Question

Let $\bar{x}$ , M and ${\sigma }^2$ be respectively the mean, mode and variance of n observations $x_1,x_2,....,x_n$ and $d_i=-x_i-a,i=1,2,....,n,$ where a is any number.
Statement I: Variance of $d_1,d_2,...d_n$ is ${\sigma }^2$.
Statement II: Mean and mode of $d_1,d_2,....d_n$ are $-\bar{x}-a$ and - M - a, respectively.

• A. Statement I and Statement II are both false
• B. Statement I and Statement II are both true
• C. Statement I is true and Statement II is false
• D. Statement I is false and Statement II is true

Answer 1

Answer: (B)

$\bar{x}=\frac{x_1+x_2+x_3+...+x_n}{n}$
${\sigma }^2=\frac{1}{n}$ $\sum _{i=1}^n$${\left(x_i-\bar{x}\right)}^2$
Mean of $d_1,d_2,d_3,.....d_n$
$=\frac{d_1+d_2+d_3+....+d_n}{n}$
$=\frac{\left(-x_1-a\right)+\left(-x_2-a\right)+\left(-x_3-a\right)+...+\left(-x_n-a\right)}{n}$
$=-\left[\frac{x_1+x_2+x_3+....+x_n}{n}\right]-\frac{na}{n}$
$=-\bar{x}-a$
Since, $d_i=-x_i-a$ and we multiply or subtract each observation by any number the mode remains the same. Hence mode of $-x_i-a$ i.e. $d_i$ and $x_i$ are same. Now variance of $d_1$, $d_2,....,d_n$
$=\frac{1}{n}\sum _{i=1}^n{\left[d_i-\left(-\bar{x}-a\right)\right]}^2$
$=\frac{1}{n}\sum _{i=1}^n{\left[-x_i-a+\bar{x}+a\right]}^2$
$=\frac{1}{n}\sum _{i=1}^n{\left(-x_i+\bar{x}\right)}^2$
$=\frac{1}{n}\sum _{i=1}^n{\left(\bar{x}-x_i\right)}^2={\sigma }^2$