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. ${\sigma }_y=\frac{a}{c}{\sigma }_x$
• B. ${\sigma }_y=-\frac{a}{c}{\sigma }_x$
• C. ${\sigma }_y=\frac{a}{c}{\sigma }_x+b$
• D. none of these

Answer 1

The correct option is B

${\sigma }_y=-\frac{a}{c}{\sigma }_x$

$Y=\frac{aX+b}{c}$

$\overline{Y}=\frac{{\sum }_{i-1}^n\frac{aX+b}{c}}{n}$

$=\frac{a\frac{{\sum }_{i-1}^{n}X+nb}{c}}{n}$

$=\frac{\frac{a}{c}{\sum }_{i-1}^{n}X}{n}+\frac{b}{c}=\frac{a\overline{X}}{c}+\frac{b}{c}$

We know :

Var (X) = $\frac{{\sum }_{i=1}^n(x_i-\overline{X}{)}^2}{n}={\sigma }^2$ \

$Var\left(Y\right)=\frac{\sum (y_i-\overline{Y}{)}^2}{n}$

$=\frac{{\sum }_{i=1}^n{(\frac{aX}{c}+\frac{b}{c}-\frac{a}{c}X-\frac{b}{c})}^2}{n}$

$=\frac{{\sum }_{i=1}^n{(\frac{aX}{c}-\frac{a}{c}\overline{X})}^2}{n}$

$={\left(\frac{a}{c}\right)}^2\frac{{\sum }_{i=1}^n(x_i-\overline{X}{)}^2}{n}$

$={\left(\frac{a}{c}\right)}^2{\sigma }^2$

SD of Y = $\left({\sigma }_y\right)$ = $\sqrt{\left(\frac{a}{c}\right){\sigma }^2}$

$=\left|\frac{a}{c}\right|\sigma ac<0$

$\Rightarrow a<0orc<0$

$\therefore \left|\frac{a}{c}\right|=\frac{a}{c}$

$\Rightarrow {\sigma }_y=-\frac{a}{c}\sigma X$