Final answer:
To find the probability of the sample mean being within $500 of the population mean, we can use the Central Limit Theorem. Calculating the standard deviation of the sample mean and finding the corresponding Z-score, we can determine the probability to be approximately 1 or 100%.
Step-by-step explanation:
To find the probability that the sample mean computed using a simple random sample of 30 employees will be within $500 of the population mean, we need to determine the standard deviation of the sample mean. Since the sample size is 30, we can assume that the sample mean will be normally distributed.
Given that the personnel director believes the sample mean will be within $500 of the population mean, we can use the Central Limit Theorem to approximate the probability. Using the formula for the standard deviation of the sample mean, which is equal to the population standard deviation divided by the square root of the sample size, we can calculate the standard deviation of the sample mean as:
σx = σ / √n = $500 / √30 ≈ $90.72
Next, we can calculate the Z-score, which represents the number of standard deviations away from the mean. We can use the formula: Z = (x - μ) / σx. In this case, since we want the sample mean to be within $500 of the population mean, we have:
500 = Z * 90.72
Solving for Z, we find Z ≈ 5.5. Using a Z-table or calculator, we can find the probability corresponding to Z = 5.5, which is approximately 1. The probability that the sample mean computed using a simple random sample of 30 employees will be within $500 of the population mean is approximately 1 or 100%.