Find the probability and interpret the results. If convenient, use technology to find the probability.
The population mean annual salary for environmental compliance specialists is about $\$ 60,500$. A random sample of 37 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $\$ 58,000$ ? Assume $\sigma=\$ 6,200$.
The probability that the mean salary of the sample is less than $\$ 58,000$ is 0.0071 . (Round to four decimal places as needed.)
Interpret the results. Choose the correct answer below.
A. Only $0.71 \%$ of samples of 37 specialists will have a mean salary less than $\$ 58,000$. This is an unusual event.
B. About $71 \%$ of samples of 37 specialists will have a mean salary less than $\$ 58,000$. This is not an unusual event.
C. About $0.71 \%$ of samples of 37 specialists will have a mean salary less than $\$ 58,000$. This is not an unusual event.
D. Only $71 \%$ of samples of 37 specialists will have a mean salary less than $\$ 58,000$. This is an unusual event.

Step 1 ：Given that the population mean annual salary for environmental compliance specialists is \(\$ 60,500\), the population standard deviation is \(\$ 6,200\), and a random sample of 37 specialists is drawn from this population.

Step 2 ：We are asked to find the probability that the mean salary of the sample is less than \(\$ 58,000\).

Step 3 ：First, we calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size. The formula for the standard error is \(SE = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation and \(n\) is the sample size.

Step 4 ：Substituting the given values, we get \(SE = \frac{6200}{\sqrt{37}}\), which simplifies to \(SE \approx 1019.27\).

Step 5 ：Next, we calculate the z-score, which is the difference between the sample mean and the population mean, divided by the standard error. The formula for the z-score is \(z = \frac{x - \mu}{SE}\), where \(x\) is the sample mean, \(\mu\) is the population mean, and \(SE\) is the standard error.

Step 6 ：Substituting the given values, we get \(z = \frac{58000 - 60500}{1019.27}\), which simplifies to \(z \approx -2.45\).

Step 7 ：Finally, we use the z-score to find the probability. The probability that the mean salary of the sample is less than \(\$ 58,000\) is approximately 0.0071.

Step 8 ：This means that only about 0.71% of samples of 37 specialists will have a mean salary less than \(\$ 58,000\). This is an unusual event.

Step 9 ：\(\boxed{\text{Final Answer: A. Only 0.71 \% of samples of 37 specialists will have a mean salary less than \$ 58,000. This is an unusual event.}}\)