The standard deviation of a varibale x is $\sigma$. The standard devaition of the varibale $\frac { ax+b }{ c }$ where a, b, c are constant is _____________________.

Let,

$\begin{aligned} y & =\frac{a u+b}{c} \\ & =\frac{a}{c} u+\frac{b}{c}\end{aligned}$

$\therefore y=A x+B\left[A=\frac{a}{c}(1)=\frac{b}{c}\right]$ …….(i)

$\therefore \bar{y}=A \bar{u}+B$……..(ii)

(i)-(ii) we get

$\begin{array}{l}\therefore y-\bar{y}=A(u-\bar{x}) \\ \Rightarrow(y-\bar{y})^{2}=A^{2}(u-\bar{u})^{2} \\ \Rightarrow n \cdot \frac{\sum(y-\bar{y})^{2}}{n}=\bar{A} n \frac{\sum(u-\bar{u})}{n} \\ \Rightarrow n \cdot \sigma_{y}^{2}=A^{2} \cdot n \sigma_{x}^{2} \\\end{array}$

$\begin{array}{l} \Rightarrow \sigma_{y}^{2}={A}² \cdot \sigma_{x}^{2} \\ \Rightarrow \sigma_{y}=|A|-\sigma_{x} \\ \therefore \sigma_{y}=\left|\frac{a}{c}\right| \sigma_{x}\left[|A|=\left|\frac{a}{c}\right|\right] \end{array}$

$\therefore$ new standard deviation $=\left|\frac{a}{c}\right| \sigma$