Assume that for the population of all unemployed individuals the population mean length of unemployment is 20 weeks and that the population standard deviation is 3 weeks. Suppose you would like to select a random sample of 59 unemployed individuals for a follow-up study. Find the probability that the mean of this sample of size is more than 20.7. Round your answer to at least three decimals.

Step 1 ：Assume that for the population of all unemployed individuals the population mean length of unemployment is 20 weeks and that the population standard deviation is 3 weeks. Suppose you would like to select a random sample of 59 unemployed individuals for a follow-up study.

Step 2 ：We are asked to find the probability that the mean of this sample of size is more than 20.7. This is a problem of sampling distributions.

Step 3 ：We know that the sampling distribution of the mean is normally distributed if the sample size is large enough (n > 30). In this case, the sample size is 59, which is greater than 30, so we can use the normal distribution to approximate the sampling distribution.

Step 4 ：The mean of the sampling distribution is the same as the population mean, which is 20 weeks. The standard deviation of the sampling distribution (also known as the standard error) is the population standard deviation divided by the square root of the sample size.

Step 5 ：We can standardize the sample mean (20.7) to a z-score by subtracting the population mean and dividing by the standard error. Then, we can find the probability that the z-score is greater than the standardized sample mean using the standard normal distribution.

Step 6 ：Let's calculate the standard error, standardize the sample mean to a z-score, and find the probability. The standard error is \( \frac{3}{\sqrt{59}} \approx 0.39 \)

Step 7 ：The z-score is \( \frac{20.7 - 20}{0.39} \approx 1.79 \)

Step 8 ：The probability that the z-score is greater than 1.79 is approximately 0.037. This means that there is about a 3.7% chance that the sample mean is greater than 20.7 weeks.

Step 9 ：Final Answer: The probability that the mean of this sample of size is more than 20.7 is approximately \(\boxed{0.037}\).