Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 36.3 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 36.3 weeks and that the population standard deviation is 7.5 weeks. Suppose you would like to select a random sample of 124 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is between 35.8 and 36.1. P(35.8 < X < 36.1) =

Find the probability that a sample of size n=124 is randomly selected with a mean between 35.8 and 36.1. P(35.8 < M < 36.1) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Solution :

Given that,

mean = = 36.3

standard deviation = = 15

a ) P (35.8 < x < 36.1 )

P ( 35.8 - 36.3 / 7.5) < ( x -  / ) < ( 36.1 - 36.3 / 7.5)

P ( -0.5 / 7.5 < z < - 0.2 / 7.5 )

P (-0.07 < z < - 0.03)

P ( z < - 0.03) - P ( z < -0.07)

Using z table

= 0.4880 - 0.4721

= 0.0159

Probability = 0.0159

b ) n = 124

=36.3

= / n = 7.5 124 = 37.94733

P (35.8 < M< 36.1 )

P ( 35.8 - 36.3 / 0.6735) < ( M - / ) < ( 36.1 - 36.3 / 0.6735)

P ( -0.5 / 0.6735 < z < - 0.2 / 0.6735 )

P (-0.74 < z < - 0.30)

P ( z < - 0.30) - P ( z < -0.74)

Using z table

= 0.3821- 0.2296

= 0.1525

Probability = 0.1525