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?

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smithbrandonr860 smithbrandonr860

10-03-2018
Mathematics

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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?

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Blitztiger Blitztiger · Answer 1 · 2018-03-20T02:56:30+01:00

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.