# Chebyshev inequality

#### nacho

##### Active member
Could I get some help with this question?

Thanks

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#### chisigma

##### Well-known member
Could I get some help with this question?
The Chebysheff inequality extablishes that for a r.v. X with mean $\mu$ and variance $\sigma^{2}$ for $k \ge 0$ is...
$\displaystyle P \{|X - \mu| \ge k \} \le \frac{\sigma^{2}}{k^{2}}\ (1)$
In Your case is $\sigma^{2}= 9$ and $k = 3$, so that the (1) supplies $\displaystyle P \le 1 \implies 1 - P \ge 0$... an 'information' we have independently from Mr Chebysheff ...
$\chi$ $\sigma$