Question

The standard deviation of $X$, denoted $SD\left(X\right)$, is given by

$SD\left(X\right)=\sqrt{Var\left(X\right)}$

Find $SD(aX+b)$ if $X$ has variance ${\sigma }^2$.

Short answer

The value of$SD(aX+b)=\left|a\right|\sigma$.

Step-by-step solution

Given Information

The standard deviation of $X$, denoted $SD\left(X\right)$.

Explanation

To find $SD(aX+b)$:

Using the definition of standard deviation and basic properties of Variance, we have that

$SD(aX+b)=\sqrt{Var(aX+b)}=\sqrt{a^{2}Var\left(X\right)}=\left|a\right|\sqrt{Var\left(X\right)}=\left|a\right|\sigma$

since we have that $Var\left(X\right)={\sigma }^2\Rightarrow SD\left(X\right)=\sigma$.