Consider the given data with frequency distribution

$$x_i$$	3	8	11	10	5	4
$$f_i$$	5	2	3	2	4	4

Match each entry in List-I to the correct entries in List-II.

List-I		List-II
(P)	The mean of the above data is	(1)	2.5
(Q)	The median of the above data is	(2)	5
(R)	The mean deviation about the mean of the above data is	(3)	6
(S)	The mean deviation about the median of the above data is	(4)	2.7
		(5)	2.4

The correct option is:

The six distinct observations are $$x_i=3,\,4,\,5,\,8,\,10,\,11$$ with corresponding frequencies $$f_i=5,\,4,\,4,\,2,\,2,\,3$$.

Total number of observations
$$N=\sum f_i = 5+4+4+2+2+3 = 20$$

Case P : Mean
The mean is $$\bar x = \dfrac{\sum f_i x_i}{N}$$.
$$\sum f_i x_i = 3\!(5)+4\!(4)+5\!(4)+8\!(2)+10\!(2)+11\!(3)=15+16+20+16+20+33=120$$
$$\bar x=\dfrac{120}{20}=6$$
So $$\bar x=6$$, which is entry (3).

Case Q : Median
The cumulative frequencies are $$5,\,9,\,13,\,15,\,17,\,20$$.
For $$N=20$$ (even), the median is the average of the $$10^{\text{th}}$$ and $$11^{\text{th}}$$ items.
Both the $$10^{\text{th}}$$ and $$11^{\text{th}}$$ items lie in the class $$x=5$$ (cumulative frequency just exceeds 9).
Hence $$\text{Median}=5$$, which is entry (2).

Case R : Mean deviation about the mean
Mean deviation about the mean is $$\dfrac{\sum f_i \lvert x_i-\bar x\rvert}{N}$$.
$$\begin{aligned} \sum f_i \lvert x_i-\bar x\rvert &= 5\lvert3-6\rvert +4\lvert4-6\rvert +4\lvert5-6\rvert +2\lvert8-6\rvert \\ &\quad +2\lvert10-6\rvert +3\lvert11-6\rvert \\ &=5(3)+4(2)+4(1)+2(2)+2(4)+3(5)\\ &=15+8+4+4+8+15=54 \end{aligned}$$
$$\text{M.D. about mean}= \dfrac{54}{20}=2.7$$, which is entry (4).

Case S : Mean deviation about the median
Mean deviation about the median is $$\dfrac{\sum f_i \lvert x_i-\text{Median}\rvert}{N}$$.
$$\begin{aligned} \sum f_i \lvert x_i-5\rvert &= 5\lvert3-5\rvert +4\lvert4-5\rvert +4\lvert5-5\rvert +2\lvert8-5\rvert \\ &\quad +2\lvert10-5\rvert +3\lvert11-5\rvert \\ &=5(2)+4(1)+4(0)+2(3)+2(5)+3(6)\\ &=10+4+0+6+10+18=48 \end{aligned}$$
$$\text{M.D. about median}= \dfrac{48}{20}=2.4$$, which is entry (5).

Thus the correct matching is
$$P\rightarrow(3),\;Q\rightarrow(2),\;R\rightarrow(4),\;S\rightarrow(5).$$
Therefore, the correct option is Option A.