11Bprob1sol

11Bprob1sol - Solutions to HW#10 1 Suppose that X is a...

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Unformatted text preview: Solutions to HW #10 1. Suppose that X is a random variable with mean μ = 121 and variance σ 2 = 15 . Use Tchebychev’s inequality to estimate the probabilities a. P ( | X- 121 | > 25) < 15 25 2 = 0 . 024 . b. P (108 ≤ X ≤ 134) > 1- 15 13 2 ≈ . 9112 . 2. Suppose that X is the number of heads observed in 20 tosses of a fair coin. a. X ∼ B (20 , . 5) . b. Since X ∼ B (20 , . 5) we know that E ( X ) = 10 and Var ( X ) = 5 , so P ( | X- 10 | ≤ 3) > 1- 5 9 = 0 . 4444 . c. Normal approximation: P ( | X- 10 | ≤ 3) = P (7 ≤ X ≤ 13) ≈ 1 √ 2 π Z z 2 z 1 e- z 2 / 2 dz , where z 1 = (7- 10- . 5) / √ 5 ≈ - 1 . 565 and z 2 = (13- 10+0 . 5) / √ 5 ≈ 1 . 565 . There is no entry in the standard normal table for 1.565, so you can use either the entry for 1.56 or 1.57, or even better, you can take the average of those two entries. I.e., you can use A (1 . 565) ≈ (0 . 4406 + 0 . 4418) / 2 = 0 . 4412 . This gives the approximation P ( | X- 10 | ≤ 3) ≈ 2 A (1 . 565) = 0 . 8824 . d. P ( | X- 10 | ≤ 3) = 13 X k =7 20 C k (0 . 5) k (0 . 5) 20- k = 927656 2 20 ≈ . 8847 ....
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11Bprob1sol - Solutions to HW#10 1 Suppose that X is a...

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