Q. If $\sigma$ is the standard deviation of a random variable $x$, then the standard deviation of the random variable $ax + b$, where $a,b \in R$ is
A) $a\sigma + b$
B) $|a|\sigma$
C) $|a|\sigma + b$
D) $a^{2}\sigma$
Explanation
(b)
Let $\ x_{1},x_{2},\ldots x_{n}$ be n values of $x.$
Then, $\sigma = \frac{1}{n}\sum_{i = 1}^{n}\left( x_{i} - \overline{x} \right)^{2}$… (i)
The variable $ax + b\$takes values $ax_{1} + b,ax_{2} + b,\ldots,ax_{n} + b$ with mean $\overline{x} + b.$
$\therefore SD\ of\ (ax + b) = \sqrt{\frac{1}{n}\sum_{i = 1}^{n}\left\{ \left( {ax}_{i} - b \right) - \left( a\overline{x} + b \right) \right\}^{2}}$
$= {\sqrt{a^{2}.\frac{1}{n}\sum_{i = 1}^{n}\left( x_{i} - \overline{x} \right)^{2}}}^{\ }$
$= |a|\sigma$
Alternate $SD(ax + b) = SD(ax) + SD(b)$
=$|a|SD(x) + 0$
=|a|$\sigma$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.