MHB Chebyshev Inequality: Get Expert Help Now

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The discussion revolves around the application of Chebyshev's inequality, which states that for a random variable X with mean μ and variance σ², the probability that X deviates from μ by at least k is bounded by σ²/k². In this specific case, the variance is given as σ² = 9 and k = 3, leading to the conclusion that the probability P is less than or equal to 1. This implies that the complement, 1 - P, is non-negative. The conversation emphasizes the importance of understanding Chebyshev's inequality for probability assessments. Overall, the thread seeks clarification on applying this statistical principle to a given problem.
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Could I get some help with this question?
Please refer to the attachment.

Thanks
 

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nacho said:
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

$\chi$ $\sigma$
 

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