Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 21.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 21.5 weeks and that the population standard deviation is 4.5 weeks. Suppose you would like to select a random sample of 87 unemployed individuals for a follow-up study. Round z-scores to two decimal places.

Find the probability that a single randomly selected value is between 20.1 and 21.3.
P(20.1 < X < 21.3) =

Find the probability that a sample of size n=87n=87 is randomly selected with a mean between 20.1 and 21.3.
P(20.1 < μμ < 21.3) =

Answer 1

Solution :

Given that ,

mean = = 21.5

standard deviation = = 4.5

a) P( 20.1 < x < 21.3 ) = P[(20.1 - 21.5)/ 4.5) < (x - ) /  < (21.3 - 21.5) / 4.5) ]

= P(-0.27 < z < -0.04)

= P(z < -0.04) - P(z <-0.27 )

Using z table,

= 0.4840 - 0.3936

= 0.0904

b) n = 87

=   = 21.5

= / n = 4.5 / 87 = 0.482

P(20.1 < < 21.3)  

= P[(20.1 - 21.5) / 0.482 < ( - ) / < (21.3 - 21.5) / 0.482 )]

= P( -2.90 < Z < -0.41)

= P(Z < -0.41) - P(Z < -2.90)

Using z table,  

= 0.3409 - 0.0019  

= 0.3390