How do you find Pr$(X>Y)$ in bivariate normal distribution given that $X$ and $Y$ are independent?

Question

Specifically, if company A's sales of widgets for the upcoming year are normally distributed with mean 10,000 and standard deviation 2,000, while company B's sales of widgets for the upcoming year are normally distributed with mean 9,000 and standard deviation 2,000, what is the probability that B's sales amount exceeds A's sales amount in the upcoming year? I can set up the integrals by finding the region over which to integrate the inequality, but obviously that method doesn't work to find a value. I have no idea how to use zscores for this problem. I'm lost..

Answer 1

Assuming independence, the question can be viewed in the following way. Calculate

$$\mathbb{P}[Y>X]=\mathbb{P}[X-Y<0]$$

as per independence the distribution $Z=X-Y$ is very easy to be calculated

$$Z\sim N(1000;2000^2+2000^2)$$

(remember that $V(X-Y)=V(X)+V(Y)$)

Easy calculate the requested probability standardizing and using Z table

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$$\mathbb{P}[Z<0]=\Phi\Bigg[\frac{-1000}{\sqrt{2\cdot2000^2}}\Bigg]=\Phi(-0.3536)\approx 36.18\%$$