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:
[tex]z = \frac{\bar x - \mu}{\sigma / \sqrt{n} } = \frac{\pm 1}{4/\sqrt{60} } \approx \pm 1.94[/tex]
[tex]P = P(-1 < \bar x - \mu < 1) = P(-1.94 < z < 1.94) = 1 - 2P(z < -1.94)[/tex]
Using any standard negative z-scores table, we can find that:[tex]P(z < -1.94) = 0.0262[/tex]
Thus, we get:
[tex]P(| \bar x - \mu | < 1) = 1 - 2 \times 0.0262 = 1 - 0.0524 = 0.9476[/tex]
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
