Final answer:
To qualify for Mensa, an individual must have an IQ in the top 2 percent of the population, which translates to an IQ score of 130 or above on the Wechsler Adult Intelligence Scale, using the average score of 100 and standard deviation of 15.
Step-by-step explanation:
Mensa membership
requires an individual to have an IQ in the top 2 percent of the population. On most IQ tests, the average IQ score is set at 100 with a standard deviation of 15 points. Therefore, to be in the top 2 percent, an individual would have to score significantly above the average.
Given a normal distribution, the top 2 percent of scores would roughly start at or above a
z-score
of 2, which equates to an IQ score of about 130 when using the average IQ score and standard deviation (100 + 2*15 = 130).
However, it is critical to confirm with Mensa the specific requirements and the exact version of the IQ test they accept, as scoring may vary between different tests.
Additionally, Mensa may also accept scores from other approved intelligence tests as part of their membership criteria.