[!TLDR] Assumed mean $A=25$, class width $h=10$; find $u_i=(x_i-A)/h$, then $\bar x=A+h\bar u$ and $\sigma^2=h^2\left[\frac{\sum f_iu_i^2}{N}-\bar u^2\right]$. [!ANSW …

Class midpoints: $x_i:5,15,25,35,45$; frequencies $f_i:4,8,20,12,6$; $N=4+8+20+12+6=50$. Choose assumed mean $A=25$ and class width $h=10$, so $u_i=\dfrac{x_i-A}{h}$ gives $u_i:-2,-1,0,1,2$. Then $f_iu_i:4(-2),8(-1),20(0),12(1),6(2)=-8,-8,0,12,12$, summing to $\sum f_iu_i=8$. So $\bar x=A+h\cdot\dfrac{\sum f_iu_i}{N}=25+10\cdot\dfrac{8}{50}=25+1.6=26.6$. Next, $f_iu_i^2:4(4),8(1),20(0),12(1),6(4)=16,8,0,12,24$, summing to $\sum f_iu_i^2=60$. So $\sigma^2=h^2\left[\dfrac{\sum f_iu_i^2}{N}-\left(\dfrac{\sum f_iu_i}{N}\right)^2\right]=100\left[\dfrac{60}{50}-\left(\dfrac{8}{50}\right)^2\right]=100[1.2-0.0256]=100(1.1744)=117.44$. (Direct-method cross-check: with $\bar x=26.6$, the deviations $x_i-26.6$ are $-21.6,-11.6,-1.6,8.4,18.4$, w …

Choose an assumed mean $A$ and class width $h$, compute $u_i=\frac{x_i-A}{h}$, form the columns $f_iu_i$ and $f_iu_i^2$, then recover $\bar x=A+h\bar u …

A common mistake is scaling the mean back correctly ($\bar x=A+h\bar u$) but forgetting to scale the variance by $h^2$ (not $h$) — the variance must be …