Question

If $a$ is added to all the observations, (where $a$ can be either positive or negative integer) then what is the relation between the old and new variance.
(Old variance $={\sigma }^2$ and New variance $={{\sigma }^{\prime }}^2$)

• A. ${\sigma }^2=a\times {{\sigma }^{\prime }}^2$
• B. ${\sigma }^2=a^2\times {{\sigma }^{\prime }}^2$
• C. ${\sigma }^2=\frac{{{\sigma }^{\prime }}^2}{a}$
• D. ${\sigma }^2={{\sigma }^{\prime }}^2$

Answer 1

The correct option is D ${\sigma }^2={{\sigma }^{\prime }}^2$

Let Mean be $u=\frac{\sum x_i}{n}$
Let $y_i=x_i+a$ for all $i=1,2,\mathrm{3....}n$.
Now new Mean, $u^{\prime }=\frac{\sum y_i}{n}$
Now new Mean, $u^{\prime }=\frac{\sum (x_i+a)}{n}$
$\therefore u^{\prime }=u+a$.
Now, variance, ${{\sigma }^{\prime }}^2=\frac{1}{n}\times \sum (y_i-u^{\prime }{)}^2$
$\therefore {{\sigma }^{\prime }}^2=\frac{1}{n}\times \sum \left(\right(x_i+a)-(u+a){)}^2$
$\therefore {{\sigma }^{\prime }}^2=\frac{1}{n}\times \sum (x_i-u{)}^2={\sigma }^2$
Thus, ${\sigma }^2={{\sigma }^{\prime }}^2$
Ans-Option $D$