Questions: 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=0.78. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p̂. A. Approximately normal; μp=0.78 and σp̂ ≈ 0.0002 B. Approximately normal, μp̂=0.78 and σp̂ ≈ 0.0003 C. Approximately normal, μp̂=0.78 and σp̂ ≈ 0.0131 (b) What is the probability of obtaining x=820 or more individuals with the characteristic? P(x ≥ 820)= (Round to four decimal places as needed.)

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=0.78. Complete parts (a) through (c) below.
(a) Describe the sampling distribution of p̂.
A. Approximately normal; μp=0.78 and σp̂ ≈ 0.0002
B. Approximately normal, μp̂=0.78 and σp̂ ≈ 0.0003
C. Approximately normal, μp̂=0.78 and σp̂ ≈ 0.0131
(b) What is the probability of obtaining x=820 or more individuals with the characteristic?
P(x ≥ 820)= (Round to four decimal places as needed.)

Transcript text: Suppose a simple random sample of size $n=1000$ is obtained from a population whose size is $\mathrm{N}=1,500,000$ and whose population proportion with a specified characteristic is $p=0.78$. Complete parts (a) through (c) below. (a) Describe the sampling distribution of $\hat{p}$. A. Approximately normal; $\mu_{p}=0.78$ and $\sigma_{\hat{p}} \approx 0.0002$ B. Approximately normal, $\mu_{\hat{p}}=0.78$ and $\sigma_{\hat{p}} \approx 0.0003$ C. Approximately normal, $\mu_{\hat{p}}=0.78$ and $\sigma_{\hat{p}} \approx 0.0131$ (b) What is the probability of obtaining $x=820$ or more individuals with the characteristic? $P(x \geq 820)=$ $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Describe the Sampling Distribution of $ \hat{p} $

The sampling distribution of the sample proportion $ \hat{p} $ can be described by its mean and standard deviation.

The mean of the sampling distribution is given by: \[ \mu_{\hat{p}} = p = 0.78 \]
The standard deviation of the sampling distribution is calculated using the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p \cdot q}{n}} \cdot \sqrt{\frac{N - n}{N - 1}} \] Substituting the values: \[ \sigma_{\hat{p}} \approx 0.0131 \]

Thus, the sampling distribution of $ \hat{p} $ is approximately normal with: \[ \mu_{\hat{p}} = 0.78 \quad \text{and} \quad \sigma_{\hat{p}} \approx 0.0131 \]

Step 2: Calculate the Probability of Obtaining $ x \geq 820 $

To find the probability of obtaining $ x = 820 $ or more individuals with the characteristic, we first calculate the probability of obtaining exactly $ x = 820 $ using the binomial probability formula: \[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \] The calculated probability for $ P(X = 820) $ is approximately $ 0.0002 $.

Next, we need to find $ P(X \geq 820) $: \[ P(X \geq 820) = 1 - P(X < 820) = 1 - P(X \leq 819) \] The probability of obtaining $ x < 820 $ (or $ x \leq 819 $) is approximately $ 0.0003 $.

Thus, the probability of obtaining $ x \geq 820 $ is: \[ P(X \geq 820) \approx 1 - 0.0003 = 0.9997 \]

Final Answer

The answers to the questions are:

The sampling distribution of $ \hat{p} $ is approximately normal with $ \mu_{\hat{p}} = 0.78 $ and $ \sigma_{\hat{p}} \approx 0.0131 $.
The probability of obtaining $ x \geq 820 $ is approximately $ 0.9997 $.

Thus, the final answers are: \[ \boxed{\text{(a) } \mu_{\hat{p}} = 0.78, \sigma_{\hat{p}} \approx 0.0131} \] \[ \boxed{P(X \geq 820) \approx 0.9997} \]

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