Given the assumption of a normal distribution with a mean of 100 and a standard deviation of 15, a score of 130 represents a z-score of 2. In different mostly older handbooks of descriptive statistics you can find a ‘from z to percentile-table’. Nowadays it’s more easy to use the computer. Using R, a free and open source statistical environment (www.r-project.org), you can use the command pnorm (130, 100, 15) and it will give 0.9772. Because yo want the proportion above that score you use 1-pnorm (130, 100, 15). Another way of writing in R and with for example 3 IQ-scores:
perc = pnorm (c (70, 100, 130), 100, 15)
(1 - perc)
gives you the above-proportion of the IQ-scores of respectively 70, 100 and 130.
