Can you derive the distribution of variation of another distribution? Let's say that if I aim at a position $a$ on a dart board, there is a normal distribution $P(X=x)$ describing the probability of the dart landing at position $x$ given aimpoint $a$.
Then, if I assume that the mean of the distribution is the aimpoint $a$ (and the std. dev. is given), I can solve for threshold $t$ such that the probability of $|a - x| < t$ is greater than 0.95 for example.
Since $|a - x| < t \rightarrow a + t > x > a - t$ I would integrate the pdf, evaluate the result on the interval, then solve for $t$ (I think).
I was just curious: Can you derive a new distribution of distances $|a - x|$ from the old distribution?
If so, would this be considered a distribution of variance?

Since you are working in two dimensions, let $\mathbf{a} = (a_1,a_2)$ denote the aim point and let $\mathbf{X} = (X_1,X_2)$ denote the point that the dart hits.  It sounds like you are using the model:
$$\mathbf{X} \sim \text{N}(\mathbf{a}, \sigma^2 \mathbf{I}).$$
Displacement from the aim point follows a scaled chi distribution with two degrees-of-freedom:
$$||\mathbf{X} - \mathbf{a}|| \sim \sigma \text{Chi}(\text{df} = 2).$$
