Question

)Baron’s reported that average number of weeks an individual is unemployed is 17.5 weeks (Baron’s, February...

)Baron’s reported that average number of weeks an individual is unemployed is 17.5 weeks (Baron’s, February 18, 2008).Suppose you will select a random sample of 50 unemployed individuals for a follow up study. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 weeks and the population standard deviation is 4 weeks. Let X= sample mean of the 50 randomly selected unemployed individuals.

(a)Suppose you want to find the probability that a single unemployed individual selected at random has been unemployed for at least 18.5 weeks. Describe why you could not find this probability based on the information given in the problem statement.

(b)Carefully describe the approximate distribution of X: What is the name of the distribution? What is its mean? What is its standard deviation? How can you know(approximate)the distribution of X?

(c)What is the approximate probability that the sample mean of the 50 randonmly selected unemployed individuals is at least 18.63 weeks?

(d)What is the probability that in a random sample of 50 unemployed individuals, the total employment time (sum of all the 50 durations) is less than 1 year? (Let 1 year = 52 weeks.)

Homework Answers

Answer #1

a) We are not given the distribution of the variable which gives the least time for which the single unemployed individual selected at random has been unemployed.This is the reason that we cannot find the required probability here.

b) As the sample size here is 50 > 30, therefore we can apply the central limit theorem to approximate the distribution of X as a normal distribution given as:

b) The probability here is computed as:

P( X >= 18.63)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.0229 is the required probability here.

d) Probability that the sum of total employment time is less than 52 weeks

means that P(X < 52/50)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0 is the required probability here.

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