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STATISTICS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSTATISTICS1 Mark
MCQ
If two variates X and Y are connected by the relation $\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}},$ where a, b, c are constants such that ac < 0, then
  • A
    $\sigma\text{Y}=\frac{\text{a}}{\text{c}}\sigma\text{X}$
  • ✓
    $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
  • C
    $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}+\text{b}$
  • D
    None of these
✓

Answer

Correct option: B.
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
We know:
$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD of Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{ac}<0$
$\Rightarrow\text{a}<0\text{ or }\text{c}<0$
$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$
$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$

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