For 36 unemployed individuals, find the probability that the average time that they found the next job is more than 30 weeks.

Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 28 weeks. Assume that the length of unemployment is normally distributed with population mean of 28 weeks and the population standard deviation of 8 weeks. Suppose you would like to select a random sample of 36 unemployed individuals for a follow-up study. Round the answers of following questions to 4 decimal places.

For 36 unemployed individuals, find the probability that the average time that they found the next job is more than 30 weeks. 

Answer 1

Hi Kaitlyn,

We are working again with sample mean, so we first need a standard error, which can be obtained via the equation:

SE=s/sqrt(n) where:

s=standard deviation

n=sample size

For this problem:

s=8

n=36

Thus:

SE=8/sqrt(36)=

8/6=

4/3 or 1.33 repeating, in decimal form

Now, we plug that standard error into the classic equation z=(x-mu)/sigma, BUT:

-Recall that we are dealing with a sample mean, so we substitute x-bar in for x, and standard error in for sigma:

Z=(X-bar-mu)/SE

For this problem:

X-bar=30

mu=28

SE=4/3

Therefore:

z=(30-28)/(4/3)

z=1.5

From z-table, look for 1.5 on the column, 0 on the row:

P(Z<1.5)=P(X<30)=0.9332

But the question asked for greater than 30 weeks, so:

P(X>30)=P(Z>1.5)=1-P(Z<1.5)

1-P(Z<1.5)=1-0.9332=

P(X>30)= 0.0668

This indicates that most unemployed people are indeed able to find a new job within 30 weeks, or about 7 months, but about 6.7% are not. I hope this helps.