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?

• Please feel free to correct the assumptions I made leading up to the question. I'm almost positive I made some fatal error in my basic understanding. Commented Aug 4, 2022 at 20:28
• Yes. When the two components of $X$ in Cartesian coordinates are independent (and with the same variance), this is a multiple of the Chi Distribution with 2 d.f.
– whuber
Commented Aug 4, 2022 at 22:57

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:
$$\mathbf{X} \sim \text{N}(\mathbf{a}, \sigma^2 \mathbf{I}).$$
$$||\mathbf{X} - \mathbf{a}|| \sim \sigma \text{Chi}(\text{df} = 2).$$