Question

The standard deviation of variate $x_i$ is $\sigma$. Then standard deviation of the variate $\frac{a{x}_i+b}{c}$, where $a,b,c$ are constants is

• A. $\left(\frac{a}{c}\right)\sigma$
• B. $\left|\frac{a}{c}\right|\sigma$
• C. $\left(\frac{a^2}{c^2}\right)\sigma$
• D. none of these

Answer 1

Answer: (B)

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

We have standard deviation for variate $x_i$ as
$\because \sigma =\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n}}$
Let $y_i=\frac{a{x}_i+b}{c}=\left(\frac{a}{c}\right)x_i+\frac{b}{c}$
So, standard deviation for $y_i$ is
${\sigma }_1=\sqrt{\frac{\sum (y_i-\overline{y}{)}^2}{n}}$ ....(1)
Now, since $y_i=\left(\frac{a}{c}\right)x_i+\frac{b}{c}$
$\Rightarrow \overline{y}=\left(\frac{a}{c}\right)\overline{x}+\frac{b}{c}$
$\Rightarrow y_i-\overline{y}=\frac{a}{c}(x_i-\overline{x})$
Substitute this value in (1), we get
${\sigma }_1=\sqrt{\frac{\sum \left[\frac{a^2}{c^2}\right(x_i-\overline{x}{)}^2]}{n}}$
$\Rightarrow {\sigma }_1=\left|\frac{a}{c}\right|\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n}}$
$\Rightarrow {\sigma }_1=\left|\frac{a}{c}\right|\sigma$