Consider the following frequency distribution:

Value	4	5	8	9	6	12	11
Frequency	5	f₁	f₂	2	1	1	3

Sum of frequencies = 19 and median = 6.

Let mean = μ, mean deviation about mean = α, mean deviation about median = β, variance = σ².

List-I	List-II
(P) 7f₁ + 9f₂ is equal to	(1) 146
(Q) 19α is equal to	(2) 47
(R) 19β is equal to	(3) 48
(S) 19σ² is equal to	(4) 145
	(5) 55

Correct Option: (P)→(5), (Q)→(3), (R)→(2), (S)→(1)

Let us first write down the given data. The frequency distribution is as follows:
Values ($x_i$): 4, 5, 8, 9, 6, 12, 11
Frequencies ($f_i$): 5, $f_1$, $f_2$, 2, 1, 1, 3

First, we sort the values in ascending order to analyze the median and find the frequencies:
Sorted values ($x_i$): 4, 5, 6, 8, 9, 11, 12
Corresponding frequencies ($f_i$): 5, $f_1$, 1, $f_2$, 2, 3, 1

We are given that the sum of the frequencies is 19. Therefore:
$\sum f_i=5+f_1+1+f_2+2+3+1=19$
$12+f_1+f_2=19\Rightarrow f_1+f_2=7$ (Equation 1)

We are also given that the median is 6.
Since the total number of observations is $N=19$ (an odd number), the median is the ${\left(\frac{19+1}{2}\right)}^{\text{th}}={10}^{\text{th}}$ observation when sorted in ascending order.
Let us find the cumulative frequencies:
Cumulative frequency for $x=4$ is 5.
Cumulative frequency for $x=5$ is $5+f_1$.
Cumulative frequency for $x=6$ is $5+f_1+1=6+f_1$.
For the median to be 6, the ${10}^{\text{th}}$ observation must be 6.
This means:
$5+f_1<10\Rightarrow f_1<5$
and
$6+f_1\ge 10\Rightarrow f_1\ge 4$
Combining these inequalities, we get $f_1=4$.

Substitute $f_1=4$ in Equation 1:
$4+f_2=7\Rightarrow f_2=3$

Now we check list item (P):
$7f_1+9f_2=7\left(4\right)+9\left(3\right)=28+27=55$
Thus, (P) matches with (5).

Now let us compute the mean ($\mu$):
$\mu =\frac{\sum f_i{x}_i}{N}$
Let us calculate $\sum f_i{x}_i$:
$\sum f_i{x}_i=4\left(5\right)+5\left(4\right)+6\left(1\right)+8\left(3\right)+9\left(2\right)+11\left(3\right)+12\left(1\right)$
$\sum f_i{x}_i=20+20+6+24+18+33+12=133$
Therefore, the mean is:
$\mu =\frac{133}{19}=7$

Next, we calculate the mean deviation about the mean ($\alpha$):
$\alpha =\frac{\sum f_i|x_i-\mu |}{N}$
Let us find the absolute deviations from the mean $\mu =7$:
For $x=4$: $|4-7|=3\Rightarrow f_i|x_i-\mu |=5\times 3=15$
For $x=5$: $|5-7|=2\Rightarrow f_i|x_i-\mu |=4\times 2=8$
For $x=6$: $|6-7|=1\Rightarrow f_i|x_i-\mu |=1\times 1=1$
For $x=8$: $|8-7|=1\Rightarrow f_i|x_i-\mu |=3\times 1=3$
For $x=9$: $|9-7|=2\Rightarrow f_i|x_i-\mu |=2\times 2=4$
For $x=11$: $|11-7|=4\Rightarrow f_i|x_i-\mu |=3\times 4=12$
For $x=12$: $|12-7|=5\Rightarrow f_i|x_i-\mu |=1��5=5$
Summing these up:
$\sum f_i|x_i-\mu |=15+8+1+3+4+12+5=48$
Thus, we have:
$\alpha =\frac{48}{19}\Rightarrow 19\alpha =48$
So, (Q) matches with (3).

Next, we calculate the mean deviation about the median ($\beta$). The median is $M=6$:
$\beta =\frac{\sum f_i|x_i-M|}{N}$
Let us find the absolute deviations from the median $M=6$:
For $x=4$: $|4-6|=2\Rightarrow f_i|x_i-M|=5\times 2=10$
For $x=5$: $|5-6|=1\Rightarrow f_i|x_i-M|=4\times 1=4$
For $x=6$: $|6-6|=0\Rightarrow f_i|x_i-M|=1\times 0=0$
For $x=8$: $|8-6|=2\Rightarrow f_i|x_i-M|=3\times 2=6$
For $x=9$:
For $x=11$: $|11-6|=5\Rightarrow f_i|x_i-M|=3\times 5=15$
For $x=12$: $|12-6|=6\Rightarrow f_i|x_i-M|=1\times 6=6$
Summing these up:
$\sum f_i|x_i-M|=10+4+0+6+6+15+6=47$
Thus, we have:
$\beta =\frac{47}{19}\Rightarrow 19\beta =47$
So, (R) matches with (2).

Finally, we calculate the variance (${\sigma }^2$):
${\sigma }^2=\frac{\sum f_i{(x_i-\mu )}^2}{N}$
Let us find the squared deviations from the mean $\mu =7$:
For $x=4$: ${(4-7)}^2=9\Rightarrow f_i{(x_i-\mu )}^2=5\times 9=45$
For $x=5$: ${(5-7)}^2=4\Rightarrow f_i{(x_i-\mu )}^2=4\times 4=16$
For $x=6$: ${(6-7)}^2=1\Rightarrow f_i{(x_i-\mu )}^2=1\times 1=1$
For $x=8$: ${(8-7)}^2=1\Rightarrow f_i{(x_i-\mu )}^2=3\times 1=3$
For $x=9$:
For $x=11$: ${(11-7)}^2=16\Rightarrow f_i{(x_i-\mu )}^2=3\times 16=48$
For $x=12$: ${(12-7)}^2=25\Rightarrow f_i{(x_i-\mu )}^2=1\times 25=25$
Summing these up:
$\sum f_i{(x_i-\mu )}^2=45+16+1+3+8+48+25=146$
Thus, we have:
${\sigma }^2=\frac{146}{19}\Rightarrow 19{\sigma }^2=146$
So, (S) matches with (1).

Combining all the matches:
(P)→(5)
(Q)→(3)
(R)→(2)
(S)→(1)