Question

A person must score in the upper 2% 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 100 and a standard deviation of 15.

what score must a person have to qualify for Mensa? If required, round your answers to nearest whole number.

Answer 1

Given,

= 100 , = 15

We convert this to standard normal as

P(X < x) = P( Z < (X - ) / )

We have to calculate x such that

P(X > x) = 0.02

P(X < x) = 1 - 0.02

P(X < x) = 0.98

P( Z < (X - ) / ) = 0.98

From Z table, z-score for the probability of 0.98 is 2.0537

(X - ) / = 2.0537

( x - 100) / 15 = 2.0537

Solve for x

x = 131