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Section 10.5 Let X be any random variable with (finite) mean and (finite) variance 2. We shall assume X is a continuous type random variable with p.d.f. f(x), but what follows applies to a discrete type random variable where integral signs are replaced by summation signs and the p.d.f. is replaced by a p.m.f. For any k 1, we observe that 2 = E[(X – )2] = (x – )2f(x) dx = – (x – )2f(x) dx {x : |x – | k} (x – )2f(x) dx {x : |x – | k} + {x : |x – | < k} k2 2 f(x) dx = {x : |x – | k} (x – )2f(x) dx k2 2 f(x) dx {x : |x – | k} We now have that 2 k2 2 P(|X – | k) which implies that P(|X – | k) 1 — k2 This is Chebyshev’s inequality and is stated in Theorem 10.5-1. We may also write P(|X – | < k) 1 1 – —2 k 1. Let X be a random selection from one of the first 9 positive integers. (a) Find the mean and variance of X. = E(X) = 5 2 20 = Var(X) = — 3 (b) Find P(|X – 5| 4) . 2 P(|X – 5| 4) = P(X = 1 X = 9) = — 9 (c) If Y is any random variable with the same mean and variance as X, find the upper bound on P(|Y – 5| 4) that we get from Chebyshev’s inequality. 20/3 5 P(|Y – 5| 4) = P[|Y – 5| 4(3/20)1/2(20/3)1/2] —— = — 16 12 k 2. Let X be a random variable with mean 100 and variance 75. (a) Find the lower bound on P(|X – 100| < 10) that we get from Chebyshev’s inequality. 1 1 P(|X – 100| < 10) = P[|X – 100| < (2/3)(53)] 1 – ——–2 = — (2/3) 4 k (b) Find what the value of P(|X – 100| < 10) would be, if X had a U(85 , 115) distribution. 20 2 P(|X – 100| < 10) = P(90 < X < 110) = — = — 30 3 3. Let Y have a b(n, 0.75) distribution. (a) Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from Chebyshev’s inequality when n = 12. When n = 12, E(Y) = 9 , and Var(Y) = 2.25, and P(|Y / n – 0.75| < 0.05) = P(|Y – 9| < 0.6) = P[|Y – 9| < (0.4)(1.5)] (A lower bound cannot be found, since 0.4 < 1.) (b) Find the exact value of P(|Y / n – 0.75| < 0.05) when n = 12. When n = 12, P(|Y / n – 0.75| < 0.05) = P(|Y – 9| < 0.6) = P(Y = 9) = 0.6488 – 0.3907 = 0.2581 (c) Find the lower bound on P(|Y / n – 0.75| < 0.05) that we get from Chebyshev’s inequality when n = 300. When n = 300, E(Y) = 225 , and Var(Y) = 56.25, and P(|Y / n – 0.75| < 0.05) = P(|Y – 225| < 15) = 1 P[|Y – 225| < (2)(7.5)] 1 – —2 = 0.75 2 (d) By using the normal approximation, show that when n = 300, then P(|Y / n – 0.75| < 0.05) = 0.9464.