[!TLDR] Weighted mean $\bar x=\frac1N\sum f_ix_i$, then $\frac1N\sum f_i|x_i-\bar x|$. [!ANSWER] Mean $=14$, Mean Deviation about the mean $=4.8$.

Here $x_i:5,10,15,20,25$ and $f_i:4,6,8,5,2$, so $N=4+6+8+5+2=25$. Now $\sum f_ix_i=5(4)+10(6)+15(8)+20(5)+25(2)=20+60+120+100+50=350$, giving $\bar x=350/25=14$. Absolute deviations $|x_i-14|$: $9,4,1,6,11$. Weighted: $f_i|x_i-\bar x|=4(9)+6(4)+8(1)+5(6)+2(11)=36+24+8+30+22=120$. Therefore $\text{M.D.}(\bar x)=120/25=4.8$. [!ANSWER] Mean $=14$; Mean Deviation about the mean $=4.8$.

Find $N=\sum f_i$ and $\bar x=\frac1N\sum f_ix_i$; form $|x_i-\bar x|$ for each distinct value, weight by $f_i$, sum, and divide by $N$.

Forgetting to weight each absolute deviation by its frequency before summing (i.e. treating every $x_i$ as occurring only once) is the most common error for discrete frequency data.