In a large population of college-educated adults, the mean iq is 112 with standard deviation 25. suppose 300 adults from this population are randomly selected for a market research campaign. the probability that the sample mean iq is greater than 115 is:

Given:

μ = 112, population mean

σ = 25, population std. deviation

The randomly selected sample population of 300 (>30) is large enough for probability testing based on the normal distribution.

x = 115, random variable.

z-score:

x = (x-μ) / σ = (115 - 112) / 25 = 0.12

Let xs = sample mean.

From standard tables, obtain

P (xs<115) = 0.548

Therefore

P (xs>115) = 1 - 0.548 = 0.452 = 45% (approx)

Answer: 45%