Question

According to a survey, about 19% of all Americans are without health insurance. Assume that this estimate is reasonably accurate. Suppose that you select a random sample of 46 Americans. (a) Calculate the expected number of Americans without health insurance in the sample. Answer: (Keep at least four decimal places.) (b) Calculate the standard deviation of the number of Americans without health insurance in the sample. Answer: (Keep at least four decimal places.)

Answer 1

Answer:

a) 8.74

b) 2.6607

Step-by-step explanation:

For each American, there are only two possible outcomes. Either they are without health insurance, or they are not. In the sample, each person is independent of any other person, which means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

According to a survey, about 19% of all Americans are without health insurance.

This means that $p = 0.19$

Suppose that you select a random sample of 46 Americans.

This means that $n = 46$

(a) Calculate the expected number of Americans without health insurance in the sample.

$E(X) = np = 46*0.19 = 8.74$

(b) Calculate the standard deviation of the number of Americans without health insurance in the sample.

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{46*0.19*0.81} = 2.6607$