Q.Find the mean deviation from the mean of the following data, using the step deviation method.
Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
No. of Students: 6, 5, 8, 15, 7, 6, 3

First find the mean using the step-deviation method, then average the absolute deviation of each class midpoint from that mean.

Class midpoints ($x$) and frequencies ($f$):

$N=\sum f = 6+5+8+15+7+6+3=50$.

Step-deviation for the mean: take assumed mean $A=35$, class width $h=10$, $d=\dfrac{x-A}{h}$:

$d: -3,-2,-1,0,1,2,3.$

$\sum fd = 6(-3)+5(-2)+8(-1)+15(0)+7(1)+6(2)+3(3) = -18-10-8+0+7+12+9=-8.$

$\bar x = A+h\cdot\frac{\sum fd}{N} = 35+10\cdot\frac{-8}{50} = 35-1.6=33.4.$

Mean deviation about $\bar x=33.4$: compute $|x-\bar x|$ for each midpoint, weight by $f$, and average:

| $x$ | 5 | 15 | 25 | 35 | 45 | 55 | 65 | …