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Let's break down the problem step-by-step.

### Given:
- The average number of weeks unemployed \( \mu = 29 \) weeks.
- The population standard deviation \( \sigma = 9 \) weeks.
- We are selecting a random sample of 37 unemployed individuals.

We are asked to round all answers to 4 decimal places.

---

### Part a) What is the distribution of \( X \)?

Here, \( X \) represents the length of unemployment for an individual, and it follows a normal distribution. Given that the population mean is \( \mu = 29 \) and the population variance is \( \sigma^2 = 9^2 = 81 \), we can write:

\[
X \sim N(29, 81)
\]

---

### Part b) What is the distribution of \( \bar{X} \)?

Here, \( \bar{X} \) represents the sample mean of 37 individuals. The distribution of the sample mean \( \bar{X} \) also follows a normal distribution by the Central Limit Theorem. The mean of \( \bar{X} \) is the same as the population mean (\( \mu = 29 \)), and the variance of \( \bar{X} \) is the population variance divided by the sample size \( n = 37 \):

\[
\text{Variance of } \bar{X} = \frac{\sigma^2}{n} = \frac{81}{37} = 2.1892
\]

Thus, the distribution of \( \bar{X} \) is:

\[
\bar{X} \sim N(29, 2.1892)
\]

---

### Part c) What is the probability that one randomly selected individual found a job in less than 28 weeks?

We are asked to find \( P(X < 28) \), where \( X \) follows \( N(29, 81) \).

First, calculate the z-score:

\[
z = \frac{28 - 29}{9} = \frac{-1}{9} = -0.1111
\]

Now, using the standard normal distribution table or a calculator, find \( P(Z < -0.1111) \). This corresponds to a probability of approximately 0.4552.

\[
P(X < 28) \approx 0.4552
\]

---

### Part d) For 37 unemployed individuals, find the probability that the **average** time they found the next job is less than 28 weeks.

We are asked to find \( P(\bar{X} < 28) \), where \( \bar{X} \sim N(29, 2.1892) \).

First, calculate the z-score:

\[
z = \frac{28 - 29}{\sqrt{2.1892}} = \frac{-1}{1.4796} \approx -0.676
\]

Now, using the standard normal distribution table or a calculator, find \( P(Z < -0.676) \). This corresponds to a probability of approximately 0.2496.

\[
P(\bar{X} < 28) \approx 0.2496
\]

---

### Part e) Is the assumption of normal necessary?

Yes, the assumption of normality is necessary to use the z-scores and calculate probabilities based on the normal distribution. However, for large sample sizes, even if the population isn't normally distributed, the Central Limit Theorem ensures that the sampling distribution of the sample mean \( \bar{X} \) approaches normality.

---

Would you like more details on any of these parts? Here are some related questions for further exploration:

1. What would the z-score be if the unemployment time was 30 weeks instead of 28?
2. How would the probability change if the population standard deviation were 10 weeks instead of 9?
3. What if the sample size were smaller (e.g., 10)? How would that affect the variance of \( \bar{X} \)?
4. Can we calculate a confidence interval for the sample mean of unemployment time? How?
5. How would the calculations change if the distribution of unemployment time was not normal?

**Tip:** Always double-check the assumptions when using normal approximations, especially for smaller sample sizes or non-normal populations.

A leading magazine reported that the average number of weeks an individual is unemployed is 29 weeks. Assume that the length of unemployment is normally distributed with a population mean of 29 weeks and a standard deviation of 9 weeks. Suppose a random sample of 37 unemployed individuals is selected for a follow-up study.

Solve a statistics problem involving a normally distributed population with an average unemployment duration of 29 weeks and a standard deviation of 9 weeks. Learn to calculate probabilities and use the Central Limit Theorem for a sample of 37 unemployed individuals. This exercise covers normal distributions, Z-scores, and statistical probability.

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