Match Statistical Measures for Discrete Frequency Distribution

To find the correct matches between List-I and List-II, let us first arrange the given observations in ascending order along with their respective frequencies:

$$\begin{array}{|c|c|c|c|c|c|c|} \hline x_i & 3 & 4 & 5 & 8 & 10 & 11 \\ \hline f_i & 5 & 4 & 4 & 2 & 2 & 3 \\ \hline \text{c.f.} & 5 & 9 & 13 & 15 & 17 & 20 \\ \hline \end{array}$$

The total frequency is:

$$N = \sum f_i = 5 + 4 + 4 + 2 + 2 + 3 = 20$$

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(P) Mean of the data ($\bar{x}$):

$$\sum f_i x_i = (3 \times 5) + (4 \times 4) + (5 \times 4) + (8 \times 2) + (10 \times 2) + (11 \times 3)$$ $$\sum f_i x_i = 15 + 16 + 20 + 16 + 20 + 33 = 120$$

Thus, the mean is:

$$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{120}{20} = 6$$

Hence, (P) $\rightarrow$ (3).

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(Q) Median of the data ($M$):

Since $N = 20$ is an even number, the median is the average of the $\left(\frac{N}{2}\right)^{\text{th}}$ and $\left(\frac{N}{2} + 1\right)^{\text{th}}$ observations, i.e., the $10^{\text{th}}$ and $11^{\text{th}}$ observations.

From the cumulative frequency table:

• The cumulative frequency just greater than or equal to $10$ and $11$ is $13$, which corresponds to the value $x_i = 5$.

Therefore:

$$\text{Median } M = \frac{5 + 5}{2} = 5$$

Hence, (Q) $\rightarrow$ (2).

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(R) Mean Deviation about the Mean ($\text{MD}(\bar{x})$):

Using $\bar{x} = 6$:

$$\begin{array}{|c|c|c|c|} \hline x_i & f_i & |x_i - \bar{x}| = |x_i - 6| & f_i |x_i - \bar{x}| \\ \hline 3 & 5 & |3 - 6| = 3 & 5 \times 3 = 15 \\ 4 & 4 & |4 - 6| = 2 & 4 \times 2 = 8 \\ 5 & 4 & |5 - 6| = 1 & 4 \times 1 = 4 \\ 8 & 2 & |8 - 6| = 2 & 2 \times 2 = 4 \\ 10 & 2 & |10 - 6| = 4 & 2 \times 4 = 8 \\ 11 & 3 & |11 - 6| = 5 & 3 \times 5 = 15 \\ \hline \text{Total} & N = 20 & & \sum f_i |x_i - \bar{x}| = 54 \\ \hline \end{array}$$

Thus:

$$\text{MD}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{54}{20} = 2.7$$

Hence, (R) $\rightarrow$ (4).

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(S) Mean Deviation about the Median ($\text{MD}(M)$):

Using $M = 5$:

$$\begin{array}{|c|c|c|c|} \hline x_i & f_i & |x_i - M| = |x_i - 5| & f_i |x_i - M| \\ \hline 3 & 5 & |3 - 5| = 2 & 5 \times 2 = 10 \\ 4 & 4 & |4 - 5| = 1 & 4 \times 1 = 4 \\ 5 & 4 & |5 - 5| = 0 & 4 \times 0 = 0 \\ 8 & 2 & |8 - 5| = 3 & 2 \times 3 = 6 \\ 10 & 2 & |10 - 5| = 5 & 2 \times 5 = 10 \\ 11 & 3 & |11 - 5| = 6 & 3 \times 6 = 18 \\ \hline \text{Total} & N = 20 & & \sum f_i |x_i - M| = 48 \\ \hline \end{array}$$

Thus:

$$\text{MD}(M) = \frac{\sum f_i |x_i - M|}{N} = \frac{48}{20} = 2.4$$

Hence, (S) $\rightarrow$ (5).

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Conclusion:

The correct mapping is:

$$(\text{P}) \rightarrow (3), \quad (\text{Q}) \rightarrow (2), \quad (\text{R}) \rightarrow (4), \quad (\text{S}) \rightarrow (5)$$

This corresponds to Option A.