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

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Solution

Q.
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:

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

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

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

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

Ans.
(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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