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. There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013). 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

Concepts and reason

A $Z -$ score indicates the number of standard deviations an element is deviated from the mean. $Z -$ scores may be positive or negative. The positive value indicates that the score is above the mean value. The negative value indicates that the score is below the mean value. The normal probability values can be determined with the help of $Z -$ score.

Fundamentals

The formula for calculating the standardized $Z -$ score is,

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

Here,

$x$ = The value of the element.

$\mu$ = The mean of the population.

$\sigma$ = The standard deviation of the population.

Let $X$ be the IQ scores follows normal distribution with mean 100 and standard deviation 15.

Consider that the person must score in the upper 2% of the population on an IQ test for membership in Mensa.

Use the Excel formula; determine the area to the left of $Z = 0.98$ is,

$\begin{array}{c}\\{Z_{0.98}} = \left( {{\rm{ = NORMSINV}}\left( {{\rm{0}}{\rm{.98}}} \right)} \right)\,\\\\ = 2.05\\\end{array}$

Determine the score must a person have to qualify for membership in Mensa.

$\begin{array}{l}\\P\left( {X > k} \right) = 0.02\\\\ \Rightarrow P\left( {X \le k} \right) = 1 - 0.02\\\\ \Rightarrow P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{k - \mu }}{\sigma }} \right) = 0.98\\\\ \Rightarrow P\left( {\frac{{X - 100}}{{15}} \le \frac{{k - 100}}{{15}}} \right) = 0.98\\\\ \Rightarrow z = \frac{{X - 100}}{{15}} = 2.05\\\\ \Rightarrow X = 2.05 \times 15 + 100\\\\ \Rightarrow X = 130.75\\\end{array}$

Ans:

The score must a person have to qualify for membership in Mensa is 130.75.