A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 17 weeks.

Question

Assume that for the population of all unemployed individuals the population mean length of unemployment is 17 weeks and that the population standard deviation is 2.5 weeks. Suppose you would like to select a random sample of 89 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 16.7.

P(X > 16.7) =

(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=89 is randomly selected with a mean greater than 16.7.

P(M > 16.7) =

(Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Note: I am going to explain this using the most common method, the z-table method, but there are methods that use technology as well.

To calculate the probability of a certain value, first find the z-score of the value.

z = (x - μ)/σ = (16.7 - 17)/2.5 = -0.12.

P(X > 16.7) = P(Z > -0.12) = 1 - P(Z < -0.12)

(Note: Z-tables only tell the less than probability, so we have to calculate using complements)

On the z-table, if we look up the row -0.1 and the column 0.02, we find the probability of P(Z < -0.12)

1 - P(Z < -0.12) = 1 - 0.4522 = 0.5478.

To calculate the probability of a mean value, first find the z-score of the mean.

z = (x_bar - μ)/(σ/√n) = (16.7 - 17) / (2.5/√98) ≈ -1.19

P(M > 16.7) ≈ P(Z > -1.19) = 1 - P(Z < -1.19)

On the z-table, if we look up the row -1.1 and the column 0.09, we find the probability of P(Z < -1.19)

1 - P(Z < -1.19) = 1 - 0.1170 = 0.8830.