chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions41.a. E[X(X-1)] = E(X 2 ) – E(X), ⇒E(X 2 ) = E[X(X-1)] + E(X) = 32.5b. V(X) = 32.5 – (5) 2 = 7.5c. V(X) = E[X(X-1)] + E(X) – [E(X)] 242. With a = 1 and b = c, E(X – c) = E(aX + b) = aE(X) + b = E(X) – c. When c = µ, E(X - µ)= E(X) - µ = µ - µ = 0, so the expected deviation from the mean is zero.43.a.k 2 3 4 5 1012k.25 .11 .06 .04 .01662 ⎡ 2 ⎤ 2µ = ∑, σ = ⎢∑x ⋅ p(x)⎥ − µ = 2.37,σ =1. 54x= 0⎣ x=0 ⎦b. x ⋅ p(x)= 2. 64Thus µ - 2σ = -.44, and µ + 2σ = 5.72,so P(|x-µ| ≥ 2σ) = P(X is lat least 2 s.d.’s from µ)= P(x is either ≤-.44 or ≥ 5.72) = P(X = 6) = .04.Chebyshev’s bound of .025 is much too conservative. For K = 3,4,5, and 10, P(|x-µ| ≥kσ) = 0, here again pointing to the very conservative nature of the bound 1 .2kc. µ = 0 and σ1= , so P(|x-µ| ≥ 3σ) = P(| X | ≥ 1)31 1 1= P(X = -1 or +1) = + = , identical to the upper bound.18 18 9d. Let p(-1) = 1 124, p ( + 1) = , p(0) = .505025106