Discussion
A test of Abstract Thinking had the norms in terms of normalized T scores (Mean = 50, SD = 10). If the subject received a T score of 60, what would be his percentile rank ?
Answer: D
A T score has a mean of 50 and a standard deviation of 10, so a T score of 60 is one standard deviation above the mean.\[ z = \frac{60 - 50}{10} = 1 \]In a normal distribution about 84 percent of the scores lie below z = +1, so the percentile rank is about 84. The value 16 is the rank for z = -1, and 60 and 66 do not correspond to a z score of 1. A quick check: in the normal curve about 68 percent of scores fall within one standard deviation of the mean, which leaves about 16 percent above +1 SD, so the person is at about the 84th percentile.
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