Probability that XI have physically measured two random variables X and Y and determined their respective mean  and standard deviation. Both variables have a gaussian distribution. Now I wish to calculate the probability that a given value of X will be less than a given value of Y.
For example, if variable X has a mean of 100 and a standard deviation of 10 and variable Y has a mean of 120 and a standard deviation of 15, what is the probability of X being less than Y given Y=120?
 A: This problem cannot be answered unless we have additional information. But it can be answered if we assume that $X$ and $Y$ are independent.  You will have to judge whether the assumption of independence is reasonable in your situation. 
If $X$ and $Y$ are independent, then the difference $W=X-Y$ is Gaussian, mean $100-120$, and variance $\sigma^2$, where $\sigma^2=10^2+15^2$.  
So (under the assumption of independence of $X$ and $Y$), we know that $W=X-Y$ is Gaussian with mean $-20$ and standard deviation $\sqrt{325}$. Now it is a calculation to find $P(W\lt 0)$.
Remark: Let  $X$ and $Y$ be independent Gaussian, with mean $\mu_X$ and $\mu_Y$, and variance $\sigma_X^2$ and $\sigma_Y^2$ respectively. Let $W=aX+bY$ where $a$ and $b$ are constants. Then $W$ is Gaussian, mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2 +b^2\sigma_Y^2$. 
A: *

*The probability that [X<Y] conditionally on [Y=120] is the probability that [X<120] conditionally on [Y=120]. 

*Assume that X and Y are independent. (The question cannot be solved without a hypothesis on the dependence of X and Y.) Then, this is the (unconditional) probability that [X<120].

*If X is normal with mean 100 and standard deviation 10, then X=100+10Z with Z standard normal, hence [X<120]=[Z<2] and P(X<Y|Y=120)=P(Z<2)=97.7%.

