Question

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

The correct option is B Statement I and Statement II are both true

Mean of $x_i$ 's $=\overline{x}=\frac{x_1+x_2+.....+x_n}{n}$

Mean of $d_i$ 's $=\frac{{(-x}_1-a)+{(-x}_2-a)+.....+{(-x}_n-a)}{n}=\frac{{-(x}_1+x_2+.....+x_n)-an}{n}=-\overline{x}-a$

Mode of $x_i$ 's is $M$

Clearly, Mode of $d_i^{\prime }s$ will be $-M-a$. .

Variance of $x_i^{\prime }s={\sigma }^2=\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n}$

Variance of $d_i^{\prime }s=\frac{{\sum }_{i=1}^n\{{-x}_i-a-(-\overline{x}-a){\}}^2}{n}=\frac{{\sum }_{i=1}^n\{{-(x}_i-\overline{x}){\}}^2}{n}={\sigma }^2$