1-the graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard diviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that test is normally distributed?

Answer 1

Convert x=score to z, the standard variable for a normally distributed experiment.

z=(x-μ)/σ
=(x-500)/100

From the normal distribution table,
http://www.math.unb.ca/~knight/utility/NormTble.htm
the z value for 90% cut-off is 1.282, so we have
1.282=(x-500)/100
x=100*1.282+500
=628