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Let $X$ have a Poisson distribution with $\mu = 100$. Use Chebyshev's inequality to determine a lower bound for $\Bbb{P}(75 < X < 125)$.

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1 Answer

The mean of X is 100 and the standard deviation of X is 10.

$P(|X-100|<2.5(10))=>1-1/2.5^2=0.84$

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P(75<X<125)=P(|X-100|<2.5(10)) –  Amr Nov 7 '12 at 11:09
    
Can you explain how? –  user48495 Nov 7 '12 at 14:15
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The variance is $10$? –  Dilip Sarwate Nov 7 '12 at 15:26
    
Sorry, I meant the standard deviation. I will edit my ansewr. –  Amr Nov 10 '12 at 12:14
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