# Probability that value from distribution A will be greater than value from distribution B

I have 2 normal probability distributions, A and B. If random values were to be chosen from each, what's the probability that the value from A will be greater than the value from B? How can I calculate that knowing the means and std of each distribution, or conceptually from the areas under the curves:

Assuming independence (because in the case of dependence, you should provide the joint PDF), linear combinations of normal RVs are still normal. So, $$A-B$$ will be a normal RV. Then, we have $$P(A>B)=P(A-B>0)$$. In order to calculate this probability we need to know the mean and the variance of $$C=A-B$$. And, it is: $$\mu_c=E[C]=E[A-B]=\mu_a-\mu_b$$ $$\sigma_C^2=\operatorname{var}(C)=\operatorname{var}(A-B)=\operatorname{var}(A)+\operatorname{var}(B)=\sigma_A^2+\sigma_B^2$$
Then, you'll have the following probability: $$P(C>0)=P\left(\frac{C-\mu_c}{\sigma_c}>\frac{0-\mu_c}{\sigma_c}\right)=P\left(Z>\frac{0-\mu_c}{\sigma_c}\right)$$ Where you can use a Z-table to figure out the approximate probability.
• There are no simple integrals to calculate CDF of normals, which is why we have numeric Z tables. Common statistical tests (like t-test, f-test etc.) are data-driven and for deciding if the two samples differ by mean/std or not, they don't calculate $P(A>B)$. – gunes Apr 26 '19 at 12:07