According to data from the United States Elections Project, only 38 percent of eligible voters voted in the 2014 elections. For random samples of size 40, which of the following best describes the sampling distribution of p, the sample proportion of people who voted in the 2014 elections?
A. The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.076.
B. The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.006.
C. The sampling distribution is skewed to the right, with mean 0.64 and standard deviation 0.006.
D. The sampling distribution is approximately normal, with mean 0.64 and standard deviation 0.076.
E. The sampling distribution is skewed to the left, with mean 0.36 and standard deviation 0.076.

Using Central Limit Theorem, it is found that the option which best describes the sampling distribution of p, the sample proportion of people who voted in the 2014 elections is:

A. The sampling distribution is approximately normal, with mean 0.36 and standard deviation 0.076.

The Central Limit Theorem states that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this problem:

The proportion is [tex]p = 0.38[/tex].

Samples of size 40, hence [tex]n = 40[/tex].

Hence, the mean and standard error are given by:

[tex]\mu = p = 0.38[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.38(0.62)}{40}} = 0.076[/tex]

Also approximately normal, hence, option A is correct.

To learn more about the Central Limit Theorem, you can take a look at https://brainly.com/question/16695444