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)

45%