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Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks (Barron's, February $18,2008 ) .$ Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 50 unemployed individuals for a follow-up study.

a. Show the sampling distribution of $\overline{x},$ the sample mean average for a sample of 50 unemployed individuals.

b. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean?

c. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1$/ 2$ week of the population mean?

a. See graph

b. 0.9232

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all right, So this question asks us about the sampling distribution on the meantime of unemployment. So the first part wants us to sketch the sampling distribution if we have a sample of 50 employees. So that's just going to look like we have our main 17.5 weeks right here. And it's just going to be a bell curve because an equals 50 that's line this up a little better, and this is normal approximately with a mean of 17.5 weeks and a standard deviation of four weeks. And remember, it's gonna be a bell curve because the central limit theorem says, if your sample size is big enough, everything starts to look normal in the sampling distribution, which in this case, our sample size is 50 which is more than enough to guarantee normality in this case, thin Part B. Once the probability that a sample of 50 is within one week of the mean so if the mean is 17.5 within one week means the probability that it's between 16.5 and 18.5. All right, one week, about one week below. And to do this, we need to know the standard error because we can't use the population standard deviation because we're working with a sample. We're not just looking at one individual employees. We're looking at multiple. So we're looking for standard air, which is the population sigma over the square root of the sample size, which in this case is four divided by square root of 50 which works out to point 5657 So now we have everything we need. So the probability that it's within one week well, this is just another area problem, which we learned from last chapter. Weaken Do normal CDF are lower limit 16.5. Our upper limit 18.5 our mean is 17.5. And our standard deviation is not our population sigma, but rather our standard error of 0.5657 And if you put that in your calculator, you'll get an answer off 0.9229 Then part See wants the probability that it's within ah half of a week. So this is the same thing as saying that it's between 17 in point. Oh, an 18 point. Oh, because I mean a 17.5. So subtracting a week we get 17 and adding half a week. We get 18 which is very similar. We can use normal CDF again are lower bound is 17. Our upper bound is 18. Our mean is still 17.5. And for a standard deviation, we still use the standard error of 0.56 57 And that gives us an answer of 0.6234 So, as you can see, due to the fact that we have a relatively large sample of 50 people, we're going toe likely be very close to the population mean if we sample here. So even just half a week we have over 62% chance of receiving a sample like that. You can see really the power that samples have at reducing variability, because we're all the way up from a problem from ah Sigma of four weeks in our population to a standard error of 0.5657 and our sample so variability

University of Michigan - Ann Arbor