Question

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

• A. $a\sum _{}a+b$
• B. $\left|\frac{\mathrm{a}}{\mathrm{c}}\right|\sigma$
• C. $\left|a\right|\underset{}{\sum a}+b$
• D. $a^2\sum _{}$

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Find the mean

If the standard deviation of variate $X$ is given by:

$\sigma \left(X\right)=\sqrt{\frac{{\sum }_{i=1}^N{\left(X_i-\overline{X}\right)}^2}{N}}$
For $(a\mathrm{X}+\mathrm{b})$:
$\begin{array}{rcl}Mean& =& \frac{\left(a{X}_1+b\right)+\left(a{X}_2+b\right)+\dots +\left(a{X}_N+b\right)}{N}\\ & =& a\frac{\left(X_1+X_2+\dots +X_N\right)}{N}+b\\ & =& a\overline{X}+b\end{array}$

Step 2: Find the standard deviation

The standard deviation is calculated as :
$\therefore \sigma (aX+b)=\sqrt{\frac{{\sum }_{i=1}^N{\left(x_i-\overline{x}\right)}^2}{N}}$

$$\begin{gathered} =\sqrt{\frac{{\left[\left(a{X}_1+b\right)-(a\overline{X}+b)\right]}^2+{\left[\left(a{X}_2+b\right)-(a\overline{X}+b)\right]}^2+\dots .+{\left[\left(a{X}_N+b\right)-(a\overline{X}+b)\right]}^2}{N}} \\ =\sqrt{\frac{{\mathrm{a}}^2\left[{\left({\mathrm{X}}_1-\overline{\mathrm{X}}\right)}^2+{\left({\mathrm{X}}_2-\overline{\mathrm{X}}\right)}^2+\dots +{\left({\mathrm{X}}_{\mathrm{N}}-\overline{\mathrm{X}}\right)}^2\right]}{\mathrm{N}}} \\ =\left|\mathrm{a}\right|\sqrt{\frac{{\displaystyle {\sum }_{\mathrm{i}=1}^N{\left(X_i-\overline{X}\right)}^2}}{{\displaystyle N}}} \\ =\left|\mathrm{a}\right|\sigma \left(\mathrm{X}\right) \end{gathered}$$

For
$\begin{array}{rcl}\frac{(a\mathrm{X}+\mathrm{b})}{\mathrm{c}}& =& \sigma \frac{(\mathrm{aX}+\mathrm{b})}{\mathrm{c}}\\ & =& \left|\frac{\mathrm{a}}{\mathrm{c}}\right|\sigma \left(\mathrm{X}\right)\end{array}$

Hence, option B is correct.