The arithmetic mean of a set of $50$ numbers is $38$. If two numbers of the set, namely $45$ and $55$, are discarded, 
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$
If the [arithmetic mean](https://artofproblemsolving.com/wiki/index.php/Arithmetic_mean) of a set of $50$ numbers is $38$, then the sum of the $50$ numbers equals $1900$.  Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$.  Therefore, the new mean is $37.5$, which is answer choice $\boxed{\textbf{(B)}}$.