Statistical Techniques for Transportation Engineering

But

${\int }_{|x-\mu |\ge k}f(x)\mathrm{d}x=P(|x-\mu |\ge k)$

Hence we get

${\sigma }^2\ge k^2(|x-\mu |\ge k)$

or

$P(|x-\mu |\ge k)\le \frac{{\sigma }^2}{k^2}$ (4.7)

(4.7)

Also we have

$P(|x-\mu |\ge k)+P(|x-\mu |\ge k)=1$

Therefore we get

$P(|x-\mu |\le k)=1=P(|x-\mu |\ge k)$

From Eq. (4.7), we get

$P(|x-\mu |\le k)\ge 1-\frac{{\sigma }^2}{k^2}$

Hence proved.

Remark

1. $P(|x-\mu |\ge k)\ge 1-{\displaystyle \frac{{\sigma }^{}}{}}$