Question

# Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume...

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 40 unemployed individuals for a follow-up study.

1. What is the probability that a simple random sample of 40 unemployed individuals will provide a sample mean within 1 week of the population mean? (Round your answer to four decimal places.)

2. What is the probability that a simple random sample of 40 unemployed individuals will provide a sample mean within 1/2 week of the population mean? (Round your answer to four decimal places.)

a)

Here, μ = 17.5, σ = 0.6325, x1 = 16.5 and x2 = 18.5. We need to compute P(16.5<= X <= 18.5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (16.5 - 17.5)/0.6325 = -1.58
z2 = (18.5 - 17.5)/0.6325 = 1.58

Therefore, we get
P(16.5 <= X <= 18.5) = P((18.5 - 17.5)/0.6325) <= z <= (18.5 - 17.5)/0.6325)
= P(-1.58 <= z <= 1.58) = P(z <= 1.58) - P(z <= -1.58)
= 0.9429 - 0.0571
= 0.8858

b)

Here, μ = 17.5, σ = 0.6325, x1 = 17 and x2 = 18. We need to compute P(17<= X <= 18). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (17 - 17.5)/0.6325 = -0.79
z2 = (18 - 17.5)/0.6325 = 0.79

Therefore, we get
P(17 <= X <= 18) = P((18 - 17.5)/0.6325) <= z <= (18 - 17.5)/0.6325)
= P(-0.79 <= z <= 0.79) = P(z <= 0.79) - P(z <= -0.79)
= 0.7852 - 0.2148
= 0.5704