A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 38 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 38 weeks and that the population standard deviation is 2.8 weeks. Suppose you would like to select a random sample of 77 unemployed individuals for a follow-up study. Find the probability that a sample of size $n=77 /$ is randomly selected with a mean greater than 38.4.

Step 1 ：We are given that the population mean (\(\mu\)) is 38 weeks, the population standard deviation (\(\sigma\)) is 2.8 weeks, the sample size (\(n\)) is 77, and we are asked to find the probability that the sample mean (\(\bar{x}\)) is greater than 38.4 weeks.

Step 2 ：We first calculate the standard deviation of the sample mean (\(\sigma_{\bar{x}}\)), which is given by the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\).

Step 3 ：Substituting the given values into the formula, we get \(\sigma_{\bar{x}} = \frac{2.8}{\sqrt{77}} = 0.3190896140869862\).

Step 4 ：We then calculate the z-score for 38.4 weeks using the formula \(z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\).

Step 5 ：Substituting the given values into the formula, we get \(z = \frac{38.4 - 38}{0.3190896140869862} = 1.2535663410560132\).

Step 6 ：Finally, we find the probability that the z-score is greater than the calculated value. This can be done using a z-table or a statistical software. The probability is approximately 0.105.

Step 7 ：Final Answer: The probability that a sample of size \(n=77\) is randomly selected with a mean greater than 38.4 is approximately \(\boxed{0.105}\).