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exam1practicesol

# exam1practicesol - STAT4101 Practice Exam 1 solutions not...

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STAT4101 Practice Exam 1 solutions not to be turned in 1a D , R , and I partition the set of Minnesotans, so by the law of total probability, P ( A ) = P ( A | D ) P ( D ) + P ( A | R ) P ( R ) + P ( A | I ) P ( I ) . We know that P ( D ) + P ( R ) + P ( I ) = 1. Let p = P ( D ). Then P ( R ) = 1 - P ( I ) - P ( D ) = 1 - 0 . 4 - p = 0 . 6 - p, so substituting this in, as well as all that we given, we have 0 . 3 = 0 . 05 p + 0 . 7(0 . 6 - p ) + 0 . 25(0 . 4) . Solving for p , we find that p = 0 . 338. So P ( D ) = p = 0 . 338 and P ( R ) = 0 . 6 - p = 0 . 262. 1b By Bayes’ Rule, P ( D | A ) = P ( A | D ) P ( D ) P ( A ) = 0 . 05(0 . 338) 0 . 3 = 0 . 056 2a No. If they were mutually exclusive, P ( A B ) = P ( A )+ P ( B ) = 1 . 1, which is impossible. 2b By independence, P ( A B ) = P ( A ) P ( B ) = 0 . 6 · 0 . 5 = 0 . 3. 2c P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 5 + 0 . 6 - 0 . 3 = 0 . 8. 2d Now P ( A B ) = P ( B | A ) P ( A ) = 0 . 3 · 0 . 6 = 0 . 18, so P ( A B ) = 0 . 5 + 0 . 6 - 0 . 18 = 0 . 92. 3a E ( X ) = x xp ( x ) = - 1(0 . 4) + 0(0 . 4) + 2(0 . 2) = 0 3b E ( X 2 ) = x x 2 p ( x ) = ( - 1) 2 (0 . 4) + 0 2 (0 . 4) + 2 2 (0 . 2) = 0 . 4 + 0 . 8 = 1 . 2, so Var( X ) = E ( X 2 ) - ( EX ) 2 = 1 . 2 - 0 2 = 1 . 2. 3c E ( X 3 ) = x x 3 p ( x ) = ( - 1) 3 (0 . 4) + 0 3 (0 . 4) + 2 3 (0 . 2) = - 0 . 4 + 1 . 6 = 1 . 2 3d E ( e tX ) = x e tx p ( x ) = 0 . 4 e - t + 0 . 4 + 0 . 2 e 2 t 3e m 0 ( t ) = - 0 . 4 e - t + 2 · 0 . 2 e 2 t m 00 ( t ) = 0 . 4 e - t + 2 2 · 0 . 2 e 2 t m 000 ( t ) = - 0 . 4 e - t + 2 3 · 0 . 2 e 2 t so E ( X 3 ) = m 000 (0) = - 0 . 4 + 8(0 . 2) = 1 . 2. 4a E (2 X - 1) = 2 EX - 1 = 2(3) - 1 = 5 4b Var(2 X - 1) = 2 2 Var( X ) = 2 2 (9) = 36 1

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STAT4101 Practice Exam 1 solutions not to be turned in 4c Var( X ) = E ( X 2 ) - ( EX ) 2 , so E ( X 2 ) = Var( X ) + (
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exam1practicesol - STAT4101 Practice Exam 1 solutions not...

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