# Bivariate Normal Probability

Assume we have a large data set of PSAT and SAT scores with bivariate normal distribution with $\rho = 0.6$. The mean and SD of the PSAT scores are (respectively) $1200$ and $100$. The mean and SD of the SAT scores are (respectively) $1300$ and $90$. What percentage of students got at least 50 points more on the SAT than on the PSAT.

Let $X$ denote PSAT scores and $Y$ denote SAT scores. The desired probability is $$P(90Y + 1300>100X +1200+50).$$

I don't understand why it shouldn't be $P(Y>X +50)$, however.

Any help would be appreciated. Thanks.

If $X$ and $Y$ represent a randomly selected PSAT and SAT score respectively, then you are correct, it should be $P(Y \geq X+50)$. However, there are standard normal random variables $Z_X \stackrel{D}{=} N(0,1)$ and $Z_Y \stackrel{D}{=} N(0,1)$ such that $$(X,Y) \stackrel{D}{=} (\sigma_X Z_X + \mu_X,\, \sigma_Y Z_Y + \mu_Y) = (100Z_X + 1200, 90Z_Y + 1300).$$ (Here $\stackrel{D}{=}$ means "equal in distribution"). It seems the solution $P(90Y+1300 > 100X+1200+50)$ is conflating the variables $X$ and $Y$ with $Z_X$ and $Z_Y$.