(11) Use the commonly stated form of Chebyshev's inequality to find an upper bound to P(X > 300) when X ~ Poisson(10). (a) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. so, we have Std(X) VIO = 0.010540925. P(X > 300) = P(|X| > 300) < 300 300 (b) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. So, we have Var(X) 3002 10 = 0.00011111. 3002 P(X > 300) = P(|X| > 300) < (c) Cannot be done, since Chebyshev's inequality can only be applied when two sided tail probabilities are sought. (d) Note that P(|X – 10| > 290) = P(X > 300) + P(X < -280), since P(X < –280) = 0. So, we have P(X > 300) = P(|X – 10| > 290) < Var(X) 10 = 0.000118906. - 2902 2902 (e) None of the above. The correct proof is (a) (b) (c) (d) (e) N/A (Select One)

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(11) Use the commonly stated form of Chebyshev's inequality to find an upper bound to P(X > 300) when X ~
Poisson(10).
(a) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. so, we have
Std(X)
V10
P(X > 300) = P(|X| > 300) <
0.010540925.
300
300
(b) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. so, we have
Var(X)
3002
10
P(X > 300) = P(|X| > 300) <
0.00011111.
%3D
3002
(c) Cannot be done, since Chebyshev's inequality can only be applied when two sided tail probabilities are sought.
(d) Note that P(|X – 10| > 290) = P(X > 300) + P(X < -280), since P(X < –280) = 0. So, we have
Var(X)
10
= 0.000118906.
2902
P(X > 300) = P(|X – 10| > 290) <
2902
(e) None of the above.
The correct proof is
(a)
(b)
(c)
(d)
N/A
(Select One)
Transcribed Image Text:(11) Use the commonly stated form of Chebyshev's inequality to find an upper bound to P(X > 300) when X ~ Poisson(10). (a) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. so, we have Std(X) V10 P(X > 300) = P(|X| > 300) < 0.010540925. 300 300 (b) Note that P(X > 300) = P(X > 300) + P(X < -300), since P(X < -280) = 0. so, we have Var(X) 3002 10 P(X > 300) = P(|X| > 300) < 0.00011111. %3D 3002 (c) Cannot be done, since Chebyshev's inequality can only be applied when two sided tail probabilities are sought. (d) Note that P(|X – 10| > 290) = P(X > 300) + P(X < -280), since P(X < –280) = 0. So, we have Var(X) 10 = 0.000118906. 2902 P(X > 300) = P(|X – 10| > 290) < 2902 (e) None of the above. The correct proof is (a) (b) (c) (d) N/A (Select One)
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