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)

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.

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions