Intelligence quotas on two different tests are normally distributed. test a has a mean of 100 and a standard deviation of 16. test b has a mean of 100 and a standard deviation of 14. use z-scores to determine which person has the higher iq: an individual who scores 127 on test a or an individual who scores 133 on test b.

The individual who scores 133 on test b has the higher iq.

Step-by-step explanation:

Let X be the random variable that represents the score for an individual who takes test a and Y the random variable that represents the score for an individual who takes test b. z-scores are in both cases defined as (x-100)/16 and (y-100)/14. The z-score for an individual who scores 127 on test a is (127-100)/16 = 1.6875 and the z-score for an individual who scores 133 on test b is (133-100)/14 = 2.3571. Because 1.6875 < 2.3571 the individual who scores 133 on test b has the higher iq.