Consider the following frequency distribution:

Value	4	5	8	9	6	12	11
Frequency	5	$$f_1$$	$$f_2$$	2	1	1	3

Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.

For the given frequency distribution, let $$\alpha$$ denote the mean deviation about the mean, $$\beta$$ denote the mean deviation about the median, and $$\sigma^2$$ denote the variance.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I                                                  List-II

(P) $$7f_1 + 9f_2$$ is equal to           (1) 146 

(Q) $$19\alpha$$ is equal to                     (2) 47

(R) $$19\beta$$ is equal to                    (3) 48

(S) $$19\sigma^2$$ is equal to                  (4) 145

                                                        (5) 55

The given information can be arranged in a two-row table.

Values (x): 4 5 8 9 6 12 11
Frequencies (f): 5 $$f_1$$ $$f_2$$ 2 1 1 3

Adding all the frequencies must give 19:

$$5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \;\Longrightarrow\; f_1 + f_2 = 7 \qquad -(1)$$

The median is stated to be 6. Since the total number of observations is 19 (an odd number), the median is the 10-th item in the ordered data.

Arranging the values in ascending order:
4 (5), 5 ($$f_1$$), 6 (1), 8 ($$f_2$$), 9 (2), 11 (3), 12 (1)

Cumulative frequency just before 6 is $$5 + f_1$$ and including the single 6 it becomes $$6 + f_1$$. To make the 10-th observation land on the value 6 we require

$$5 + f_1 \lt 10 \le 6 + f_1 \;\Longrightarrow\; f_1 = 4$$

Using $$f_1 + f_2 = 7$$ from $$(1)$$ gives $$f_2 = 3$$.

The completed frequency table is

4(5), 5(4), 6(1), 8(3), 9(2), 11(3), 12(1)   with   $$\sum f = 19$$.

1. Mean $$\mu$$
$$\sum fx = 4\cdot5 + 5\cdot4 + 6\cdot1 + 8\cdot3 + 9\cdot2 + 11\cdot3 + 12\cdot1 = 133$$
$$\mu = \frac{133}{19} = 7$$

2. Mean deviation about the mean $$\alpha$$

$$\sum f|x-\mu| = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48$$
$$\alpha = \frac{48}{19}\;\Longrightarrow\; 19\alpha = 48$$

3. Mean deviation about the median $$\beta$$ (median = 6)

$$\sum f|x-6| = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47$$
$$\beta = \frac{47}{19}\;\Longrightarrow\; 19\beta = 47$$

4. Variance $$\sigma^{2}$$

$$\sum f(x-\mu)^2 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146$$
$$\sigma^{2} = \frac{146}{19}\;\Longrightarrow\; 19\sigma^{2} = 146$$

5. Expression $$7f_1 + 9f_2$$
$$7f_1 + 9f_2 = 7\cdot4 + 9\cdot3 = 28 + 27 = 55$$

Matching with List-II
(P) $$7f_1 + 9f_2 = 55$$ → (5)
(Q) $$19\alpha = 48$$ → (3)
(R) $$19\beta = 47$$ → (2)
(S) $$19\sigma^{2} = 146$$ → (1)

Thus the correct matching is
(P)→(5), (Q)→(3), (R)→(2), (S)→(1)

Option C which is: (P)→(5), (Q)→(3), (R)→(2), (S)→(1)