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

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$a σ$

$\frac{a}{c} σ$

$|\frac{a}{c}| σ$

$\frac{a σ + b}{c}$

answer is C.

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Detailed Solution

Let $Y = \frac{a X + b}{c}$

Let $x_{1}, x_{2}, . . . ., x_{n}$ be the values taken by variable $X .$ Then, the corresponding values of variable $Y$ are given by

$\begin{matrix} y_{i} = \frac{a x_{i} + b}{c}, i = 1, 2, \dots, n \\ \Rightarrow \frac{1}{n} \sum_{i = 1}^{n} y_{i} = \frac{a}{c} (\frac{1}{n} \sum_{i = 1}^{n} x_{i}) + \frac{b}{c} \end{matrix}$

$\begin{array}{ll} \Rightarrow \bar{Y} = \frac{a}{c} \bar{X} + \frac{b}{c} \\ ∴ y_{i} - \bar{Y} = \frac{a}{c} (x_{i} - \bar{X}); i = 1, 2, \dots, n \\ \Rightarrow \frac{1}{n} \sum_{i = 1}^{n} {(y_{i} - \bar{Y})}^{2} = \frac{a^{2}}{c^{2}} \{\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2}\} \\ \Rightarrow σ_{Y}^{2} = \frac{a^{2}}{c^{2}} σ^{2} \Rightarrow σ_{Y} = |\frac{a}{c}| σ \end{array}$