[!TLDR] Find $N=\sum f_i$, the weighted mean $\bar x=\frac1N\sum f_ix_i$, then use $\sigma^2=\frac1N\sum f_ix_i^2-\bar x^2$. [!ANSWER] Mean $=5.5$, …

Here $x_i:2,4,6,8,10$ and $f_i:3,5,7,4,1$, so $N=3+5+7+4+1=20$. Now $\sum f_ix_i = 2(3)+4(5)+6(7)+8(4)+10(1)=6+20+42+32+10=110$, giving $\bar x=\dfrac{110}{20}=5.5$. Using the shortcut identity, $\sum f_ix_i^2=2^2(3)+4^2(5)+6^2(7)+8^2(4)+10^2(1)=12+80+252+256+100=700$, so $\sigma^2=\dfrac{700}{20}-5.5^2=35-30.25=4.75$. (As a cross-check by the direct method: the deviations $x_i-\bar x$ are $-3.5,-1.5,0.5,2.5,4.5$, with squares $12.25,2.25,0.25,6.25,20.25$; weightin …

Find $N=\sum f_i$ and $\bar x=\frac1N\sum f_ix_i$; then use either the direct method $\sigma^2=\frac1N\sum f_i(x_i-\bar x)^2$ or the shortcu …

A common error is dividing by the number of distinct values (here, $5$) instead of the total frequency $N=\sum f_i$ ( …