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. there are 110,000 mensa members in 100 countries throughout the world (mensa international website, january 8, 2013). 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 (0 decimals)?

Answer 1

An IQ of 132 on a normal distribution would be the 98th percentile. 

Answer 2

131 (approx)

Step-by-step explanation:

Given:

Mean = 100

The critical z score = 2.06 (Using z value calculator)

Standard deviation = 15

Alpha value = 2% = 2/100 = 0.02

Right tailed area = 0.98

Computation:

Qualifying score = Mean + (critical z score × standard deviation)

Qualifying score = 100 + (2.06 × 15)

Qualifying score = 100 + (2.06 × 15)

Qualifying score = 100 + 30.9

= 130.9

= 131 (approx)