Question

Assume that IQ scores are normally distributed, with a standard deviation of 18 points and a mean of 100 points. If 115 people are chosen at random, what is the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points? (Round your answer to four decimal places.)

Answer 1

To find the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points, we'll use the Central Limit Theorem. For a sample size of n = 115, the standard deviation of the sample mean is σ/√n.

First, calculate the standard deviation of the sample mean:
σ_sample = σ/√n = 18/√115 ≈ 1.674

Now, we'll find the z-scores for the range of 98 to 102 (100 ± 2):
z1 = (98 - 100)/1.674 ≈ -1.196
z2 = (102 - 100)/1.674 ≈ 1.196

Next, we'll use a z-table or calculator to find the probability between these z-scores:
P(-1.196 < z < 1.196) ≈ 0.2324 + 0.2324 = 0.4648

So, the probability that the sample mean will not differ from the population mean by more than 2 points is approximately 0.4648, or 46.48%.