Question

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. The top 2.5 percent of the applicants would have a score of at least (choose the nearest integer)

Answer 1

Solution:-

Given that,

mean = = 460

standard deviation = = 80

Using standard normal table,

P(Z > z) = 2.5%

= 1 - P(Z < z) = 0.025

= P(Z < z) = 1 - 0.025

= P(Z < z ) = 0.975

= P(Z < 1.96 ) = 0.975  

z = 1.96

Using z-score formula,

x = z * +

x = 1.96 * 80 + 460

x = 616.8

score = 617