Question

In: Statistics and Probability

Use the following information for this and the next question. In 2019, the mean duration of...

Use the following information for this and the next question. In 2019, the mean duration of unemployment for a person is 21.6 weeks. Assume that the population standard deviation is 5 weeks. You would like to conduct a follow-up study, so you select a sample of 40 unemployed people. You need to look into the sampling distribution of the mean number of weeks of unemployment.

What is the mean and standard deviation of that distribution? What is the probability that in a sample of 40 unemployed people, the mean number of weeks of unemployment is within 1 week of the population mean?

Expert Solution

Solution :

Given that,

mean = = 21.6

standard deviation = = 5

n = 40

= = 21.6

= / n = 5 / 40 = 0.791

P(20.6 < < 22.6)

= P[(20.6 - 21.6) / 0.791< ( - ) / < (22.6 - 21.6) / 0.791)]

= P(-1.26 < Z < 1.26)

= P(Z < 1.26) - P(Z < -1.26)

Using z table,

= 0.8962 - 0.1038

= 0.7924

orchestra answered 1 year ago

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