Consider the given data with frequency distribution x i begin array llllll 3 & 8 & 11 & 10 & 5 & 4end array f i begin array llllll 5 & 2 & 3 & 2 & 4 & 4end array Match each entry in List-I to the correct entries in List-II. The correct option is: 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 eman 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

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

( P ) rightarrow(3) \ Q) rightarrow (2)\\ ( R ) rightarrow(1) \ S) rightarrow (5)

( P ) rightarrow(2) \ Q) rightarrow (3)\\ ( R ) rightarrow(4) \\ ( S ) rightarrow(1)

( P ) rightarrow(3) (Q) rightarrow (3) ( R ) rightarrow(5) ( S ) rightarrow(5)

Mathematics · Statistics

JEE Advanced 2023 — Paper 1 — Question 14

Consider the given data with frequency distribution

$\mathrm{x}_{\mathrm{i}} \quad \begin{array}{llllll}3 & 8 & 11 & 10 & 5 & 4\end{array}$

$\mathrm{f}_{\mathrm{i}} \quad \begin{array}{llllll}5 & 2 & 3 & 2 & 4 & 4\end{array}$

Match each entry in List-I to the correct entries in List-II. The correct option is:

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 eman 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

Option A:
(P) $\rightarrow$ (3) $(\mathrm{Q}) \rightarrow(2)$$(\mathrm{R}) \rightarrow(4)$$(\mathrm{S}) \rightarrow(5)$
Correct
Option B:
$(\mathrm{P}) \rightarrow(3)$ \(Q) $\rightarrow$ (2)\ $(\mathrm{R}) \rightarrow(1)$ \(S) $\rightarrow$ (5)
Option C:
$(\mathrm{P}) \rightarrow(2)$ \(Q) $\rightarrow$ (3)\ $(\mathrm{R}) \rightarrow(4)$ \ $(\mathrm{S}) \rightarrow(1)$
Option D:
$(\mathrm{P}) \rightarrow(3)$ (Q) $\rightarrow$ (3) $(\mathrm{R}) \rightarrow(5)$$(\mathrm{S}) \rightarrow(5)$

Answer: A

Step-by-step solution

$\mathrm{x}_{\mathrm{i}}$	$\mathrm{f}_{\mathrm{i}}$	$\mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}$	$\mathrm{f}_{\mathrm{i}}\left\\|\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{x}}\right\\|$	$\mathrm{f}_{\mathrm{i}}\left\\|\mathrm{X}_{\mathrm{i}}-\mathrm{M}\right\\|$
3	5	15	15	10
4	4	16	8	4
5	4	20	4	0
8	2	16	4	6
10	2	20	8	10
11	3	33	15	18
	$\Sigma \mathrm{f}_{\mathrm{i}} =20$	$\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=120$	sum $=54$	sum $=48$

(P) ${Mean}(\bar{x})=\frac{120}{20}=6$

(Q) Median $=\frac{\left(10^{\text {th }}+11^{\text {th }}\right) \text { observation }}{2}=5$

(R) M.D. $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{54}{20}=2.7$

(S) M.D. (M) $=\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{48}{20}=2.4$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Statistics
Topic: Measures of Central Tendency