Question

What is the probability that a simple random sample of 60 unemployed individuals will provide a sample mean within 1 week of the population mean? (Round your answer to four decimal places.)

Answer 1

The question is not complete so I will answer it with an example and a few assumptions. Follow the steps to find the answer to your question.

Important to note:

Your question is a z-scores problem

Assume that for the population of unemployed individuals the population standard deviation is 4 weeks.

Thus, we need to find the z-value.

The z-value is the sample mean decreased by the population mean, divided by the standard deviation that we assumed. So, we have:

$z = \frac{\bar x - \mu}{\sigma / \sqrt{n} } = \frac{\pm 1}{4/\sqrt{60} } \approx \pm 1.94$

$P = P(-1 < \bar x - \mu < 1) = P(-1.94 < z < 1.94) = 1 - 2P(z < -1.94)$

Using any standard negative z-scores table, we can find that:$P(z < -1.94) = 0.0262$

Thus, we get:

$P(| \bar x - \mu | < 1) = 1 - 2 \times 0.0262 = 1 - 0.0524 = 0.9476$

Therefore, the probability that a simple random sample of 60 unemployed individuals will provide a sample mean within 1 week of the population mean is 0.9476

Answer:

0.9476