Question

Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 17.5 and that the population standard deviation is 4 weeks. Suppose you would like to select a sample of 50 unemployed individuals for a follow-up study. What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean?

Answer 1

Answer:

$P(16.5$

And using the normal standard table or excel we find the probability:

$P(-1.768< Z< 1.768) = P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the avergae number of weeks an individual is unemployed of a population, and for this case we know the distribution for X is given by:

$X \sim N(17.5,4)$  

Where $\mu=17.5$ and $\sigma=4$

Since the distribution for X is normal then, the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu= 17.5, \frac{\sigma}{\sqrt{n}}= \frac{4}{\sqrt{50}}=0.566)$

We select a sample of n =50 people. And we want to find the following probability

$P(16.5$

And using the normal standard table or excel we find the probability:

$P(-1.768< Z< 1.768) = P(Z$