Question

Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks. Suppose you would like to select a sample of 55 unemployed individuals for a follow-up study.
A) show the sampling distribution of x, the sample mean average for a sample of 50 unemployment individuals.B) What is the probability that a simple random sample of 50 unemployment individuals will provide a sample mean within one week of the population mean?C) What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within a half week of the population mean?

Answer 1

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

$\mu_s=\mu=18.5$

The standard deviation of the sampling distribution is:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849$

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179$

The probability it then:

$P(|X_s-\mu_s|$

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589$

$P(|X_s-\mu_s|$