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The S.D of a variate x is S. The S.D of the variate \frac{ax+b}{c}  where a,b c are constant is

  • Option 1)

    \left ( \frac{a}{c} \right )\sigma

  • Option 2)

    |\frac{a}{c}|\sigma

  • Option 3)

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

  • Option 4)

    None of These

 

Answers (1)

best_answer

Use the concept

Standard Deviation -

If x1, x2...xn are n observations then square root of the arithmetic mean of 

\sigma = \sqrt{\frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}}

\bar{}

- wherein

where \bar{x} is mean

 

 Let y= \frac{ax+b}{c}

Then \bar{y} = \frac{1}{c}(a\bar{x}+b)

\Rightarrow y - \bar{y}=\frac{a}{c} (x-\bar{x})

\frac{1}{n}\sum (y-\bar{y})^{2} = \frac{a^{2}}{c^{2}}\cdot \frac{1}{n}\sum (x-\bar{x})^{2}

S.D of y=\sqrt{\frac{a^{2}}{c^{2}}\frac{1}{n}\sum (x-\bar{x})^{2}}

=\sqrt{\frac{a^{2}}{c^{2}}\sigma ^{2}} =\left | \frac{a}{c} \right | \sigma


Option 1)

\left ( \frac{a}{c} \right )\sigma

This solution is incorrect.

Option 2)

|\frac{a}{c}|\sigma

This solution is correct.

Option 3)

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

This solution is incorrect.

Option 4)

None of These

This solution is incorrect.

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