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I was wondering, what the average IQ at Mensa is. Mensa is a group of people with an IQ of at least 130. And the IQ is normally distribed with $\mu = 100$ and $\sigma = 15$.

My idea was this:

To get the mean of a function in interval $[a,b]$ I have to calculate

$$\bar{f}(x) = \frac{1}{b-a} \int_a^b f(x)\; dx$$

So the mean $p$ is

$$p = \lim_{b \to \infty} \frac{1}{b-130} \int_{130}^{b} \frac{1}{2 \pi} e^{-\frac{1}{2} \left(\frac{x-100}{15}\right)^2}dx$$

And then I just have to calculate, which IQ corresponds to this $p$.

Is my idea correct? How do I solve this integral and calculate the limit?

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There is a cutoff on the right too! Perhaps at $b = 150$. Not everybody is attracted by the idea of spending time with people who are proud of their IQ of 130. But this will not affect the average much. – TonyK Apr 11 '14 at 14:46
Calculation methods aside, what we are missing here is that not everybody who has an $IQ \ge 132$ will want to join Mensa. Since the IQ distribution is for the general population. So it is not be correct to use this distribution to estimate Mensa's average IQ. I would also assume that Mensa's IQ distribution should be proportionally a lot denser close to the limit 132, and drops down more quickly than the distribution for the general population as we move higher in the range. – DATAOCD Apr 11 '14 at 14:54
Technically, the average IQ of a population is, by definition, 100. – OrangeDog Apr 11 '14 at 16:51
Many people who could qualify for MENSA membership are either not snobby enough, or too snobby, to join. (I'm being a bit hard on them. Out in Middle Nowheresville, MENSA can be a valid way to find others who share your interests. Closer to a population center, there are better ways to do so and MENSA does degenerate to puffery.) – keshlam Apr 11 '14 at 16:56
There is an interesting math problem here independent of one's opinion of MENSA. – Michael Joyce Apr 11 '14 at 17:05

2 Answers 2

up vote 12 down vote accepted

The $x$ corresponding to the mean value of a density $f$ is not (in general) equal to the mean $x$.

Instead, you want to compute the mean of $x$, weighted by $f$, i.e., \begin{equation} \frac{\int_{130}^\infty xf(x)dx}{\int_{130}^\infty f(x)dx}. \end{equation}

This can also be interpreted probabilistically: we are looking for the expectation $E[X \mid X\geq 130]$, which is obtained by the integral \begin{equation} E[X \mid X\geq 130] = \int_{130}^{\infty} x p(x) dx, \end{equation} where $p$ is the conditional density of the probability distribution, obtained by \begin{equation} p(x) = \frac{f(x)}{\int_{130}^\infty f(t) dt},~x\geq 130 \end{equation}

Note that this answer is focused on the mathematical content, i.e., not questioning the assumptions. For various reasons I suspect this method does not capture the average IQ of Mensa (e.g., probability of joining Mensa might vary as a function of IQ even above the threshold).

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If the distribution is continuous uniform, then your idea is correct. But since IQ test scores typically follow what is known as a normal (or Gaussian) distribution, then you should use the left censored and shifted variable method. The expected value using this method can be obtained by $$ \text{E}[(X-d)_+]=\int_d^\infty (x-d)f(x)\ dx. $$ In your case $d=130$ and $X\sim\mathcal{N}(100,15)$. Actually, the minimum accepted IQ score to be a Mensan on the Stanford-Binet Intelligence Scales is $132$.


If you consider this case using the conditional distribution, then for a given value of $d$ with $\Pr[X>d]>0$, in your case $\Pr[X>d]=0.02$ because a Mensa member is a person who scores at or above the 98th percentile on certain standardized IQ or other approved intelligence tests, then for $Y=X-d$ given that $X>d$, its expected value is $$ \text{E}[Y|X>d]=\frac{\int_d^\infty (x-d)f(x)\ dx}{1-\Pr[X\le d]}. $$

Your problem is similar to the problem: Mensa (The High IQ Society) that I posted on

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Um, I think the question is about $E(X \mid X \geq d)$ (see my answer). Your integral counts all subjects below 130 as zeros, while actually they should be ignored when we are computing the average of those exceeding the threshold. Is there some connection of the left-censored shifted mean and the conditional mean that I'm not aware of (perhaps specific to Gaussians?)? – Juho Kokkala Apr 11 '14 at 14:44
There is no observed data nor unknown parameters in this question, and thus this is not an 'inference' problem at all. Using conditional distributions is not limited to Bayesians if the random variables involved are well defined. – Juho Kokkala Apr 11 '14 at 18:14
@Integrals Whether the 'solution' above is in agreement with how some book defines left censored random variables is entirely irrelevant as to whether the above (and/or said book) is correct or not. – wolfies Apr 11 '14 at 19:43
@Integrals Consider a simpler population with only 4 people, IQs $(115, 125, 135, 145)$ (keep the Mensa limit at 130 and assume everyone eligible joins). Is the average IQ of Mensa members in this population $E[(X-d)_+]=(0+0+5+15)/4 = 5$ or $E[X\mid X\geq d] = (135 + 145)/2 = 140$? – Juho Kokkala Apr 11 '14 at 20:09
I was not aware we were supposed to be "on somebody's side" or "defending users", naively I figured we were here to ask and to answer (if possible, accurately) mathematical questions. Users pointing mistakes in one's answers are actually rendering a service to the OP, not "attacking" them... – Did Apr 15 '14 at 7:57

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