Your friend scores a 22.5 on a normally distributed new logic test with a mean of 20 and a standard deviation of 2.5, and you're confident that 13.59% of people are more logical than your friend but not as logical as you. if you're really as smart as you think you are, what would you score on the test?

The z-score for the point x = 22.5 is: z = (x - m) / s = (22.5 - 20) / 2.5 = 1. The probability that z > 1, determined from z-tables, is p (z > 1) = 0.1587. If 13.59% of people score higher than the friend, but not higher than you, this means that only 0.1587 - 0.1359 = 0.0228 of people scored higher than you. Checking the z-table for the z-score for which p (z > z_cutoff) = 0.0228, z_cutoff = 2. Therefore this z_cutoff of 2 corresponds to x = m + zs = 20 + 2(2.5) = 25, which would be your score on the test.

1.5 pounds because you put 9over 6 and then divide by 6