Question

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

• A. $a\sigma$
• B. $\frac{a}{c}\sigma$
• C. $\left|\frac{a}{c}\right|\sigma$
• D. $\frac{a\sigma +b}{c}$

Answer 1

The correct option is C $\left|\frac{a}{c}\right|\sigma$

If the S.D. of variate $X$ is .
$\sigma \left(X\right)=\sqrt{\frac{{\sum }_{i=1}^N(X_i-\overline{X}{)}^2}{N}}$

For $(aX+b)$ ,
Mean, $\frac{(a{X}_1+b)+(a{X}_2+b)+...+(a{X}_N+b)}{N}$

$=a\frac{(X_1+X_2+...+X_N)}{N}+b$
$=a\overline{X}+b$

$\sigma (aX+b)=\sqrt{\frac{{\sum }_{i=1}^N(x_i-\overline{x}{)}^2}{N}}$

$=\sqrt{\frac{{\left[\right(a{X}_1+b)-(a\overline{X}+b\left)\right]}^2+{\left[\right(a{X}_2+b)-(a\overline{X}+b\left)\right]}^2+....+{\left[\right(a{X}_N+b)-(a\overline{X}+b\left)\right]}^2}{N}}$

$=\sqrt{\frac{a^2[{(X_1-\overline{X})}^2+{(X_2-\overline{X})}^2+...+{(X_N-\overline{X})}^2]}{N}}$

$=\left|a\right|\sqrt{\frac{{\sum }_{i=1}^N{(X_i-\overline{X})}^2}{N}}$
$=\left|a\right|\sigma \left(X\right)$

For $\frac{(aX+b)}{c}$

$\sigma \frac{(aX+b)}{c}=\left|\frac{a}{c}\right|\sigma \left(X\right)$