Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test and their summary scores indicate that their mean IQ score is 133. IQ scores are normally distributed and have a mean of 100 and a standard deviation of 15. If nine people are randomly selected from the population, find the probability that their mean score is at least 133.

To find the probability that the mean IQ score of nine people is at least 133, we can use the central limit theorem. According to the central limit theorem, regardless of the shape of the population distribution, the distribution of sample means will be approximately normal if the sample size is large enough (in this case, n = 9).

First, we need to calculate the standard deviation of the sample mean. The standard deviation of the sample mean (also known as the standard error) can be calculated using the formula:

Standard Error (SE) = standard deviation / square root of sample size

In this case, the standard deviation is given as 15 (from the population) and the sample size is 9. So, the standard error is:

SE = 15 / sqrt(9) = 15 / 3 = 5

Next, we need to standardize the sample mean using the z-score formula:

z = (sample mean - population mean) / standard error

In this case, the sample mean is 133 (from the problem statement) and the population mean is 100 (given). Using the standard error calculated above (SE = 5), we can calculate the z-score:

z = (133 - 100) / 5 = 6.6

Now, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. Since we want the probability that the mean score is at least 133, we need to find the area to the right of the z-score (6.6).

Using a standard normal distribution table or a calculator, we can find that the probability associated with a z-score of 6.6 is close to 1.

Therefore, the probability that the mean IQ score of nine people is at least 133 is approximately 1.