Average IQ of Mensa 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?
 A: 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 $132$.
ADDENDUM :
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 Brilliant.org.
A: 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).  
