Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 17.2 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.2 weeks and that the population standard deviation is 8.1 weeks. Suppose you would like to select a random sample of 190 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is between 15.6 and 18.8. P(15.6 < X < 18.8) =

Find the probability that a sample of size n = 190 is randomly selected with a mean between 15.6 and 18.8. P(15.6 < M < 18.8) = Enter your answers as numbers accurate to 4 decimal places.

Answer 1

a)

Given,

= 17.2 , = 8.1

We convert this to standard normal as

P( X < x) = P( Z < x - / )

So,

P( 15.6 < X < 18.8) = P( X < 18.8) - P( X < 15.6)

= P( Z < 18.8 - 17.2 / 8.1) - P( Z < 15.6 - 17.2 / 8.1)

= P( Z < 0.1975) - P( Z < -0.1975)

= 0.5783 - 0.4217

= 0.1566

b)

Using central limit theorem,

P( < x) = P( Z < x - / / sqrt(n) )

So,

P( 15.6 < < 18.8) = P( < 18.8) - P( < 15.6)

= P( Z < 18.8 - 17.2 / (8.1 / sqrt(190) ) ) - P( Z < 15.6 - 17.2 / (8.1 / sqrt(190) ) )

= P( Z < 2.7228) - P( Z < -2.7228)

= 0.9968 - 0.0032

= 0.9935