Final answer:
To join MENSA, an individual scored within the top 2% of the Wechsler Adult Intelligence Scale must achieve a score of approximately 161.3, which is calculated using the mean IQ of 110, a standard deviation of 25, and the z-score for the 98th percentile. Therefore correct option is B
Step-by-step explanation:
To qualify for MENSA membership, an individual must score within the top 2% of IQ scores. Given that the IQ scores are normally distributed with a mean (μ) of 110 and a standard deviation (σ) of 25 for the 20- to 34-year-old age group on the Wechsler Adult Intelligence Scale, we need to find the score that corresponds to the 98th percentile.
To find this score, we typically use a z-table or a calculator that can compute the inverse cumulative distribution function for the normal distribution. The z-score that corresponds with the top 2% of the distribution is approximately 2.05. We can then use the formula for converting a z-score to an actual score on the IQ test which is X = μ + zσ. Plugging in the values, we get:
X = 110 + (2.05)(25)
X = 110 + 51.25
X = 161.25
Therefore, an individual would need to earn a score of approximately 161.3 to qualify for MENSA membership, which corresponds to option B.