Question

Barron's reported that the average number of weeks an individual is unemployed is 15 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 15 weeks and that the population standard deviation is 6 weeks. Suppose you would like to select a sample of 65 unemployed individuals for a follow-up study. Use z-table.

a. Show the sampling distribution of X bar, the sample mean average for a sample of  unemployed individuals.

E(Xbar)	(to 1 decimal)
	(to 2 decimals)

b. What is the probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1 week of the population mean?

(to 4 decimals)

c. What is the probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1/2 week of the population mean?

(to 4 decimals)

Answer 1

Solution :

Given that,

mean = = 15

standard deviation = = 6

a) n = 65

= = 15

= / n = 6 / 65 = 0.74

b) 15  ± 1 = 14, 16

P(14 < < 16)  

= P[(14 - 15) / 0.74 < ( - ) / < (16 - 15) / 0.74)]

= P( -1.35 < Z < 1.35)

= P(Z < 1.35) - P(Z < -1.35)

Using z table,  

= 0.9115 - 0.0885  

= 0.8230

c) 15  ± 0.5 = 14.5, 15.5

P(14.5 < < 15.5)  

= P[(14.5 - 15) / 0.74 < ( - ) / < (15.5 - 15) / 0.74)]

= P( -0.68 < Z < 0.68)

= P(Z < 0.68) - P(Z < -0.68)

Using z table,  

= 0.7517 - 0.2483

= 0.5034