Consider the given data with frequency distribution

x_i = 3, 8, 11, 10, 5, 4 f_i = 5, 2, 3, 2, 4, 4

Match each entry in List-I to the correct entries in List-II.

The correct option is (P ) → (3), (Q) → (2), (R) → (4), (S) → (5).

Let us analyze the given frequency distribution table:

From the question, the given values of observations x_i and their corresponding frequencies f_i are:

Arranging the data in ascending order of x_i along with their frequencies f_i and cumulative frequencies c.f.:

• x₁ = 3, f₁ = 5, c.f. = 5
• x₂ = 4, f₂ = 4, c.f. = 9
• x₃ = 5, f₃ = 4, c.f. = 13
• x₄ = 8, f₄ = 2, c.f. = 15
• x₅ = 10, f₅ = 2, c.f. = 17
• x₆ = 11, f₆ = 3, c.f. = 20

Total frequency, $N=\sum f_i=5+4+4+2+2+3=20$.

Step 1: Calculate Mean ($\bar{x}$) - Entry (P)

Sum of products $\sum f_i{x}_i$ is:

$\sum f_i{x}_i=(3\times 5)+(4\times 4)+(5\times 4)+(8\times 2)+(10\times 2)+(11\times 3)$

$\sum f_i{x}_i=15+16+20+16+20+33=120$

Therefore, the mean is:

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

Hence, (P) → (3).

Step 2: Calculate Median (M) - Entry (Q)

Since total frequency N = 20 (an even number), the median is the average of the (N/2)^th and (N/2 + 1)^th observations, which are the 10^th and 11^th observations.

Looking at the cumulative frequencies:

• Cumulative frequency up to x = 4 is 9.
• Cumulative frequency up to x = 5 is 13.

Thus, both the 10^th and 11^th observations lie in the group x = 5.

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

Hence, (Q) → (2).

Step 3: Calculate Mean Deviation about the Mean - Entry (R)

We calculate $|x_i-\bar{x}|$ for each observation with $\bar{x}=6$:

• x = 3: |3 - 6| = 3, f_i|x_i - 6| = 5 × 3 = 15
• x = 4: |4 - 6| = 2, f_i|x_i - 6| = 4 × 2 = 8
• x = 5: |5 - 6| = 1, f_i|x_i - 6| = 4 × 1 = 4
• x = 8: |8 - 6| = 2, f_i|x_i - 6| = 2 × 2 = 4
• x = 10: |10 - 6| = 4, f_i|x_i - 6| = 2 × 4 = 8
• x = 11: |11 - 6| = 5, f_i|x_i - 6| = 3 × 5 = 15

$\sum f_i|x_i-\bar{x}|=15+8+4+4+8+15=54$

Mean deviation about mean = $\frac{\sum f_i|x_i-\bar{x}|}{N}=\frac{54}{20}=2.7$

Hence, (R) → (4).

Step 4: Calculate Mean Deviation about the Median - Entry (S)

We calculate $|x_i-M|$ for each observation with M = 5:

• x = 3: |3 - 5| = 2, f_i|x_i - 5| = 5 × 2 = 10
• x = 4: |4 - 5| = 1, f_i|x_i - 5| = 4 × 1 = 4
• x = 5: |5 - 5| = 0, f_i|x_i - 5| = 4 × 0 = 0
• x = 8: |8 - 5| = 3, f_i|x_i - 5| = 2 × 3 = 6
• x = 10: |10 - 5| = 5, f_i|x_i - 5| = 2 × 5 = 10
• x = 11: |11 - 5| = 6, f_i|x_i - 5| = 3 × 6 = 18

$\sum f_i|x_i-M|=10+4+0+6+10+18=48$

Mean deviation about median = $\frac{\sum f_i|x_i-M|}{N}=\frac{48}{20}=2.4$

Hence, (S) → (5).

Conclusion:

Matching the items from List-I to List-II:

(P) → (3)
(Q) → (2)
(R) → (4)
(S) → (5)