Question

Statistics

Posted about 2 months ago

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A politician takes an SRS of 75 citizens in a county to see what proportion of citizens sampled are satisfied with their standard of living. Suppose that $80 \%$ of the 118,000 citizens in the county are satisfied with their standard of living.
What are the mean and standard deviation of the sampling distribution of the proportion of citizens who are satisfied with their standard of living?
Choose 1 answer:
(A) $\mu_{\hat{p}}=(75)(0.8)$
$\sigma_{\hat{p}}=\sqrt{75(0.8)(1-0.8)}$
(B) $\mu_{\hat{p}}=0.8$
$\sigma_{\hat{p}}=\sqrt{75(0.8)(1-0.8)}$
(c) $\mu_{\hat{p}}=(75)(0.8)$
$\sigma_{\hat{p}}=\sqrt{\frac{0.8(1-0.8)}{75}}$
(D) $\mu_{\hat{p}}=0.8$
$\sigma_{\hat{p}}=\sqrt{\frac{0.8(1-0.8)}{75}}$
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Answer from Sia

Posted about 2 months ago

Solution by Steps

step 1

To find the mean of the sampling distribution of the proportion of citizens who are satisfied with their standard of living, we use the formula for the mean of the sampling distribution of a proportion: $\mu_{\hat{p}} = p$, where $p$ is the population proportion

step 2

Given that $p = 0.8$, we have $\mu_{\hat{p}} = 0.8$

step 3

To find the standard deviation of the sampling distribution of the proportion, we use the formula: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $n$ is the sample size

step 4

Given that $p = 0.8$ and $n = 75$, we have $\sigma_{\hat{p}} = \sqrt{\frac{0.8(1-0.8)}{75}}$

Answer

D

Key Concept

Mean and standard deviation of the sampling distribution of a proportion

Explanation

The mean of the sampling distribution of a proportion is equal to the population proportion, and the standard deviation is calculated using the formula $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$.

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