The average grade in a statistics course has been 71, with a standard deviation of 10.5. If a random sample of 60 is selected from this population, what is the probability that the average grade is more than 75?

Given:
Mean ( \mu ) = 71
Standard deviation ( \sigma ) = 10.5
Sample size (n) = 60
We need to find the probability that the average grade is more than 75, which can be represented as P(\bar{X} > 75) .

We know that the sampling distribution of the sample mean ( \bar{X} ) is approximately normally distributed when the sample size is large (Central Limit Theorem). The mean of the sampling distribution is the same as the population mean, which is \mu = 71 . The standard deviation of the sampling distribution is given by the formula: \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} .

Now, we can calculate the standard deviation of the sampling distribution:
\sigma_{\bar{X}} = \frac{10.5}{\sqrt{60}} \approx \frac{10.5}{7.75} \approx 1.355

Next, we need to calculate the z-score for a sample mean of 75:
z = \frac{75 - 71}{1.355} \approx \frac{4}{1.355} \approx 2.95

Using a standard normal table or calculator, we can find the probability that a z-score is greater than 2.95, which is approximately 0.0016.

Therefore, the probability that the average grade is more than 75 is \boxed{0.0016} .