Question

If the standard deviation of a variable X is σ, then the standard deviation of variable $\frac{aX+b}{c}$ is
(a) a σ

(b) $\frac{a}{c}\mathrm{\sigma }$

(c) $\left|\frac{a}{c}\right|\mathrm{\sigma }$

(d) $\frac{a\mathrm{\sigma }+b}{c}$

Answer 1

(c) $\left|\frac{a}{c}\right|\sigma$


$$\begin{gathered} Y=\frac{aX+b}{c} \\ \overline{Y}=\frac{\sum y_i}{n}=\frac{{\displaystyle \frac{a\sum X+nb}{c}}}{n} \\ =\frac{a\sum X}{nc}+\frac{nb}{nc} \\ =\frac{a\overline{X}}{c}+\frac{b}{c} \end{gathered}$$


$$\begin{gathered} \mathrm{Var}\left(X\right)=\frac{\sum {\left(x_i-\overline{X}\right)}^2}{n} \\ ={\sigma }^2 \\ \mathrm{Var}\left(Y\right)=\frac{\sum {\left(y_i-\overline{Y}\right)}^2}{n} \\ =\frac{\sum {\left({\displaystyle \frac{aX}{c}}+{\displaystyle \frac{b}{c}}-{\displaystyle \frac{a}{c}}\overline{X}-{\displaystyle \frac{b}{c}}\right)}^2}{n} \\ =\frac{\sum {\left({\displaystyle \frac{aX}{c}}-{\displaystyle \frac{a}{c}}\overline{X}\right)}^2}{n} \\ ={\left(\frac{a}{c}\right)}^2\frac{\sum {\left(x_i-\overline{X}\right)}^2}{n} \\ ={\left(\frac{a}{c}\right)}^2{\sigma }^2 \\ \mathrm{SD}\left(\sigma \right)=\sqrt{{\left(\frac{a}{c}\right)}^2{\sigma }^2} \\ =\left|\frac{a}{c}\right|\sigma \end{gathered}$$