Question 541054
on a standarized test with a mean of 500 and and a standard deviation of 100 what would be the cutoff score for selecting the top 10% of the applicants 
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Find the z-value with a left-tail of 90%
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invNOrm(0.90) = 1.2816
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cutoff = z*s + u
cutoff = 1.281556*100 + 500 = 628.1556
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Cheers,
Stan H.
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