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My Question is:

Given that Chebyshev's inequality is $$P(|X-µ|≥ kσ)≤\frac{1}{k^2}$$ Find a lower bound on the probability that X is within two standard deviations from its mean.

I've found an upper bound for this when the probability is more than one standard deviation from its mean but cant seem to see this to crack this question.

Any help will be appreciated.

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Hint: Consider the event: $|X - \mu| < 2 \sigma$. How is $P(|X - \mu| < 2 \sigma)$ related to $P(|X - \mu| \geq 2\sigma)$?

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  • $\begingroup$ would you get $1-\frac{1}{2^2}=0.75$?? $\endgroup$ – user384716 Dec 9 '16 at 12:51
  • $\begingroup$ @OPFragster, Yep! $\endgroup$ – Marcus M Dec 9 '16 at 14:01

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