Let X be a discrete random variable with the following probability distribution: X = 0 | P(X = 0) = .1 X = 1 | P(X = 1) = .15 X = 2 | P(X = 2) = .2 X = 3 | P(X = 3) X = 4 | P(X= 4) = .2 X = 5 | P(X= 5) = .06 X = 6 | P(X= 6) = .04 (1) (2) %3D (3) %3D .25 (4) %3D (5) %3D 1.) Compute the Expected Value E(X) of X. 2.) Compute the Variance Var(X) of X. 3.) Compute the upper bound for the probability P(|X – E(X)| > 30)

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Chapter1: Combinatorial Analysis
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Let X be a discrete random variable with the following probability distribution:
X = 0 | P(X = 0) = .1
х 3D1 | Р(Х — 1) %3 .15
(1)
(2)
X = 2 | P(X= 2) = .2
X = 3 | P(X = 3) = .25
X = 4 | P(X= 4) = .2
X = 5 | P(X = 5) = .06
X = 6 | P(X = 6) = .04
(3)
%3D
(4)
(5)
(6)
(7)
%3D
1.) Compute the Expected Value E(X) of X.
2.) Compute the Variance Var(X) of X.
3.) Compute the upper bound for the probability P(|X – E(X)| > 30)
Transcribed Image Text:Let X be a discrete random variable with the following probability distribution: X = 0 | P(X = 0) = .1 х 3D1 | Р(Х — 1) %3 .15 (1) (2) X = 2 | P(X= 2) = .2 X = 3 | P(X = 3) = .25 X = 4 | P(X= 4) = .2 X = 5 | P(X = 5) = .06 X = 6 | P(X = 6) = .04 (3) %3D (4) (5) (6) (7) %3D 1.) Compute the Expected Value E(X) of X. 2.) Compute the Variance Var(X) of X. 3.) Compute the upper bound for the probability P(|X – E(X)| > 30)
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