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Unformatted text preview: 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 ) = = 40, SD ( X ) = = npq = 4 . 9 Since > 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 ) = (1 . 12) ( . 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 cx 2 dx = c 1 3 x 3  1 = 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 x (3 x 2 ) dx = 3 4 x 4  1 = 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 x 2 (3 x 2 ) dx = 3 5 x 5  1 = 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 f ( x ) dx = Z x 3 x 2 dx = x 3  x = x 3 1 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 Z x 1 2 e 1 x e 2 y dydx = Z 1 e 1 x (1 e 2 x ) dx = 1 1 1 + 2 = 2 1 + 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....
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This note was uploaded on 02/12/2012 for the course UGBA 101A taught by Professor Mccullough during the Spring '08 term at University of California, Berkeley.
 Spring '08
 MCCULLOUGH

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