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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

     
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Solution

 \[\sigma_Y = - \frac{a}{c} \sigma_X\]

\[Y = \frac{aX + b}{c}\]

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

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

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

\[ = \frac{aX}{c} + \frac{b}{c}\]

\[\text{ We know: }  \]

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

\[ = \sigma^2 \]

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

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

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

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

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

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

\[ = \left| \frac{a}{c} \right|\sigma\]

\[ac < 0\]

\[ \Rightarrow a < 0 \text{ or }  c < 0 \]

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

\[\Rightarrow \sigma_Y = - \frac{a}{c} \sigma_X\]

 

  Is there an error in this question or solution?
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APPEARS IN

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