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Chebyshev inequality

nacho

Active member
Sep 10, 2013
156
Could I get some help with this question?
Please refer to the attachment.

Thanks
 

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chisigma

Well-known member
Feb 13, 2012
1,704
Could I get some help with this question?
Please refer to the attachment.

Thanks
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 (Emo)...

Kind regards

$\chi$ $\sigma$