Question

in order for you to become a member of Mensa, a worldwide organization with approximately 100,000 members, your IQ must be in the top 2%. The word mensa is Latin for “table,”and was chosen to denote a group or round table of people with equal ability. In 1996, Mensa, which was founded by two British barristers, celebrated its 50th birthday. American Mensa Ltd., which was founded in 1960 has almost 50,000 members. Marilyn vos Savant, who is reputed to have the highest registered IQ, is a member. Assuming that IQ scores have an approximately normal distribution with a mean and standard deviation of 100 and 15, respectively, answer the following question

Answer 1

a. What IQ must one have in order to become a member of Mensa?

Let $X=$ IQ score: $X\sim N(\mu,\sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

Given that $\mu=100, \sigma=15$

$$P(X\geq X^*)=0.02$$

$$0.02=1-P(X<X^*)=1-P(Z<{X^*-100 \over15})$$

$${X^*-100 \over15}\approx2.0537$$

$$X^*\approx130.8$$

One must have IQ at least 131 in order to become a member of Mensa.

b. What percent of all Americans have an IQ of at least 145?

$$P(X\geq 145)=1-P(X<145)=1-P(Z<{145-100 \over15})=$$

$$=1-P(Z<3)\approx0.00135$$

Only $0.135\%$ of all Americans have an IQ of at least 145.

c. What percent of all members of Mensa have an IQ of at least 145?

We determine that 2% of all Americans can be the members of Mensa.

We determine that only $0.135\%$ of all Americans have an IQ of at least 145.

$$\begin{array}{cc} 2 & 100\% \\ 0.135 & y\% \end{array}$$ $${0.135 \over 2}\times100\%=6.75\%$$

Only $6.75\%$ of all members of Mensa have an IQ of at least 145.

d. If Mensa decided to become more exclusive, and accepted only the top 1% instead of the top 2% as members, what IQ would one need in order to become a member of Mensa?

$$P(X\geq X^*)=0.01$$

$$0.01=1-P(X<X^*)=1-P(Z<{X^*-100 \over15})$$

$${X^*-100 \over15}\approx2.32635$$

$$X^*\approx134.9$$

Now one must have IQ at least 135 in order to become a member of Mensa.