A large number of applicants for admission to graduate study in business are given an aptitude test with a resulting mean score of 460 with a standard deviation of 80. The top 2.5 percent of the applicants would have a score of at what?

A. 606
B. 617
C. 600
D. 646

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  1. 97.5% must be below
    or in your "tables" find the probability closest to
    You should have a z-score of 4.9625 depending on the accuracy of your tables

    then (x - 460)/80 = 1.9625
    x-460 = 157
    x = 157+460 = 617

    OR, use this wonderful webpage:

    Click on the second option: value under an area

    enter 0.975 for "Area" --- (from 100% - 2.5% = 97.5%)
    enter 460 for mean
    enter 80 for SD
    click on "below" and recalculate to get

    616.832 , round off to 617

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  2. 617

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