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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Statistics

If SD of $X$ is $ \sigma $ then SD of $ \large\frac{aX+b}{c}$ ( where $a,b,c$ are constants ) is

$\begin {array} {1 1} (A)\;\large\frac{a}{c} \sigma & \quad (B)\;|\large\frac{a}{c}| \sigma \\ (C)\;|\large\frac{c}{a}| \sigma & \quad (D)\;\large\frac{c}{a} \sigma \end {array}$

 

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1 Answer

Let $ Y = \large\frac{aX+b}{c}$
$ \Rightarrow \overline Y = \large\frac{a\overline X +b}{c}$
$ \Rightarrow Y-\overline Y = \large\frac{a}{c} ( X - \overline X)$
$ \large\frac{1}{N} $$ \Sigma ( Y - \overline Y )^2 = \large\frac{a^2}{c^2}$ $ \large\frac{1}{N}$$ \Sigma ( X - \overline X )^2$
$ \sigma_y^2=\large\frac{a^2}{c^2} \sigma^2$
$ \sigma_y= |\large\frac{a}{c} | \sigma$
Ans : (B)
answered Feb 1, 2014 by thanvigandhi_1
 

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