Let $$M$$ denote the median of the following frequency distribution.

Then $$20M$$ is equal to :

Total frequency $N = 28$. Median position $= \frac{N}{2} = 14$.

The value 14 falls in the class 8 - 12.

• $l$ (lower limit) $= 8$

• $cf$ (cumulative frequency before) $= 12$

• $f$ (frequency of class) $= 10$

• $h$ (width) $= 4$

 Calculate $M$ and $20M$

$$M = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 8 + \left( \frac{14 - 12}{10} \right) \times 4 = 8 + \frac{2}{10} \times 4 = 8 + 0.8 = 10.4$$

• $20M = 20 \times 10.4 = \mathbf{208}$ (Option D)