Difference in two Normal Distributions? I am unsure what I should do if I have two Normally Distributed variables with known parameters and I want to find the probability that one of these variables is greater than the other. Should I use the distribution of differences of the two normal variables, as is shown here? 
 A: Do this. This is not my own answer from from math.stackexchange, but it may be useful to duplicate it here: Probability of a point taken from a certain normal distribution will be greater than a point taken from another?. 
Quoting Shai Covo:
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,
$$
{\rm P}(X_1  > X_2 ) = {\rm P}(X_1  - X_2  > 0) = 1 - {\rm P}(X_1  - X_2  \le 0).
$$
Now, by independence, $X_1 - X_2$ is normally distributed with mean
$$
\mu := {\rm E}(X_1 - X_2) = \mu_1 - \mu_2
$$
and variance
$$
\sigma^2 := {\rm Var}(X_1 - X_2) = \sigma_1^2 + \sigma_2^2.
$$
Hence, 
$$
\frac{{X_1  - X_2  - \mu}}{{\sigma}} \sim {\rm N}(0,1),
$$
and so
$$
{\rm P}(X_1  - X_2  \le 0) = {\rm P}\bigg(\frac{{X_1  - X_2  - \mu }}{\sigma } \le \frac{{0 - \mu }}{\sigma }\bigg) = \Phi \Big(  \frac{-\mu }{\sigma }\Big),
$$
where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Thus,
$$
{\rm P}(X_1  > X_2 )  = 1 - {\rm P}(X_1  - X_2  \le 0) = 1 - \Phi \Big(  \frac{-\mu }{\sigma }\Big).
$$
