Question

Suppose a person must score in the upper 2.5% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributed with a mean of 80 and a standard deviation of 22, what score must a person get to qualify for Mensa?

Answer 1

Given that,

mean = = 80

standard deviation = =22

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 standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.96*22+80

x= 123.12

x=123