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) $19\alpha^{2}$ is equal to | (4) 145 | | | (5) 55 |

$\begin{array}{c|ccccccc} \text{Value} & 4 & 5 & 8 & 9 & 6 & 12 & 11 \\ \hline \text{Frequency} & 5 & f_1 & f_2 & 2 & 1 & 1 & 3 \end{array}$ $\sum f_i = 19, \qquad \text{median} = 6$ $\alpha = \text{mean deviation about mean}, \quad \beta = \text{mean deviation about median}$ $\begin{array}{ll} \text{List-I} & \text{List-II} \\ (P)\ 7f_1 + 9f_2 & (1)\ 146 \\ (Q)\ 19\alpha & (2)\ 47 \\ (R)\ 19\beta & (3)\ 48 \\ (S)\ 19\alpha^{2} & (4)\ 145 \\ & (5)\ 55 \end{array}$ $\sum f_i = 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19$ $f_1 + f_2 = 7$ $\text{ascending } x:\ 4,\ 5,\ 6,\ 8,\ 9,\ 11,\ 12$ $\text{cumulative}:\ 5,\ 5+f_1,\ 6+f_1,\ \dots$ $5 + f_1 < 10 \le 6 + f_1 \Rightarrow f_1 = 4,\quad f_2 = 3$ $\begin{array}{c|ccccccc} x & 4 & 5 & 6 & 8 & 9 & 11 & 12 \\ \hline f & 5 & 4 & 1 & 3 & 2 & 3 & 1 \end{array}$ $(P)\ 7f_1 + 9f_2 = 7(4) + 9(3) = 55 \ \Rightarrow\ (5)$ $\sum f_i x_i = 4(5)+5(4)+6(1)+8(3)+9(2)+11(3)+12(1)$ $= 20+20+6+24+18+33+12 = 133$ $\bar x = \dfrac{133}{19} = 7$ $(Q)\ \alpha = \dfrac{1}{19}\sum f_i |x_i - \bar x|$ $19\alpha = 5(3)+4(2)+1(1)+3(1)+2(2)+3(4)+1(5) = 48 \ \Rightarrow\ (3)$ $(R)\ \beta = \dfrac{1}{19}\sum f_i |x_i - 6|$ $19\beta = 5(2)+4(1)+1(0)+3(2)+2(3)+3(5)+1(6) = 47 \ \Rightarrow\ (2)$ $(S)\ \sigma^{2} = \dfrac{1}{19}\sum f_i (x_i - \bar x)^{2}$ $19\sigma^{2} = 5(9)+4(4)+1(1)+3(1)+2(4)+3(16)+1(25) = 146 \ \Rightarrow\ (1)$ $(P)-(5),\quad (Q)-(3),\quad (R)-(2),\quad (S)-(1)$ $\boxed{\text{Answer: option (3)}}$