JEE Advanced Maths PYQ – Statistics (Quartile, Dispersion, Variance & SD) | Solutions

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Topic Question Set

Q 1 :

Consider the following frequency distribution: [2025]

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) $19 α^{2}$ is equal to (4) 145

55

(P)–(5), (Q)–(3), (R)–(2), (S)–(4)
(P)–(5), (Q)–(2), (R)–(3), (S)–(1)
(P)–(5), (Q)–(3), (R)–(2), (S)–(1)
(P)–(3), (Q)–(2), (R)–(5), (S)–(4)

1

2

3

4

(3)

$Given sum of frequencies = 19$

$12 + f_{1} + f_{2} = 19$

$f_{1} + f_{2} = 7$

$∵$

$x_{i}$	$f_{i}$	$C . F .$
4	5	5
5	$f_{1}$	$5 + f_{1}$
6	1	$5 + f_{1} + 1$
8	$f_{2}$	$6 + f_{1} + f_{2}$
9	2	$8 + f_{1} + f_{2}$
11	3	$11 + f_{1} + f_{2}$
12	1	$12 + f_{1} + f_{2}$

$∵$ $Median = 10^{th} observation$

$∴$ $5 + f_{1} + 1 = 10$

$f_{1} = 4$ $∴$ $f_{2} = 3$

$∵$ $\bar{x} = \frac{\sum x_{i} f_{i}}{\sum f_{i}}$ $= \frac{20 + 20 + 6 + 24 + 18 + 33 + 12}{19}$ $= 7$

$x_{i}$	$f_{i}$	$f_{i} \| x_{i} - \bar{x} \|$
4	5	15
5	4	8
6	1	1
8	3	3
9	2	4
11	3	12
12	1	5

$∴$ $\sum f_{i}$ $| x_{i} - \bar{x} |$ $= 48$

$∴$ $α = \frac{\sum f_{i} | x_{i} - \bar{x} |}{\sum f_{i}} = \frac{48}{19}$

$x_{i}$	$f_{i}$	$f_{i}$ $\| x_{i} - M \|$
4	5	10
5	4	4
6	1	0
8	3	6
9	2	6
11	3	15
12	1	6

$∴$ $\sum f_{i}$ $| x_{i} - M |$ $= 47$

$∴$ $β = \frac{47}{19}$

$∵$ $σ^{2} = \frac{\sum f_{i} {(x_{i} - \bar{x})}^{2}}{\sum f_{i}}$

$= \frac{5 \times 9 + 4 \times 4 + 1 \times 1 + 3 \times 1 + 2 \times 4 + 3 \times 16 + 1 \times 25}{19}$

$σ^{2} = \frac{146}{19}$

Q 2 :

Consider the given data with frequency distribution [2023]

$x_{i}$ 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

The correct option is:

(P) → (3), (Q) → (2), (R) → (4), (S) → (5)
(P) → (3), (Q) → (2), (R) → (1), (S) → (5)
(P) → (2), (Q) → (3), (R) → (4), (S) → (1)
(P) → (3), (Q) → (3), (R) → (5), (S) → (5)

1

2

3

4

(1)

Given

$x_{i}$	3	4	5	8	10	11
$f_{i}$	5	4	4	2	2	3

$x_{i}$	$f_{i}$	$x_{i} f_{i}$	$C . F .$	$\| x_{i} - Mean \|$	$f_{i} \| x_{i} - Mean \|$	$\| x_{i} - Median \|$	$f_{i} \| x_{i} - Median \|$
3	5	15	5	3	15	2	10
4	4	16	9	2	8	1	4
5	4	20	13	1	4	0	0
8	2	16	15	2	4	3	6
10	2	20	17	4	8	5	10
11	3	33	20	5	15	6	18
	$\sum f_{i} = 20$	$\sum x_{i} f_{i} = 120$			$\sum f_{i}$ $\| x_{i} - Mean \|$ $= 54$		$\sum f_{i}$ $\| x_{i} - Median \|$ $= 48$

$(P) Mean = \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{120}{20} = 6$

$(Q) Median = {(\frac{N}{2})}^{th} obs. = {(\frac{20}{2})}^{th} obs. = 10^{th} obs. = 5$

$(R) Mean deviation about mean = \frac{\sum f_{i} | x_{i} - Mean |}{\sum f_{i}} = \frac{54}{20} = 2.70$

$(S) Mean deviation about median = \frac{\sum f_{i} | x_{i} - Median |}{\sum f_{i}} = \frac{48}{20} = 2.40$

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