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

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

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.  = E(X) =  2 = Var(X) = 5 20 — 3 Find P(|X – 5|  4). 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. P(|X – 5|  4) =P(X = 1  X = 9) = 2 — 9 P(|Y – 5|  4) =P[|Y – 5|  4(3/20) 1/2 (20/3) 1/2 ]  20/3 5 —— =— 1612 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. P(|X – 100| < 10) = P[|X – 100| < (2/  3)(5  3)]  1 1 – ——– = (2/  3) 2 Find what the value of P(|X – 100| < 10) would be, if X had a U(85, 115) distribution. 1 — 4 P(|X – 100| < 10) = 20 — = 30 P(90 < X < 110) = 2 — 3 k 

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. Find the exact value of P(|Y / n – 0.75| < 0.05) when n = 12. When n = 12, E(Y) =, and Var(Y) =, and92.25 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.) 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. By using the normal approximation, show that when n = 300, then P(|Y / n – 0.75| < 0.05) = 0.9464. When n = 300, E(Y) =, and Var(Y) =, and22556.25 P(|Y / n – 0.75| < 0.05) =P(|Y – 225| < 15) = P[|Y – 225| < (2)(7.5)]  1 1 – — = 0.75 2 2