Question

question 9 (1 point)

solve the problem.

the probability distribution shown below describes a population of measurements.

\\(\

$$\begin{array}{c|ccc} x & 0 & 2 & 4 \\\\ \\hline p(x) & 1/3 & 1/3 & 1/3 \\end{array}$$

\\)

suppose that we took repeated random samples of \\(n = 3\\) observations from the population described above. find the expected value of the sampling distribution of the sample mean.

( ) 1
( ) 4
( ) 0
( ) 3
( ) 2

question 10 (1 point)

solve the problem.

a random sample of \\(n = 100\\) measurements is drawn from a binomial population with probability of success .26. give the mean and the standard deviation of the sampling distribution of the sample proportion, \\(\hat{p}\\).

( ) .26; .011
( ) .26; .439
( ) .26; .044
( ) .26; .025

Explanation:

Calculate the population mean

Using the Discrete Probability Distribution knowledge point

$$ \mu = \sum x \cdot p(x) = 0 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} + 4 \cdot \frac{1}{3} = 2 $$

Find the expected value of the sample mean

Using the Sampling Distribution knowledge point

$$ E(\bar{x}) = \mu_{\bar{x}} = \mu = 2 $$

Calculate the mean of the sample proportion

For Question 10, the mean of the sampling distribution of the sample proportion \(\hat{p}\) is equal to the population proportion \(p\).
Given:

$$ p = 0.26 $$

Therefore:

$$ \mu_{\hat{p}} = p = 0.26 $$

Calculate the standard deviation of the sample proportion

The standard deviation (standard error) of the sampling distribution of the sample proportion \(\hat{p}\) is:

$$ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.26 \cdot (1 - 0.26)}{100}} = \sqrt{\frac{0.26 \cdot 0.74}{100}} = \sqrt{\frac{0.1924}{100}} = \sqrt{0.001924} \approx 0.04386 $$

Rounding to three decimal places gives \(0.044\).

Answer:

Question 9

• 1
• 4
• 0
• 3
• 2 (Correct answer)

Question 10

• .26; .011
• .26; .439
• .26; .044 (Correct answer)
• .26; .025