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hw11_sol - IEOR 172 Probability and Risk Analysis for...

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IEOR 172: Probability and Risk Analysis for Engineers, Fall 2007 Homework 11 Solution Chapter 8 Question 2 (a) P { X 85 } ≤ E[ X ] / 85 = 15 / 17 (b) P { 65 X 85 } = 1 - P {| X - 75 | > 10 } ≥ 1 - 25 / 100 (c) P {| n i =1 X i /n - 75 | > 5 } ≤ 25 25 n so need n = 10. Question 4 (a) P { 20 i =1 X i > 15 } ≤ 20 / 15 (b) P ( 20 X i =1 X i > 15 ) = P ( 20 X i =1 X i > 15 . 5 ) P ( Z > 15 . 5 - 20 20 ) = P { Z > - 1 . 006 } . 8424 Question 13 (a) P { ¯ X > 80 } = P n ¯ X - 74 14 / 5 > 15 / 7 o P { Z > 2 . 14 } ≈ . 0162 (b) P { ¯ Y > 80 } = P n ¯ Y - 74 14 / 8 > 24 / 7 o P { Z > 3 . 43 } ≈ . 0003 (c) Using that SD ( ¯ Y - ¯ X ) = q 196 / 64 + 196 / 25 3 . 30 we have P { ¯ Y - ¯ X > 2 . 2 } = P { ( ¯ Y - ¯ X ) / 3 . 3 > 2 . 2 / 3 . 3 } P { Z > . 67 } ≈ . 2514 . (d) same as in (c) Question 18 Let Y i denote the additional number of fish that need to be caught to obtain a new type when there are at present i distinct types. Then Y i is geometric with paramter 4 - i 4 . E[ Y ] = E " 3 X i =0 Y i # = 1 + 4 3 + 4 2 + 4 = 25 3 Var( Y ) = Var 3 X i =0 Y i ! = 4 9 + 2 + 12 = 130 9 Hence, P | Y - 25 3 | > 25 3 s 1300 9 1 10 1
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and so we can take a = 25 - 1300 3 , b = 25+ 1300 3 . Also, P Y - 25 3 > a 130 130 + 9 a 2 = 1 10 when a = 1170 3 . Hence P ( Y > 25 + 1170 3 ) 1 .
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