[!TLDR] Use class midpoints as $x_i$; weighted mean then $\frac1N\sum f_i|x_i-\bar x|$. [!ANSWER] Mean …

The class midpoints are $x_i:5,15,25,35,45$ with frequencies $f_i:4,6,10,14,6$, so $N=4+6+10+14+6=40$. Now $\sum f_ix_i=5(4)+15(6)+25(10)+35(14)+45(6)=20+90+250+490+270=1120$, giving $\bar x=1120/40=28$. Absolute deviations $|x_i-28|$: $23,13,3,7,17$. Weighted: $f_i|x_i-\bar x|=4(23)+6(13)+10(3)+14(7) …

Find the class midpoints and the frequency-weighted mean, then weight each absolute deviation …

Arithmetic slips while forming and summing five $f_i|x_i-\bar x|$ products are the most common source of error; recomputing the mean by t …