Chapter 8 - Pg 427 Question#2 a P{X 85 E[X 85 = 15/17...

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Pg 427 Question #2 a) P{X ≥ 85} ≤ E[X] / 85 = 15/17 ≈ 0.8824 b) P{65 ≤ X 85} = 1 – P{ | X – 75 | > 10} ≥ 1 – 25/100 = ¾ c) P{ = - n i i n X 1 75 > 5} ≤ 25/25n So we need n = 10 to ensure with probability @ least 0.9 that the class average would be within 5 of 75. 1
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Pg 427 Question #3 Let Z be a distributed N(0,1). So, P{ = - n i i n X 1 75 > 5} ≈ P{ | Z | > n 0.5 } ≤ 0.1 So we need n = 3 in this case. 2
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Pg 427 Question #5 Let X i denote the i th roundoff error. So E = 50 1 i i X = 0 and Var = 50 1 i i X = 50. Since we have a uniform distribution, Var(X 1 ) = 50/12 since 0.5+X ~ Uniform(0,1). So, we can see that Var(X) = Var(1/2 + X) = 1/12. Thus it is clear: P{ = n i i X 1 > 3} ≈ P { | N(0,1) | > 3(12/50) 0.5 } using the C.L.T. = 2P{N(0,1) > 1.47} = 0.1416. 4
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Pg 427 Question #6 Let X i denote the outcome of the i th role then E[X i ] = 3.5 and Var[X i ] = 35/12 and thus P{ = 300 79 1 i i X } = P{ } 5 . 300 79 1 = i i X P{Φ(z) ≤ 12 / 35 79 ) 5 . 3 ( 79 5 . 300 x - } = P{Φ(z) ≤ 1.58} = 0.9429 7. 5
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8. Pg 427 Question #9 One needs to use the fact that a Γ(n, 1) is the sum of n independent exponential random
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