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From the following table
x 2 3 4 5 6 7
Fi 5 4 7 6 3 2

COLUMN-I

COLUMN-II
a) Mean p) 1.19
b) Median q) 4.15
c) Mean deviation from Mean r) 4
d) Mean deviation from median s) 1.21

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$a \to q, b \to p, c \to r, d \to s$

$a \to s, b \to r, c \to q, d \to p$

$a \to q, b \to r, c \to s, d \to p$

$a \to s, b \to r, c \to p, d \to q$

answer is C.

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Detailed Solution

$a) N= \sum f i, \bar{x} = \frac{\sum f i x_{i}}{\sum f i} = \frac{112}{27} = 4.15$
b) Median M=4

$\begin{array}{l} c) \frac{\sum f i | x_{i} - \bar{x} |}{\sum f i} = \frac{32.75}{27} = 1.21 \\ d) \frac{\sum f i | x_{i} - M |}{\sum f i} = \frac{32}{27} = 1.19 \end{array}$

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