X = 55 + 1.645 * 10 X = 71.45
Ex am ple 6
The Jones twins are highly competitive. The first twin is in a group for mathematics which has been given a test for which the results are distributed normally with a mean of 65 and a standard deviation of 5. The first twin’s score is 73. The second twin is in a different group which does a totally different type of mathematics test and whose results are distributed normally with a mean of 35 and a standard deviation of 11. The second twin scores 43. Which twin has done relatively better in their test? First twin: Z-score =
73 − 65 = 1.6 standard deviations 5
Second twin: Z-score =
43 − 35 = 0.73 standard deviations 11
Using tables P (Z ≥ 1.6 ) = 1 − .9452 = .0548 = .055 = 5.5% which means that this twin is in the top 5.5% of the class. Using tables P (Z ≥ 0.73 ) = 1 − .7673 = .2327 = .234 = 23.4% which means that this twin is in the top 23.4% of the class Thus the first twin has performed relatively better to the peer group than the second twin.
Figure 2.18: The shaded part of the curve shows the probability of first twin with a score of 73.
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