# If Two Variates X and Y Are Connected by the Relation Y = a X + B C , Where A, B, C Are Constants Such that Ac < 0, Then - Mathematics

MCQ

If two variates X and Y are connected by the relation $Y = \frac{a X + b}{c}$ , where abc are constants such that ac < 0, then

#### Options

•   $\sigma_Y = \frac{a}{c} \sigma_X$

• $\sigma_Y = - \frac{a}{c} \sigma_X$

•  $\sigma_Y = \frac{a}{c} \sigma_X + b$

• none of these

#### Solution

$\sigma_Y = - \frac{a}{c} \sigma_X$

$Y = \frac{aX + b}{c}$

$Y = \frac{\sum^n_{i = 1} \frac{aX + b}{c}}{n}$

$= \frac{\frac{a \sum^n_{i = 1} X + nb}{c}}{n}$

$= \frac{\frac{a}{c} \sum^n_{i = 1} X}{n} + \frac{b}{c}$

$= \frac{aX}{c} + \frac{b}{c}$

$\text{ We know: }$

$Var (X) = \frac{\sum^n_{i = 1} \left( x_i - X \right)^2}{n}$

$= \sigma^2$

$Var(Y) = \frac{\sum^n_{i = 1} ( y_i - Y )^2}{n}$

$= \frac{\sum^n_{i = 1} \left( \frac{aX}{c} + \frac{b}{c} - \frac{a}{c}X - \frac{b}{c} \right)^2}{n}$

$= \frac{\sum^n_{i = 1} \left( \frac{aX}{c} - \frac{a}{c}X \right)^2}{n}$

$= \left( \frac{a}{c} \right)^2 \frac{\sum^n_{i = 1} \left( x_i - X \right)^2}{n}$

$= \left( \frac{a}{c} \right)^2 \sigma^2$

$SD \text{ of } Y ( \sigma_y ) = \sqrt{\left( \frac{a}{c} \right)^2 \sigma^2}$

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

$ac < 0$

$\Rightarrow a < 0 \text{ or } c < 0$

$\therefore \left| \frac{a}{c} \right| = - \frac{a}{c}$

$\Rightarrow \sigma_Y = - \frac{a}{c} \sigma_X$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Q 14 | Page 51