Question

Scores on Modern Language Aptitude Test (MLAT) are normally distributed with a mean of 68 and a variance of 6. What is the lowest score that will place a student in the top 1.50% of the distribution? Round your final answer to two decimal places.

Answer 1

Given that,

mean = = 68

standard deviation = =6=2.4495

Using standard normal table,

P(Z > z) =1.50 %

= 1 - P(Z < z) = 0.015

= P(Z < z ) = 1 - 0.015

= P(Z < z ) = 0.985

= P(Z < 2.17) = 0.985

z = 2.17 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 2.17 *2.4495+68

x= 73.3154

x=73.32