SOLVED - Question involving the normal distribution I have some variable that's normally distributed with mean A and standard deviation B (variance B^2). I make two (independent) observations. What's the probability that these observations will deviate from each other by at least a standard deviation B?

The difference is normally distributed as well, with $\mu_{x_1-x_2}=\mu_{x_1}-\mu_{x_2}=\mu_y=0$ and $\sigma^2_{x_1-x_2}=\sigma_{x_1}^2+\sigma_{x_2}^2=\sigma_y^2=2B^2$.

Therefore $X_1-X_2=Y\sim \mathcal N(0, 2B^2)$. Now we calculate the probability that the deviation of the absolute value of $Y$ is more than $B$. For this purpose we can standardize Y:

$Z=\frac{Y-\mu_y}{\sigma_y}=\frac{Y-0}{\sqrt 2\cdot B}$

Since $Z$ is symmetric around $0$ we have $P(|Z|>B)=2-2\Phi(B)$.

Thus $P(|Z|>B)=2-2\Phi\left(\frac{B}{\sqrt 2\cdot B} \right)= 2-2\Phi\left(\frac{1}{\sqrt 2} \right)=2-2\Phi\left(0.707107\right)$

Using a calculator for thes (standard) normal distribution we get $\Phi\left(0.707107\right)=0.76$

$P(|Z|>B)=2-2\cdot 0.76=2-1.52=0.48$