
Andy C. answered 02/18/18
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The Z score for TOM is (19-17)/5 = 2/5 = 0.4
So Tom scores 0.4 standard deviations above the mean
The Z score for JAN is (1175 - 1100)/155 = 75/155 = 0.48
SO Jan scores 0.48 standard deviations above the mean
Jan's score is slightly better