To solve this problem, we need to find the probability that the sample mean of a group of nine graduates will be over $45,000, given that the population mean salary is $40,000 with a standard deviation of $5000. We use the sampling distribution of the sample mean for this.

The steps are as follows:

1. Determine the mean and standard deviation of the sample mean:

   • The population mean ($\mu$) is $40,000.
   • The population standard deviation ($\sigma$) is $5000.
   • The sample size ($n$) is 9.

2. Calculate the standard error of the mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{5000}{\sqrt{9}} = \frac{5000}{3} \approx 1666.67$

3. Find the z-score for the sample mean of $45,000: $z = \frac{\bar{x} - \mu}{\text{SEM}} = \frac{45000 - 40000}{1666.67} \approx \frac{5000}{1666.67} \approx 3$

4. Determine the probability corresponding to the z-score:

   • The z-score of 3 corresponds to a cumulative probability of 0.9987 (using standard normal distribution tables).

5. Calculate the probability of the sample mean being over $45,000: $P(\bar{X} > 45000) = 1 - P(Z \leq 3) = 1 - 0.9987 = 0.0013$

Therefore, the probability that a group of nine graduates will have a group average salary over $45,000 is approximately $0.0013$ or 0.13%.

Would you like more details or have any questions?

Here are 5 related questions you might want to ask next:

1. How do you find the z-score for a given value?
2. What does the standard error of the mean represent?
3. How do you use the standard normal distribution table?
4. What are the properties of a normal distribution?
5. How do you calculate probabilities for non-standard normal distributions?

Tip: When dealing with sample means, remember that the standard error of the mean decreases as the sample size increases, making the estimate more precise.