# what is the formula for P(|a-b| > c) in a normal distribution?

Given two random samples a and b of a normal distribution, how can I calculate the probability that their difference |a-b| is bigger (or lesser) than c?

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Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $X_1 - X_2 \sim {\rm N}(\mu_1 - \mu_2 , \sigma_1^2 + \sigma_2^2)$. Set $Y = X_1 - X_2$, $\mu = \mu_1 - \mu_2$, and $\sigma^2 = \sigma_1^2 + \sigma_2^2$. Then, for any $c > 0$, $$P(|Y| \le c) = P( - c \le Y \le c) = P\bigg(\frac{{ - c - \mu }}{{\sigma }} \le \frac{{Y - \mu }}{{\sigma }} \le \frac{{c - \mu }}{{\sigma }}\bigg).$$ Noting that $(Y - \mu)/\sigma \sim {\rm N}(0,1)$, we are actually done.