat{p} \leq 0.73)\)?** - \[P(\hat{p} \leq 0.73) = \quad \text
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