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)

$\bar{x}=\frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}$
$\sigma^{2}=\frac{1}{n}$ $\displaystyle \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 $d_{1}$, $d_{2},...., d_{n}$
$= \frac{1}{n} \sum\limits^{n}_{i = 1} \left[d_{i} - \left(-\bar{x} - a\right)\right]^{2}$
$= \frac{1}{n} \sum\limits^{n}_{i = 1} \left[-x_{i} - a + \bar{x} + a\right]^{2}$
$= \frac{1}{n} \sum\limits^{n}_{i = 1}\left(-x_{i} + \bar{x}\right)^{2}$
$= \frac{1}{n} \sum\limits^{n}_{i = 1}\left(\bar{x} - x_{i}\right)^{2} = \sigma^{2}$