Question

what score on the wechsler adult intelligence scale would an individualaged 20 to 34 have to earn to qualify for mensa membership?round your answer to 2 decimal places.

Answer 1

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.

Answer 2

To qualify for Mensa membership, an individual aged 20 to 34 would need to earn a score of 149.1 or higher on the Wechsler Adult Intelligence Scale.

Step-by-step explanation:

Mensa is an organization that requires its members to have an IQ score in the top 2% of the population. To determine the minimum IQ score needed to qualify for Mensa membership, we need to find the threshold value at the 98th percentile of the IQ distribution. Given that a typical adult has an average IQ score of 105 with a standard deviation of 20, we can use the z-score formula to calculate the threshold value:

Z = (X - µ) / σ

Where X is the score we want to find, µ is the mean, and σ is the standard deviation. Rearranging the formula to solve for X, we have:

X = Z * σ + µ

Since this question is asking for the score to two decimal places, we would need to use a standard normal distribution table or a calculator to find the z-score that corresponds to the 98th percentile, which is approximately 2.05. Plugging in the values, we get:

X = 2.05 * 20 + 105 = 149.1

Therefore, an individual aged 20 to 34 would need to earn a score of 149.1 or higher on the Wechsler Adult Intelligence Scale to qualify for Mensa membership.