Properties of Frequency Distribution with Given Median and Total Frequency

To solve the problem, we first arrange the values of the frequency distribution in ascending order:

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Value } (x_i) & 4 & 5 & 6 & 8 & 9 & 11 & 12 \\ \hline \text{Frequency } (f_i) & 5 & f_1 & 1 & f_2 & 2 & 3 & 1 \\ \hline \text{Cumulative Frequency } (cf) & 5 & 5+f_1 & 6+f_1 & 6+f_1+f_2 & 8+f_1+f_2 & 11+f_1+f_2 & 12+f_1+f_2 \\ \hline \end{array}$$

Step 1: Find the values of $f_1$ and $f_2$

The total frequency is given as $N = 19$:

$$12 + f_1 + f_2 = 19 \implies f_1 + f_2 = 7$$

Since the total number of observations $N = 19$ is odd, the median corresponds to the $\left(\frac{19+1}{2}\right)^{\text{th}} = 10^{\text{th}}$ observation. We are given that the median is $6$. Therefore, the $10^{\text{th}}$ observation must correspond to $x_i = 6$, which implies:

$$5 + f_1 < 10 \le 6 + f_1$$ $$f_1 < 5 \quad \text{and} \quad f_1 \ge 4 \implies f_1 = 4$$

Since $f_1 + f_2 = 7$, we obtain:

$$f_2 = 7 - 4 = 3$$

Now, we calculate the entry for (P):

$$7f_1 + 9f_2 = 7(4) + 9(3) = 28 + 27 = 55 \implies \mathbf{(P) \rightarrow (5)}$$

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Step 2: Calculate the Mean ($\bar{x}$)

The mean $\bar{x}$ is given by:

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4(5) + 5(4) + 6(1) + 8(3) + 9(2) + 11(3) + 12(1)}{19}$$ $$\bar{x} = \frac{20 + 20 + 6 + 24 + 18 + 33 + 12}{19} = \frac{133}{19} = 7$$

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Step 3: Mean Deviation about the Mean ($\alpha$)

The mean deviation about the mean is:

$$\alpha = \frac{1}{19}\sum f_i |x_i - \bar{x}|$$ $$19\alpha = \sum f_i |x_i - 7|$$ $$19\alpha = 5|4-7| + 4|5-7| + 1|6-7| + 3|8-7| + 2|9-7| + 3|11-7| + 1|12-7|$$ $$19\alpha = 5(3) + 4(2) + 1(1) + 3(1) + 2(2) + 3(4) + 1(5)$$ $$19\alpha = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48 \implies \mathbf{(Q) \rightarrow (3)}$$

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Step 4: Mean Deviation about the Median ($\beta$)

The median is $M = 6$. The mean deviation about the median is:

$$\beta = \frac{1}{19}\sum f_i |x_i - 6|$$ $$19\beta = \sum f_i |x_i - 6|$$ $$19\beta = 5|4-6| + 4|5-6| + 1|6-6| + 3|8-6| + 2|9-6| + 3|11-6| + 1|12-6|$$ $$19\beta = 5(2) + 4(1) + 1(0) + 3(2) + 2(3) + 3(5) + 1(6)$$ $$19\beta = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47 \implies \mathbf{(R) \rightarrow (2)}$$

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Step 5: Variance ($\sigma^2$)

The variance is given by:

$$\sigma^2 = \frac{1}{19}\sum f_i (x_i - \bar{x})^2$$ $$19\sigma^2 = \sum f_i (x_i - 7)^2$$ $$19\sigma^2 = 5(4-7)^2 + 4(5-7)^2 + 1(6-7)^2 + 3(8-7)^2 + 2(9-7)^2 + 3(11-7)^2 + 1(12-7)^2$$ $$19\sigma^2 = 5(9) + 4(4) + 1(1) + 3(1) + 2(4) + 3(16) + 1(25)$$ $$19\sigma^2 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146 \implies \mathbf{(S) \rightarrow (1)}$$

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Conclusion

Matching List-I with List-II:

• $(\text{P}) \rightarrow (5)$
• $(\text{Q}) \rightarrow (3)$
• $(\text{R}) \rightarrow (2)$
• $(\text{S}) \rightarrow (1)$

Thus, the correct option is C.