Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Approximately what percent of the population have IQ scores between 100 and 130?
z = ( 122-100 ) / 15 = 22/15 = 1.467 Then we have to use a z-table for a normal distribution. The percent of the population that has an IQ less than she does is C ) 93%
