JEE Advanced 2023 — Paper 1 — MATH-IV solution

JEE Advanced2023Paper 1MATH-IV

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.

List-I	List-II
(P) The mean of the above data is	(1) 2.5
(Q) The median of the above data is	(2) 5
(R) The mean deviation about the mean of the above data is	(3) 6
(S) The mean deviation about the median of the above data is	(4) 2.7
	(5) 2.4

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

Solution

Total frequency $\sum f_{i} = 5 + 2 + 3 + 2 + 4 + 4 = 20$ .

Compute $\sum x_{i} f_{i} = 3 (5) + 8 (2) + 11 (3) + 10 (2) + 5 (4) + 4 (4) = 15 + 16 + 33 + 20 + 20 + 16 = 120$ .

(P) Mean $= \frac{120}{20} = 6$ . $\to$ (3).

(Q) Median: Sort values: $3, 4, 5, 8, 10, 11$ with frequencies $5, 4, 4, 2, 2, 3$ . Cumulative frequencies: $5, 9, 13, 15, 17, 20$ . The $10$ th and $11$ th observations both fall in the class with cumulative frequency $13$ , i.e. value $5$ . Median $= 5$ . $\to$ (2).

(R) MD about mean $= \frac{\sum f _{i} ∣ x _{i} - 6∣}{20} = \frac{5 ( 3 ) + 4 ( 2 ) + 4 ( 1 ) + 2 ( 2 ) + 2 ( 4 ) + 3 ( 5 )}{20} = \frac{15 + 8 + 4 + 4 + 8 + 15}{20} = \frac{54}{20} = 2.7$ . $\to$ (4).

(S) MD about median $= \frac{\sum f _{i} ∣ x _{i} - 5∣}{20} = \frac{5 ( 2 ) + 4 ( 1 ) + 4 ( 0 ) + 2 ( 3 ) + 2 ( 5 ) + 3 ( 6 )}{20} = \frac{10 + 4 + 0 + 6 + 10 + 18}{20} = \frac{48}{20} = 2.4$ . $\to$ (5).

Mapping: P $\to$ 3, Q $\to$ 2, R $\to$ 4, S $\to$ 5. The correct option is (A).

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