Consider the given data with frequency distribution cases x _ i & 3...

Arranging the given data in ascending order

we get,

$\begin{cases}{x}_{i} & 3 & 4 & 5 & 8 & 10 & 11 \\ {f}_{i} & 5 & 4 & 4 & 2 & 2 & 3\end{cases}$

Now finding the mean of the above data we get,

Mean $= \dfrac{3 \times 5 + 8 \times 2 + 11 \times 3 + 10 \times 2 + 5 \times 4 + 4 \times 4}{5 + 2 + 3 + 2 + 4 + 4}$

$= \dfrac{15 + 16 + 33 + 20 + 20 + 16}{20} = \dfrac{120}{20} = 6$

Now, median $= \dfrac{1}{2}\left({10}^{th} + {11}^{th}observation\right)$

$= \dfrac{1}{2}\left(5 + 5\right) = 5$

Hence, Mean deviation about mean will be,

$= \dfrac{3 \times 5 + 2 \times 4 + 1 \times 4 + 2 \times 2 + 4 \times 2 + 5 \times 3}{20}$

$= \dfrac{54}{20} = 2 . 7$

And Mean deviation about median

$= \dfrac{2 \times 5 + 1 \times 4 + 0 + 3 \times 2 + 5 \times 2 + 6 \times 3}{20}$

$= \dfrac{4 . 8}{20} = 2 . 4$

$P \to 3; Q \to 2; R \to 4; S \to 5$

$∴$ Option A is correct.