In the problem we have that $A \sim N(7, 11/60)$ and $B \sim N(7.3, 7/20)$ and the question is what is the probability that $A$ gives a higher value that $B.$ Since the textbook we have for the course doesn't include information about this type of question, I figured that it might just required to do manipulation with the normal law formula. So I came up with this :
\begin{equation*} \int_{-\infty}^{\infty}\Big(\frac{1}{\frac{11}{60}\sqrt{2\pi}} \exp\{-0.5 \frac{t-7}{\frac{11}{60}}^2 \}\Big) \Big(\int_{-\infty}^{t}\frac{1}{\frac{7}{20}\sqrt{2\pi}}\exp\{ -0.5\frac{x-7.3}{\frac{7}{20}}^2 \}dx\Big) dt \end{equation*}
What I thought could be a way to solve this was by multiplying the probability to get a lower value of $B$ (right part of the integral) by the probability of $A$ (left part of the integral) for each value of $A.$ In theory this might work, but I'm unable to actually calculate this with either my calculator or Wolfram Alpha. Is there something I'm overlooking in this problem ?