Chebyshev inequality

[nacho]

Could I get some help with this question?
Please refer to the attachment.

Thanks

 

[chisigma]

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 ...

Kind regards

$\chi$ $\sigma$