sec_3_6_and_4_1_sol

sec_3_6_and_4_1_sol - APPM 4/5570 Solutions to Problems...

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Unformatted text preview: APPM 4/5570 Solutions to Problems from Sections 3.6 and 4.1 82. P ( X = x ) = e- x x ! = e- . 2 . 2 x x ! , x = 0 , 1 , 2 , . . . a. P ( X = 1) = e- . 2 . 2 1 1! = e- . 2 (0 . 2) . 1637 b. P ( X 2) = 1- P ( X < 2) = 1- [ P ( X = 0) + P ( X = 1)] = 1- e- . 2 h . 2 0! + . 2 1 1! i . 01752 c. The probability that a disk does not contain a missing pulse is e- . 2 . 2 0! . 8187307531 . By independence, this probability for two disks is (0 . 8187307531)(0 . 8187307531) . 6703 86. P ( X = x ) = e- x x ! = e- 5 5 x x ! , x = 0 , 1 , 2 , . . . a. P ( X = 4) = e- 5 5 4 4! . 1755 b. P ( X 4) = 1- P ( X < 4) = 1- [ P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3)] = 1- e- 5 h 5 0! + 5 1 1! + 5 2 2! + 5 3 3! i = 1- e- 5 h 1 + 5 + 25 2 + 125 6 i . 7350 c. E 3 4 X = 3 4 E [ X ] = 3 4 5 = 15 4 3 . 75 . 88. Assuming independence of diodes, if we let X be the number that fail, then X bin (200 , . 01). (That is, X is binomial with parameters n = 200 and...
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sec_3_6_and_4_1_sol - APPM 4/5570 Solutions to Problems...

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