Question

A person must score in the upper 2% of the population on an 1Q test to qualify for membership in Mensa, the international high IQ society. If IQ scores are. normaly distributed with a mean of 100 and a standard deviation of 15 . What scoce must a person have to quaiify for Mensa? If required, round your answers to nearest whole number.

Answer 1

To qualify for Mensa, a person needs an IQ score in the upper 2% of the population. Given a mean IQ of 100 and a standard deviation of 15, a z-score of around 2.05 is required. The corresponding IQ score, when rounded to the nearest whole number, is 131.

Step-by-step explanation:

To qualify for membership in Mensa, a person must score in the upper 2% of the IQ distribution. Given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, we can use a z-table to find the z-score corresponding to the 98th percentile since the top 2% would be everything above the 98th percentile.

To find the z-score corresponding to the 98th percentile, we look up the value that has approximately 0.98 to its left on the z-table. This value is around 2.05.

Using the z-score formula, z = (X - mean) / standard deviation, we can solve for X, which represents the IQ score for Mensa qualification:
2.05 = (X - 100) / 15
X = 2.05 * 15 + 100
X ≈ 130.75

Since IQ scores are typically rounded to the nearest whole number, a person would need an IQ score of approximately 131 to qualify for Mensa.