JEE Advanced 2025 Paper 1, Mathematics Section 4 Q1: Measure of Dispersion

JEE Advanced2025Paper 1Mathematics Section 4

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 $α$ denote the mean deviation about the mean, $β$ denote the mean deviation about the median, and $σ^{2}$ 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) $7 f_{1} + 9 f_{2}$ is equal to	(1) $146$
(Q) $19 α$ is equal to	(2) $47$
(R) $19 β$ is equal to	(3) $48$
(S) $19 σ^{2}$ is equal to	(4) $145$
	(5) $55$

A
(P) $\to$ (5); (Q) $\to$ (3); (R) $\to$ (2); (S) $\to$ (4)
B
(P) $\to$ (5); (Q) $\to$ (2); (R) $\to$ (3); (S) $\to$ (1)
C
(P) $\to$ (5); (Q) $\to$ (3); (R) $\to$ (2); (S) $\to$ (1)
D
(P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (5); (S) $\to$ (4)

Solution

Step 1: Find $f_{1}, f_{2}$ . Sum of frequencies: $5 + f_{1} + f_{2} + 2 + 1 + 1 + 3 = 19 \Rightarrow f_{1} + f_{2} = 7.$

The data sorted by value is $4, 5, 6, 8, 9, 11, 12$ with frequencies $5, f_{1}, 1, f_{2}, 2, 3, 1$ . With $19$ observations the median is the $10$ th value. For the median to equal $6$ , the cumulative frequency must reach $10$ at the value $6$ . Cumulative up to $5$ : $5 + f_{1}$ . Up to $6$ : $6 + f_{1}$ . We need $5 + f_{1} < 10 \leq 6 + f_{1}$ , giving $f_{1} = 4$ and hence $f_{2} = 3$ .

So $7 f_{1} + 9 f_{2} = 28 + 27 = 55.$ (P) $\to$ (5).

Step 2: Compute mean. Sum $= 4 (5) + 5 (4) + 6 (1) + 8 (3) + 9 (2) + 11 (3) + 12 (1) = 20 + 20 + 6 + 24 + 18 + 33 + 12 = 133.$ Mean $\overset{x}{ˉ} = 133/19 = 7.$

Step 3: Mean deviation about the mean. $∣ x_{i} - 7∣$ values are $3, 2, 1, 1, 2, 4, 5$ for $x_{i} = 4, 5, 6, 8, 9, 11, 12.$ Weighted sum:

$5 (3) + 4 (2) + 1 (1) + 3 (1) + 2 (2) + 3 (4) + 1 (5) = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48.$

So $19 α = 48.$ (Q) $\to$ (3).

Step 4: Mean deviation about the median. $∣ x_{i} - 6∣$ values are $2, 1, 0, 2, 3, 5, 6.$ Weighted sum:

$5 (2) + 4 (1) + 1 (0) + 3 (2) + 2 (3) + 3 (5) + 1 (6) = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47.$

So $19 β = 47.$ (R) $\to$ (2).

Step 5: Variance. $(x_{i} - 7)^{2}$ values are $9, 4, 1, 1, 4, 16, 25.$ Weighted sum:

$5 (9) + 4 (4) + 1 (1) + 3 (1) + 2 (4) + 3 (16) + 1 (25) = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146.$

So $19 σ^{2} = 146.$ (S) $\to$ (1).

$∴$ The correct option is (C).

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JEE Advanced 2025 Paper 1, Mathematics Section 4 Q1: Measure of Dispersion

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