Step 1 :Given that the population mean annual salary for environmental compliance specialists is \$62,500, a random sample of 34 specialists is drawn from this population. We are asked to find the probability that the mean salary of the sample is less than \$58,500. The population standard deviation is given as \$6,000.
Step 2 :We can use the Z-score formula to find the Z-score of the sample mean, and then use the standard normal distribution to find the probability. The Z-score formula is: \[Z = \frac{X - \mu}{\sigma / \sqrt{n}}\] where \(X\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
Step 3 :Substituting the given values into the formula, we get: \[Z = \frac{58500 - 62500}{6000 / \sqrt{34}} \approx -3.887301263230201\]
Step 4 :After finding the Z-score, we can use the standard normal distribution to find the probability. The probability that the mean salary of the sample is less than \$58,500 is approximately 0.0000506824591217036, or 0.00506824591217036%.
Step 5 :This is a very small probability, which means it is an extremely unusual event. Therefore, the correct answer is: \[\boxed{\text{B. Only 0.01\% of samples of 34 specialists will have a mean salary less than \$58,500. This is an extremely unusual event.}}\]