X is distributed so that: $$X=\begin{cases} \frac{ x-80}{400}& 80\le x \le 100\\ \frac{120-x}{400} & 100\le x\le 120\\ 0 &\text{otherwise} \end{cases} $$ $Var(X)= E((X)^2)-(E(X))^2$ and $E(X)=100$ by symmetry, so what has me confused is how to go about finding $E((X)^2)$ because of the 'split' in the p.d.f.
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Where $\mathbf{1}$ is the indicator function, $$f(x)=\frac{1}{400} (x-80) \mathbf{1}_{80\leq x\leq 100}+\frac{1}{400} (120-x) \mathbf{1}_{100\leq x\leq 120}$$ Moments: $$E(X^n)=\frac{2^{2 n+1} 5^n \left(2^{2 n+3}+2^{n+1} 3^{n+2}-5^{n+2}\right)}{(n+1) (n+2)}$$ Mean: $n=1$, $E(X)=100$
So the variance $V(X) =E(X^2)-E(X)^2=\frac{200}{3}$
The expectation is an integral, if you have a pdf which is 'splitted', you can take the sum of the integral such that $$E(X^2) = \int_{range1} x^2f(x)_{part1}dx + \int_{range2} x^2f(x)_{part2}dx $$
I hope that anwsers your question