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? [15 pts.]

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Answer 1

Solution:

Given: IQ scores are normally distributed with a mean of 80 and a standard deviation of 22.

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.

We have to find score so that a person get to qualify for Mensa.

That is find x value such that:

P( X > x) = 2.5%

P( X > x) = 0.0250

thus find z value such that:

P( Z> z) = 0.0250

that is:

P( Z < z) = 1 - P( Z > z)

P( Z < z) = 1 - 0.0250

P( Z < z) = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : z = 1.96

Now use following formula to find x value:

A person to get to qualify for Mensa must score IQ = 123.12