Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks. 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 sample of 45 unemployed individuals for a follow-up study.
(a) Show the sampling distribution of x, the sample mean average for a sample of 45 unemployed individuals.
A bell-shaped curve is above a horizontal axis labeled weeks.
A bell-shaped curve is above a horizontal axis labeled weeks.
A bell-shaped curve is above a horizontal axis labeled weeks.
A bell-shaped curve is above a horizontal axis labeled weeks.
(b) What is the probability that a simple random sample of 45 unemployed individuals will provide a sample mean within 1 week of the population mean? (Round your answer to four decimal places.)
(c) What is the probability that a simple random sample of 45 unemployed individuals will provide a sample mean within 1/2 week of the population mean? (Round your answer to four decimal places.)
a) Sampling distribution of x:
µₓ = 17.5
σₓ = σ/√n = 4/√45 = 0.6
b) Probability that sample mean will be within 1 week of the population mean, P(|X̅-µ|< 1) =
= P(-1 < (X̅-µ) < 1)
= P( (-1)/(4/√45) < (X-µ)/(σ/√n) < (1)/(4/√45) )
= P(-1.6771 < z < 1.6771)
= P(z < 1.6771) - P(z < -1.6771)
Using excel function:
= NORM.S.DIST(1.6771, 1) - NORM.S.DIST(-1.6771, 1)
= 0.9065
c) Probability that sample mean will be within 1/2 week of the population mean, P(|X̅-µ|< 0.5) =
P(|X̅-µ|< 0.5) =
= P(-0.5 < (X̅-µ) < 0.5)
= P( (-0.5)/(4/√45) < (X-µ)/(σ/√n) < (0.5)/(4/√45) )
= P(-0.8385 < z < 0.8385)
= P(z < 0.8385) - P(z < -0.8385)
Using excel function:
= NORM.S.DIST(0.8385, 1) - NORM.S.DIST(-0.8385, 1)
= 0.5983
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