Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 25 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 25 weeks and that the population standard deviation is 2 weeks. Suppose you would like to select a random sample of 36 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is less than 24.
P(X < 24) =

Find the probability that a sample of size n=36n=36 is randomly selected with a mean less than 24.
P(M < 24) =

Answer 1

Solution:-

a) The probability that a single randomly selected value is less than 24 is 0.3085.

Mean = 25, S.D = 2

x = 24

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z = -0.50

P(z < -0.50) = 0.3085

b) The probability that a sample of size n=36n=36 is randomly selected with a mean less than 24 is 0.0013.

Mean = 25, S.D = 2, n = 36

x = 24

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z = -3.0

P(z < -3.0) = 0.0013