If the mean IQ is 100 with a standard deviation of 15, what is the approximate probability that a randomly selected sample of nine people will have a mean IQ greater than 110?

To find the probability that a randomly selected sample of nine people will have a mean IQ greater than 110, we can use the Central Limit Theorem. This theorem states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (and with an IQ distribution we can assume it is normal to begin with).

First, let's define the parameters: - The population mean (μ) is 100. - The population standard deviation (σ) is 15. - The sample size (n) is 9. - We are looking for the probability that the sample mean (X̄) is greater than 110.

We can use the Z-score formula for a sample mean: Z = (X̄ - μ) / (σ/√n)

Substituting the values we have: Z = (110 - 100) / (15/√9) Z = (110 - 100) / (15/3) Z = 10 / 5 Z = 2

Looking this Z-value up in standard normal distribution tables, or using a calculator, we can find the probability to the left of Z. However, since we want the area to the right (meaning the probability of having a sample mean greater than 110), we will need to subtract that value from 1.

The probability (P) to the left of a Z-score of 2 is approximately 0.9772.

Thus, the probability to the right is: P(X̄ > 110) ≈ 1 - P(Z < 2) P(X̄ > 110) ≈ 1 - 0.9772 P(X̄ > 110) ≈ 0.0228

So, there is an approximate 2.28% chance that a randomly selected sample of nine people will have a mean IQ greater than 110.