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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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The mean deviation for normal curve measured from mean value is

$\begin {array} {1 1} (A)\;\large\frac{3\sigma}{5} & \quad (B)\;\large\frac{4\sigma}{5} \\ (C)\;\large\frac{2\sigma}{3} & \quad (D)\;\large\frac{6\sigma}{3}\end {array}$

 

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The mean deviation from mean m is
$ = \large\frac{1}{\sigma\: \sqrt 2 \: \pi}\:\int_{-\infty}^{\infty}\: 1\: x-m/e^{-\large\frac{1}{2}} \bigg( \large\frac{x-m}{\sigma} \bigg)^2dx$
Put $ \large\frac{x-m}{\sigma} = z\: \: \: \: \: dx=\sigma\: dz$
$ = \large\frac{\sigma}{\sqrt {2\pi}} \int_{-\infty}^{\infty} | Z| \: e^{-\large\frac{1}{2}z^2}d\:z$
$ = \large\frac{2\sigma}{\sqrt {2\pi}} \int_{0}^{\infty} Z \: e^{\large\frac{1}{2}z^2}d\:z$
$ \sigma \sqrt{ \large\frac{2}{\pi}}$
$ \large\frac{4}{5} \sigma \: $ ( approx)
Ans : (B)
answered Jan 15, 2014 by thanvigandhi_1
 

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