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Stat134FinalSols - b The number of hearts in the hand is...

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Solutions to Stat 134 Old Finals Final#1 1) a) 13 4 by using 13 indicators, 1 for each card in the hand. b) The number of hearts in the hand is hypergeometric, so ( 13 3 )( 39 10 ) ( 52 13 ) 2) Using indicators I i , one for each card, E (number of records) is E ( 10 X i =1 I i ) = 10 X i =1 E ( I i ) = 10 X i =1 1 i where 1 i is the probability that the i th card is the largest of the cards so far, in other words, the probability that the i th card is a record. 3) a) binomial(100,.4) b) E ( X ) = μ = 40, SD ( X ) = σ = npq = 4 . 9 Since σ > 3, normal approximation works well, and P ( X = 45) = P (44 . 5 X 45 . 5) = P ( 44 . 5 - 40 4 . 9 X - 40 4 . 9 45 . 5 - 40 4 . 9 ) = Φ(1 . 12) - Φ( . 92) c) negative binomial(4,.4) This is waiting until the 4th head. d) The negative binomial(4,.4) can be thought of as a sum of 4 geometric(.4) random variables, each of which has expected value 1 . 4 = 2 . 5, so E ( Z ) = 10. 4) a) Z 1 0 cx 2 dx = c 1 3 x 3 | 1 0 = c 3 = 1 since the integral of the density over entire range must equal 1. Thus c = 3. b) Using the density function, E ( X ) = Z 1 0 x (3 x 2 ) dx = 3 4 x 4 | 1 0 = 3 4 c) SD ( X ) = q var ( X ) = q E ( X 2 ) - [ E ( X )] 2 so we need only find E ( X 2 ). E ( X 2 ) = Z 1 0 x 2 (3 x 2 ) dx = 3 5 x 5 | 1 0 = 3 5 and so SD ( X ) = q 3 5 - 9 16 = q 3 80 = . 194. d) The cdf of X is F ( x ) = P ( X x ) = Z x 0 f ( x ) dx = Z x 0 3 x 2 dx = x 3 | x 0 = x 3 1
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e) Since g ( x ) is strictly increasing, f Y ( y ) = 3 x 2 . 5 x - . 5 = 6 y 5 for 0 y 1 , 0 else. 5) a) Z 0 Z x 0 λ 1 λ 2 e - λ 1 x e - λ 2 y dydx = Z 0 λ 1 e - λ 1 x (1 - e - λ 2 x ) dx = 1 - λ 1 λ 1 + λ 2 = λ 2 λ 1 + λ 2 b) P ( 1 2 < X Y < 2) = P ( . 5 Y < X < 2 Y ). Figuring out what region this is in the plane, integrate the jt density over this region. Z 0 Z 2 y . 5 y λ 2 e - λx - λy dxdy = Z 0 λe - λy ( e - λ. 5 y - e - λ 2 y ) dy = - 2 3 e - λ 1 . 5 y + 1 3 e - λ 3 y | 0 = 1 3 c) P ( T < 3) = P (2 or more hits by time 3). Using the Poisson, this is 1 - e - 6 - 6 e - 6 = 1 - 7 e - 6 .
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