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. 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

Answer 1

Answer:

A person must score at least 130.825 to qualify for Mensa

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 100, \sigma = 15$

Top 2%

Scores of x and higher, in which X is found when Z has a pvalue of 0.98. So it is X when Z = 2.055.

$Z = \frac{X - \mu}{\sigma}$

$2.055 = \frac{X - 100}{15}$

$X - 100 = 2.055*15$

$X = 130.825$

A person must score at least 130.825 to qualify for Mensa