Barron\'s reported that the average number of weeks an individual is unemployed

Question

Barron's reported that the average number of weeks an individual is unemployed is 17 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17 weeks and that the population standard deviation is 3 weeks. Suppose you would like to select a sample of 60 unemployed individuals for a follow-up study. Use z-table. a. Show the sampling distribution of , the sample mean average for a sample of 60 unemployed individuals. (to 1 decimal) E(E) (to 2 decimals)

Explanation / Answer

Answer: - Given data

Average number of weeks = 17 weeks

Population mean length = 17 weeks

Population standard deviation = 3 weeks

a) As per sampling distributuion of xbar (Expected value of xbar) E(xbar) = population mean

So, E(xbar) = 17.0 weeks

Standard deviation of xbar is calculated using the formula = standard deviation / n

Where n = sample size = 60

= 3 / 60 = 0.39

b) probability of providing sample mean of 1 week P(16<= x<=18)

In order to find probability first we need to find the z value, formula to calculate z value is = (sample mean - population mean)/standard deviation of xbar

Z = (18-17)/(3/60) = 2.58 , Probability for z value is 0.9951

Z = (16-17)/(3/60) = -2.58 , Probability for z value is 0.0049

Probsbility is obtained as = 0.9951 - 0.0049= 0.9902

c) probability of providing sample week of half week P(16.5<=x<=17.5)

Z = (17.5-17)/(3/60) = 1.29 , probability for z value is 0.9015

Z = (16.5-17)/(3/60) = -1.29 , probability for z value is 0.0985

Probability is obtained as = 0.9015 - 0.0985 = 0.8030

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