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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Aug 4, 2022 at 20:28
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    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Aug 4, 2022 at 22:57

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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:

$$||\mathbf{X} - \mathbf{a}|| \sim \sigma \text{Chi}(\text{df} = 2).$$

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  • $\begingroup$ Thank you for entertaining my curiosity! $\endgroup$ Commented Aug 5, 2022 at 14:36

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