A distribution is supplied as CLASS MARKS rather than as intervals: marks 8, 14, 20, 26, 32 with frequencies 3, 8, 14, 10, 5. Find the mean.

The class marks may be used directly — there is no need to rebuild the intervals (they would be 5-11, 11-17, ... since $h=14-8=6).\sum f_i{x}_i=24+112+280+260+160=836$ and $\sum f_i=40$, so the mean is $\frac{836}{40}=20.9$. "20" assumes the mean must be the middle class mark because the table looks nearly symmetric; "17.9" "corrects" every class mark down to its lower boundary $x_i-3$; "167.2" divides by 5, the number of classes.