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We are given:
- Mean salary of MBA graduates 10 years after graduation: $\mu = 138,000$
- Standard deviation of salaries: $\sigma = 45,000$
- Sample size: $n = 12$

### a. What is the distribution of $ X $?
$ X $ represents the salary of a single randomly selected graduate.

- Since individual salaries are normally distributed, $ X \sim N(\mu, \sigma^2) $.
- Therefore, $ X \sim N(138,000, 45,000^2) $.

**Answer:** $ X \sim N(138,000, 45,000^2) $.

### b. What is the distribution of $ \bar{X} $?
$ \bar{X} $ is the mean salary for a sample of 12 graduates.

- The mean of the sample means is the same as the population mean: $ \mu_{\bar{X}} = \mu = 138,000 $.
- The standard deviation of the sample means (standard error) is $ \frac{\sigma}{\sqrt{n}} = \frac{45,000}{\sqrt{12}} $.
- Therefore, \(

Business Weekly conducted a survey of graduates from 30 top MBA programs. Assume the mean annual salary for graduates 10 years after graduation is $138,000. Assume the standard deviation is $45,000. Suppose you take a simple random sample of 12 graduates.

Analyze the normal distribution of salaries for MBA graduates with a mean of $138,000 and a standard deviation of $45,000. This problem involves calculating probabilities for both individual and sample means, focusing on the properties of the normal distribution and sampling distributions.

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