Final answer:
The question involves using the Central Limit Theorem to find the probability that a sample mean is less than 60.28. We calculate the z-score using the population mean, standard deviation, and sample size, and then find the corresponding probability from a standard normal distribution table or with software.
Step-by-step explanation:
The question asks about finding the probability that a sample mean is less than a specific value, given the population mean, standard deviation, and sample size. This is a classic application of the Central Limit Theorem (CLT). First, we find the mean of the sampling distribution, which is equal to the population mean, and then calculate the standard error of the mean (SEM), which is the population standard deviation divided by the square root of the sample size.
Once we have the SEM, we can calculate the z-score for the sample mean of 60.28. The z-score formula is (X - μ) / (SEM), where X is the sample mean, μ is the population mean, and SEM is the standard error of the mean. Using the provided values, the z-score is (60.28 - 62) / (22.32/√53).
After calculating the z-score, we look up this value in a standard normal distribution table or use software to find the probability that a z-score is less than the calculated value. This reflects the probability that the sample mean is less than 60.28. It's important to recognize that the CLT allows us to use the normal distribution to approximate the sampling distribution of the mean, even if the original population is not normally distributed, provided the sample size is sufficiently large (typically n > 30).