[!TLDR] Class midpoints as $x_i$; find the weighted mean, then both $\frac1N\sum f_i|x_i-\bar x|$ and $\sqrt{\frac1N\sum f_i(x_i-\bar x)^2}$. [!ANSWER] Mean $=25$, Mean Deviation abou …

Class midpoints: $x_i:5,15,25,35,45$; frequencies $f_i:4,8,16,8,4$; $N=4+8+16+8+4=40$. $\sum f_ix_i=5(4)+15(8)+25(16)+35(8)+45(4)=20+120+400+280+180=1000$, so $\bar x=1000/40=25$. Mean deviation: absolute deviations $|x_i-25|$ are $20,10,0,10,20$; weighted, $f_i|x_i-\bar x|=4(20)+8(10)+16(0)+8(10)+4(20)=80+80+0+80+80=320$; so $\text{M.D.}(\bar x)=320/40=8$. Variance and SD: squared deviations $(x_i-25)^2$ are $400,100,0,100,400$; weighted, $f_i(x_i-\bar x)^2=4(400)+8(100)+16(0)+8(100)+4(400)=1600+800+0+800+1600=4800$; so $\sigma^2=4800/40=120$, and $\sigma=\sqrt{120}=2\sqrt{30}\approx10.95$. Comment: the mean deviation ($8$) is generally smaller than the standard deviation ($\approx10.95$) because it averages the raw (linear) distances from the mean, while the standard deviation squares each distance first — giving disproportionately more weight to the marks farthest …

Find the class midpoints and the frequency-weighted mean once, then compute the mean deviation (frequency-weighted average of absolute deviations) and the variance/standard deviation (frequen …

Computing the mean twice by two slightly different roundings (once for the mean-deviation part, once for the variance part) can introduce small inconsistencies …