For a grouped distribution a student takes $a=45$ and $h=10$ and finds the products $f_i{u}_i$ to be -16, -9, 0, 12 and 13. What is the mean?

$\sum f_i{u}_i=-16-9+0+12+13=0$, so the correction $h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)=10\times 0=0$ and the mean is exactly $a=45$. Notice that $\sum f_i$ is not even needed once $\sum f_i{u}_i=0$. "0" reports $\sum f_i{u}_i$ instead of the mean; "55" and "35" add or subtract $h$ out of habit, even though the correction is zero.