Calculate Variance with Binomial Expansion IQ is normally distributed with mean 100 and standard deviation 15. An IQ of 130 or above is considered gifted, and 150 and above is considered genius. There are 350 million people living in the United States.
Let $X$ be the IQ of a person randomly selected from the population. Compute $\operatorname{Var}(X^2).$ 
So for this question binomial expansion must be used but I am not sure how to incorporate that into my solution.
 A: I think you want a Normal distribution for IQ, in which case we can write $X=\mu+\sigma Z$ with $\mu:=100,\,\sigma:=15,\,Z\sim N(0,\,1)$. Then$$\operatorname{Var}X^2=EX^4-(EX^2)^2=E(\mu^4+6\mu^2\sigma^2Z^2+\sigma^4Z^4)-(\mu^2+\sigma^2)^2=4\mu^2\sigma^2+2\sigma^4=9101250.$$
A: You have
\begin{align}
15^2 & = \operatorname{Var}(X) = \operatorname E\big((X-100)^2\big) \\
& = \operatorname E(X^2) - 200 \operatorname E(X) +100^2 \\
& = \operatorname E(X^2) - 100^2
\end{align}
Therefore
$$
\operatorname E(X^2 ) = 100^2 + 15^2 = 32\,500.
$$
\begin{align}
\operatorname{Var}(X^2) & = \operatorname E\big( (X^2 - 32\,500)^2\big) \\
& = \operatorname E(X^4) - 2\times32\,500\operatorname E(X^2) + 32\,500^2 \\
& = \operatorname E(X^4) - 2\times32\,500\times32\,500  + 32\,500^2 \\
& = \operatorname E(X^4) - 32\,500^2.
\end{align}
Now write $X = 100 + 15Z,$ so that $Z\sim\operatorname N(0,1).$ Then
$$
X^4 = 100^4 + (4\times 100^3 (15Z)) + (6\times 100^2 (15Z)^2) + (4\times100(15Z)^3) + (15Z)^4.
$$
The expected value of $Z$ and of $Z^3$ is $0,$ (that last because the distribution is symmetric about the origin). And that of $Z^2$ is $1.$ So you have
$$
\operatorname E(X^4) = 100^4 + (6\times100^2\times15^2) + 15^4\operatorname E(Z^4).
$$
That last expected value is $3.$
A: For mean $\mu$ and standard deviation $\sigma$, the normal distribution is
$$
\rho_{\mu,\sigma}(x)=\frac1{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\tag1
$$
where
$$
\int_{-\infty}^\infty x^{2k}\rho_{0,\sigma}(x)\,\mathrm{d}x=\sigma^{2k}(2k-1)!!\tag2
$$
and
$$
\int_{-\infty}^\infty x^{2k+1}\rho_{0,\sigma}(x)\,\mathrm{d}x=0\tag3
$$
Thus, for $\mu=100$ and $\sigma=15$,
$$
\begin{align}
\mathrm{E}\!\left[X^2\right]
&=\int_{-\infty}^\infty x^2\rho_{100,15}(x)\,\mathrm{d}x\\
&=\int_{-\infty}^\infty(100+x)^2\rho_{0,15}(x)\,\mathrm{d}x\\[3pt]
&=100^2+15^2\\[12pt]
&=10225\tag4
\end{align}
$$
and
$$
\begin{align}
\mathrm{E}\!\left[X^4\right]
&=\int_{\mathbb{R}}x^4\rho_{100,15}(x)\,\mathrm{d}x\\
&=\int_{\mathbb{R}}(100+x)^4\rho_{0,15}(x)\,\mathrm{d}x\\[3pt]
&=100^4+6\cdot100^2\cdot15^2+3\cdot15^4\\[12pt]
&=113651875\tag5
\end{align}
$$
therefore,
$$
\begin{align}
\mathrm{Var}\!\left[X^2\right]
&=\mathrm{E}\!\left[X^4\right]-\mathrm{E}\left[X^2\right]^2\\[3pt]
&=113651875-10225^2\\[6pt]
&=9101250\tag6
\end{align}
$$
