97.5% must be below
or in your "tables" find the probability closest to
0.975
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:
http://davidmlane.com/hyperstat/z_table.html
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
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
2 answers
617
1 answer