[!TLDR] Weighted mean, then $\sigma^2=\frac1N\sum f_ix_i^2-\bar x^2$. [!ANSWER] Mean $=12$, Variance $=2 …

Here $x_i:4,8,12,16,20$ and $f_i:3,6,6,6,3$, so $N=3+6+6+6+3=24$. Now $\sum f_ix_i=4(3)+8(6)+12(6)+16(6)+20(3)=12+48+72+96+60=288$, giving $\bar x=288/24=12$. Deviations $x_i-12$: $-8,-4,0,4,8$, squares $64,16,0,16,64$. Weighted: $f_i(x_i-\bar x)^2=3(64)+6(16)+6(0)+6(16)+3(64)=192+96+0+96+192=576$. Therefore $\sigma …

Find $N=\sum f_i$ and $\bar x=\frac1N\sum f_ix_i$; form $f_i(x_i-\bar x)^2$ for each distinct value, s …

Forgetting to weight the squared deviations by frequency (i.e. treating repeated values as a single observation) unders …