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Unformatted text preview: STAT416: Probability — Spring 2011 Assignments 1 Homework #8 (Solution) March 31, 2011 1 Total possible points is 50 1. Ex. 6.1 (10 pts) . Let X = number of heads, then X ∼ B ( n, . 5). n = 10 , P ( X = 5) = 10 5 1 2 10 = 0 . 2461 . n = 30 , P ( X = 15) = 30 15 1 2 30 = 0 . 1445 . n = 50 , P ( X = 25) = 50 25 1 2 50 = 0 . 1123 . In general, n, P X = n 2 = n n 2 1 2 n → , as n → ∞ . 2. Ex. 6.2 (5 pts) . P ( H ) = 0 . 4, P ( H ) = 0 . 5, P ( H ) = 0 . 6. Let X = total number of heads obtained, then X = 0 , 1 , 2 , 3. P ( X = 0) = P ( TTT ) = 0 . 6 × . 5 × . 4 = 0 . 12 . P ( X = 1) = P ( TTH )+ P ( THT )+ P ( HTT ) = 0 . 6 × . 5 × . 6+0 . 6 × . 5 × . 4+0 . 4 × . 5 × . 4 = 0 . 38 . P ( X = 2) = P ( THH ) + P ( HTH ) + P ( HHT ) = 0 . 38 . P ( X = 3) = P ( HHH ) = 0 . 4 × . 5 × . 6 = 0 . 12 . So, P ( X = 0) = P ( X = 3) , and P ( X = 1) = P ( X = 2) . The distribution is not binomial....
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This note was uploaded on 04/23/2011 for the course STAT 416 taught by Professor Staff during the Spring '08 term at Purdue University.
 Spring '08
 Staff
 Probability

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