Question

$x_1,x_2,\dots ,x_n$ 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) $\because \bar{x}=\frac{x_1+x_2+x_3+\dots +x_n}{n}$
$\Rightarrow n\bar{x}=x_1+x_2+x_3+\dots +x_n$
$\Rightarrow \left(x_1-\bar{x}\right)+\left(x_2-\bar{x}\right)+\dots +\left(x_n-\bar{x}\right)=0$
$\Rightarrow \sum _{i=1}^n\left(x_i-\bar{x}\right)=0$
(b) $\because$ Variance $\left({\sigma }^2\right)=\frac{1}{n}\sum _{i=1}^n{\left(x_i-\bar{x}\right)}^2=$ 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.
$\therefore (a)\to (iii),(b)\to (v),(c)\to (iv),(d)\to (ii)$