The S.D. of a variate x is sigma. The S.D. of | vedclass

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Question

(b) Let $y = \frac{{ax + b}}{c}$ i.e., $y = \frac{a}{c}x + \frac{b}{c}$

i.e., $y = Ax + B$, where $A = \frac{a}{c}$,$B = \frac{b}{c}$

 $\ba

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $.

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

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