In two-dimensional ($\mathbb R^2$) real space, if I have two vectors $x \sim N(3,1)$ independent of another vector $y \sim N(0,1)$, then what's the probability using R that the distance from origin to $(x,y) > 3$ units?
I am a beginning graduate stats student. This is me practicing, this $Q$. isn't graded homework. I am stuck on how to make $(x,y)$ and then how to find the probability.
My questions / work: Let $z = (0,0)$.
- To find the distance between $z$ and $(x,y)$, do I find $\|x\|$ and $\|y\|$ first, THEN calculate $\|z - (x,y)\|$? $\|x\| = \sqrt{2^2+1^2} = \sqrt{5}$ and $\|y\|=\sqrt{0^2+1^2} = 1$, then distance between origin and $(x,y)$ is $\sqrt{5+1} = \sqrt{6}$.
OR
Do I subtract the vectors first, since the difference of 2 vectors is the difference of their elements, such that $\|x-y\| = (2,1) - (0,1) = (2,0)$. So $(x,y)$ = $\sqrt{2^2 + 0^2} = 2$.
- Whatever $(x,y)$ is, how do I find the probability that the distance is greater than 3? Is it normcdf? Something else?