Random Variable - Probability and Statistic Suppose that $f(x)$ is a continuous and symmetric pdf, where symmetry is the property that $f(x) = f(-x)$ for all $x$. Show that $P(-a ≤ X ≤ a) = 2F(a) - 1$
Does anyone have any idea what this is even asking? I have honestly no idea what to do.  

We have
\begin{align*}
P(-a\leq X\leq a) &= \int_{-a}^af_X(x)\,dx\\
&=2\int_0^af_X(x) \,dx\tag 1\\
&=2\left[F_X(a)-F_X(0)\right]\\
&=2\left[F_X(a)-\frac{1}{2}\right]\tag 2\\
&=2F_X(a)-1
\end{align*}
where in $(1)$ is true by symmetry, and in $(2)$, since $f_X$ is symmetric, the median is at $0$; hence $F_X(0) = 1/2$.
