Question

$x_{1}, x_{2}, \ldots, x_{n}$ are $n$ observations with mean $\bar{x}$ and standard deviation $\sigma .$ Match the items of List-I with those of List-II

List-I		List-II A ( ) i n ix
A	$\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)$	i	Median
B	Variance $\left(\sigma^{2}\right)$	ii	Coefficient of variation
C	Mean deviation	iii	Zero
D	Measure used to find the homogeneity of given two series	iv	Mean of the absolute deviations from any measure of central tendency
		v	Mean of the squares of the deviations from mean

The correct answer is

• A. $A \to i, B \to v, C \to ii, D \to iii$
• B. $A \to i, B \to iv, C \to iii, D \to ii$
• C. $A \to iii, B \to v, C \to iv, D \to ii$
• D. $A \to iii, B \to v, C \to ii, D \to i$

Answer 1

Answer: (C)

(a) $\because \bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}$
$\Rightarrow n \bar{x}=x_{1}+x_{2}+x_{3}+\ldots+x_{n}$
$\Rightarrow \left(x_{1}-\bar{x}\right)+\left(x_{2}-\bar{x}\right)+\ldots+\left(x_{n}-\bar{x}\right)=0$
$\Rightarrow \displaystyle \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)=0$
(b) $\because$ Variance $\left(\sigma^{2}\right)=\frac{1}{n} \displaystyle \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=$ Mean of the squares of the deviations from mean.
(c) $\because$ Mean deviation of observations is mean of the absolute deviations from any measure of central tendency.
(d) $\because$ Coefficient of variation is used to find the homogeneity of given two series.
$\therefore ( a ) \rightarrow( iii ),( b ) \rightarrow( v ),( c ) \rightarrow( iv ),( d ) \rightarrow( ii )$