Passage: The frequency distribution of marks obtained by 100...

Total number of students is $100$.

$10 + 20 + f + 40 = 100 \Rightarrow f = 30$

The mid-points $x_i$ of the classes are $5, 15, 25, 35$.

The arithmetic mean $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$

$\bar{x} = \dfrac{10 \times 5 + 20 \times 15 + 30 \times 25 + 40 \times 35}{100}$

$\bar{x} = \dfrac{50 + 300 + 750 + 1400}{100} = \dfrac{2500}{100} = 25$

The absolute deviations $|x_i - \bar{x}|$ are $|5 - 25| = 20$, $|15 - 25| = 10$, $|25 - 25| = 0$, and $|35 - 25| = 10$.

Mean deviation about the mean $= \dfrac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$

$= \dfrac{10 \times 20 + 20 \times 10 + 30 \times 0 + 40 \times 10}{100}$

$= \dfrac{200 + 200 + 0 + 400}{100} = \dfrac{800}{100} = 8$

Answer: $8$