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Section 10.5

Section 10.5

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Section 10.5

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  1. 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– )2f(x) dx {x : |x– | k} {x : |x– | <k} (x– )2f(x) dx k2 2 f(x) dx = k2 2 f(x) dx {x : |x– | k} {x : |x– | k} {x : |x– | k}

  2. We now have that 2  k2 2 P(|X– | k) which implies that 1 P(|X– | k)  — k2 This is Chebyshev’s inequality and is stated in Theorem 10.5-1. We may also write 1 P(|X– | <k)  1 – — k2

  3. 1. (a) (b) (c) Let X be a random selection from one of the first 9 positive integers. Find the mean and variance of X. 20 — 3  = E(X) = 2 = Var(X) = 5 Find P(|X – 5|  4) . 2 — 9 P(|X– 5|  4) = P(X = 1  X = 9) = 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 —— = — 16 12 P(|Y– 5|  4) = P[|Y– 5|  4(3/20)1/2(20/3)1/2]  k 

  4. 2. (a) (b) Let X be a random variable with mean 100 and variance 75. Find the lower bound on P(|X – 100| < 10) that we get from Chebyshev’s inequality. 1 1 – ——– = (2/3)2 1 — 4 P(|X– 100| < 10) = P[|X– 100| < (2/3)(53)]  k  Find what the value of P(|X – 100| < 10) would be, if X had a U(85 , 115) distribution. 20 — = 30 2 — 3 P(|X – 100| < 10) = P(90 < X< 110) =

  5. 3. (a) (b) Let Y have a b(n, 0.75) distribution. 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) = , and Var(Y) = , and 9 2.25 P(|Y / n– 0.75| < 0.05) = P(|Y– 9| < 0.6) = (A lower bound cannot be found, since 0.4 < 1.) P[|Y– 9| < (0.4)(1.5)]  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

  6. (c) (d) 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) = , and Var(Y) = , and 225 56.25 P(|Y / n– 0.75| < 0.05) = P(|Y– 225| < 15) = 1 1 – — = 0.75 22 P[|Y– 225| < (2)(7.5)]  By using the normal approximation, show that when n = 300, then P(|Y / n – 0.75| < 0.05) = 0.9464.

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