Estimate Percentile Rank from a Bivariate Normal with know correlation coefficient The problems goes:
Suppose the scores of an exam follow Normal distribution and the correlation coefficient $\rho=0.8$ for exam1 and exam2. If in exam1 a student's score percentage rank (PR) is 90% (that is, his/her grade is above 90% of students.), estimate his/her PR in exam2. 
This problem seems so interesting so I really want to know if anyone can provide some insight to solve it.
 A: Possible approach / not a full answer
I interpret the OP as saying the exam scores $X_1, X_2$ are jointly normal with correlation $\rho = 0.8$.  Then scoring PR $90\%$ in exam $1$ is equivalent to saying
$$\Phi(X_1) = 0.9\,\,\,\,\,\,\,\,\,\,\,\text{ i.e. }\,\,\,\,\,\,\,\,\,\,\, X_1 = \Phi^{-1}(0.9)$$
where $\Phi(x)$ is the CDF of a standard normal variable $N(0,1)$.  As is well known $\Phi$ has no closed-form, but the numerical value of $\Phi^{-1}(0.9) \approx 1.28$ can be looked up easily.  Then, conditioned on $X_1 = \Phi^{-1}(0.9)$, the distribution of $X_2$ is now a shifted normal r.v.; according to wikipedia it is:
$$X_2 \sim N(\rho\, \Phi^{-1}(0.9), 1 - \rho^2)$$
And the PR of exam $2$ would be the random variable $\Phi(X_2)$.
Now you are asking to estimate this second PR.  There are all kinds of estimators, but in this case the mathematically simplest estimator might be just to look at $\Phi(E[X_2]) = \Phi(\rho\, \Phi^{-1}(0.9)) \approx 84.7\%$.
However I must add that there are many other estimators, e.g. the unbiased $E[\Phi(X_2)]$ (not the same as $\Phi(E[X_2])$ at all since $\Phi$ is non-linear and we're near one of its tails), the max likelihood estimator, etc.
