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The $S.D.$ of a variate $x$ is $\sigma$. The $S.D.$ of the variate $\frac{{ax + b}}{c}$ where $a, b, c$ are constant, is
$\left( {\frac{a}{c}} \right)\,\sigma $
$\left| {\frac{a}{c}} \right|\,\sigma $
$\left( {\frac{{{a^2}}}{{{c^2}}}} \right)\,\sigma $
None of these
Solution
(b) Let $y = \frac{{ax + b}}{c}$ i.e., $y = \frac{a}{c}x + \frac{b}{c}$
i.e., $y = Ax + B$, where $A = \frac{a}{c}$,$B = \frac{b}{c}$
$\bar y = A\bar x + B$
$y – \bar y = A(x – \bar x)$ ==> ${(y – \bar y)^2} = {A^2}{(x – \bar x)^2}$
==> $\sum {(y – \bar y)^2} = {A^2}\sum {(x – \bar x)^2}$
==> $n.\sigma _y^2 = {A^2}.n\sigma _x^2$ ==> $\sigma _y^2 = {A^2}\sigma _x^2$
==> ${\sigma _y} = \,|A|{\sigma _x}$ ==> ${\sigma _y} = \,\left| {\frac{a}{c}} \right|{\sigma _x}$
Thus, new $S.D$. $ = \left| {\frac{a}{c}} \right|\,\sigma $.
