Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 30 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 30 weeks and that the population standard deviation is 2 weeks. Suppose you would like to select a random sample of 59 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is greater than 30.2. P(X > 30.2) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a sample of size n = 59 is randomly selected with a mean greater than 30.2. P(M > 30.2) = (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Mean = = 30

Standard deviation = = 2

We have to find P(X > 30.2)

For finding this probability we have to find the z score.

That is we have to find P(Z > 0.1)

P(Z > 0.1) = 1 - P(Z < 0.1) = 1 - 0.5398 = 0.4602

( From z table)

Sample size = n = 59

We have to find P( M > 30.2)

For finding this probability we have to find z score.

That is we have to find P(Z > 0.77)

P(Z > 0.77) = 1 - P(Z < 0.77) = 1 - 0.7794 = 0.2206

( From z table)