If a test-taker earns a z-score of +2 on a test, approximately what
percentage of other test-takers obtained higher scores, assuming the
distribution of test scores is normal?

Would the percentage be 2%, 14%, 16% or 25%?

A z score of +2 implies that the test-taker scored 2 standard
deviations better than the mean (see
http://www.camcode.com/help/distributions/normal_distribution.htm for
a more thorough description of this).  You can look up the percentage
on each side of any z value in an appropriate statistical table - the
lower table ("Far right probabilities") at
http://www.math.unb.ca/~knight/utility/NormTble.htm is helpful for
solving this problem.  It tells you that 0.02275 = 2,275 % of the
test-takers scored better than this particular test-taker.

The question asks for the approximate percentage - the best answer would be 2 %.

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