What's wrong with this equation involving cumulative distribution functions? Suppose $X$ is a continuous random variable with finite mean $\mu = 1$ and standard deviation $\sigma$. Suppose also that it's pdf is symmetric about $x=\mu$. Show that $P(|X-\mu|\le 2\sigma) = 2P(X\le\mu + 2\sigma) - 1$.
My solution is as follows. Let $Z = X - \mu$ (which has pdf $f_Z(z) = f_X(z+\mu)$ and cdf $F_Z(z) = F_X(z+\mu)$). Then
$$
\begin{aligned}
P(|X-\mu|\le 2\sigma) & = P(|Z|\le2\sigma) \\
& = P(-2\sigma\le Z\le 2\sigma) \\
& = F_Z(2\sigma) - F_Z(-2\sigma) \\
& = 2F_Z(2\sigma)\quad\text{since $F_Z$ is symmetric about $x=0$} \\
& = 2F_X(2\sigma + \mu) = 2P(X\le\mu + 2\sigma)
\end{aligned}
$$
So somehow my answer is off by a $-1$ term. Where did I miss it? I've scanned over the steps, and I can't find anything off.

You missed it in going from the third line to the fourth. The symmetry means (also) that the areas (between the $x=0$ axis and the curve) are equal about $\mu$. Hence the $-1$.
