Question

If two variates X and Y are connected by the relation $Y=\frac{aX+b}{c}$, where a, b, c are constants such that ac < 0, then
(a) ${\mathrm{\sigma }}_Y=\frac{a}{c}{\mathrm{\sigma }}_X$

(b) ${\mathrm{\sigma }}_Y=-\frac{a}{c}{\mathrm{\sigma }}_X$

(c) ${\mathrm{\sigma }}_Y=\frac{a}{c}{\mathrm{\sigma }}_X+b$

(d) none of these

Answer 1

$\left(b\right){\sigma }_Y=-\frac{a}{c}{\sigma }_X$

$$\begin{gathered} Y=\frac{aX+b}{c} \\ \overline{Y}=\frac{{\displaystyle \sum _{i=1}^n\frac{aX+b}{c}}}{n} \\ =\frac{{\displaystyle \frac{a{\displaystyle \sum _{i=1}^n}X+nb}{c}}}{n} \\ =\frac{{\displaystyle \frac{a}{c}}{\displaystyle \sum _{i=1}^n}X}{n}+\frac{b}{c} \\ =\frac{a\overline{X}}{c}+\frac{b}{c} \\ \mathrm{We}\mathrm{know}: \\ Var\left(X\right)=\frac{{\displaystyle \sum _{i=1}^n}{\left(x_i-\overline{X}\right)}^2}{n} \\ ={\sigma }^2 \\ Var\left(Y\right)=\frac{{\displaystyle \sum _{i=1}^n}(y_i-\overline{Y}{)}^2}{n} \\ =\frac{{\displaystyle \sum _{i=1}^n}{\left({\displaystyle \frac{aX}{c}}+{\displaystyle \frac{b}{c}}-{\displaystyle \frac{a}{c}}\overline{X}-{\displaystyle \frac{b}{c}}\right)}^2}{n} \\ =\frac{{\displaystyle \sum _{i=1}^n}{\left({\displaystyle \frac{aX}{c}}-{\displaystyle \frac{a}{c}}\overline{X}\right)}^2}{n} \\ ={\left(\frac{a}{c}\right)}^2\frac{{\displaystyle \sum _{i=1}^n}{\left(x_i-\overline{X}\right)}^2}{n} \\ ={\left(\frac{a}{c}\right)}^2{\sigma }^2 \\ \mathrm{SD}\mathrm{of}Y\left({\sigma }_y\right)=\sqrt{{\left(\frac{a}{c}\right)}^2{\sigma }^2} \\ =\left|\frac{a}{c}\right|\sigma \end{gathered}$$


$$\begin{gathered} ac<0 \\ \Rightarrow a<0\mathrm{or}c<0 \\ \therefore \left|\frac{a}{c}\right|=-\frac{a}{c} \end{gathered}$$

$\Rightarrow {\sigma }_Y=\mathbf{-}\frac{a}{c}{\sigma }_X$