Suppose that 39% of the area under a standard normal curve is to the right of a
-score. How much area is between the mean and that
-score?
If 39% of the area is to the right of the z-score, then 100% - 39% = 61% of the area is to the left of the z-score.
Since the normal distribution is symmetric, we can conclude that 61%/2 = 30.5% of the area is between the mean and that z-score.
Therefore, the area between the mean and that z-score is 30.5%.