# Comparing two normal distributions

Given a normal distribution $$X$$~$$N(60,9^2)$$ with a random variable $$A$$ and a normal distribution $$Y$$~$$N(50,7^2)$$ with a random variable $$B$$, how do I go about finding the probability $$P(B>A)$$?

(Given that A and B are independent events).

If ther are dependent you cannot do this. If they are independent then $$C=B-A$$ has normal distribution with mean $$50-60$$ and variance $$9^{2}+7^{2}$$. You can compute $$P(C>0)$$ by integrating the density function from $$0$$ to $$\infty$$.

• So we get to something like this? $f(x) = \begin{cases} 0, & \text{if$A≥x≥B$} \\ c, & \text{if$A≤x≤B$} \end{cases}$ – sdds Apr 17 at 8:01
• The answer is $\int_0^{\infty} \frac 1 {\sqrt {260 \pi}} e^{-(x+10)^{2}/{260}}dx$. – Kavi Rama Murthy Apr 17 at 8:04

Hint: If $$A$$ and $$B$$ are independent, you have that $$A-B \sim \mathcal{N}(10,7^2+9^2)$$ and $$P(B>A) = P(A-B<0)$$.

If they are dependent and are jointly normally distributed, you need to use the joint distribution.