STAT 333 midterm1sol

# STAT 333 midterm1sol - Stat 333 : Midterm 1 Solutions -...

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Unformatted text preview: Stat 333 : Midterm 1 Solutions - Fall 2011 1. [11 points] Sheldon decides to base each decision in his life on the roll of a fair die (the rolls are independent of each other). For each of 7 days, he goes to the Cheesecake Factory where he purchases one of 6 types of cheesecake. Each day, he rolls the die and purchases cheesecake i if the outcome is i , i = 1 , 2 ,..., 6 . Define T as the number of types of cheesecake that are not purchased in these seven days. (For example, if the 6 types are pumpkin, walnut, chocolate, plain, marble, and mango, and he purchases only the plain type for each of the 7 days, then T = 5 .) The aim of this problem is to find the expectation and variance of T . Solution: This question is identical to one of the practice problems, and one done in the review session. (a) [1 points] Find the probability that cheesecake i is not purchased in these 7 days, i = 1 , 2 ,..., 6 . P(cheesecake i is not purchased in 7 days) = P(all 7 cheesecakes purchased are of the other 5 types) = ( 5 6 ) 7 (b) [2 points] Find the probability that cheesecakes i and j are not purchased in these 7 days where i 6 = j and i,j = 1 , 2 ,..., 6 . P(cheesecakes i and j are not purchased in 7 days) = P(all 7 cheesecakes purchased are of the other 4 types) = ( 4 6 ) 7 (c) [2 points] Express T as the sum of appropriately defined indicator random variables. For i = 1, 2, ..., 6, let X i = 1 , if the i-th cheesecake is not purchased in 7 days , otherwise Then T = ∑ 6 i =1 X i (d) [3 points] Find E ( T ) . E [ T ] = E [ 6 X i =1 X i ] = 6 X i =1 E [ X i ] = 6 X i =1 P ( X i = 1) = 6 X i =1 5 6 7 = 1 . 674 (e) [3 points] Find V ar ( T ) . Var( X i ) = ( 5 6 ) 7 (1- ( 5 6 ) 7 ) , Cov( X i ,X j ) = ( 4 6 ) 7- ( 5 6 ) 14 So Var( T ) = ∑ 6 i =1 Var( X i ) + 2 ∑∑ i<j Cov( X i ,X j ) = 6 * ( 5 6 ) 7 (1- ( 5 6 ) 7 ) + 2 * 15 * ( ( 4 6 ) 7- ( 5 6 ) 14 ) = 0 . 6264 1 2. [10 points] In Mystic Falls, three people, Klaus, Damon and Stefan are searching for Elena, independently of the other. Both Stefan and Damon must find Elena before Klaus does; otherwise, if Klaus finds her before either Stefan or Damon (or both), she will die. The times taken to find Elena are denoted as • X 1 for Stefan, where X 1 ∼ Exp ( 1 6 ) , • X 2 for Damon, where X 2 ∼ Exp ( 1 5 ) , • X 3 for Klaus, where X 3 has a p.d.f f X 3 ( x ) , x ≥ . The aim of this problem is to find the probability that Elena does not die. (a) [2 points] Find the joint distribution function of X 1 and X 2 . That is, find an expression for P ( X 1 ≤ x 1 ,X 2 ≤ x 2 ) where x 1 ,x 2 ≥ . Solution: P ( X 1 ≤ x 1 ,X 2 ≤ x 2 ) = P ( X 1 ≤ x 1 ) · P ( X 2 ≤ x 2 ) (by independence of X 1 and X 2 ) = (1- e- x 1 / 6 ) · (1- e- x 2 / 5 ) ,x 1 ,x 2 ≥ (b) [4 points] Express the probability that Elena does not die as an expectation with respect to X 3 . That is, find g ( X 3 ) such that P ( Elena does not die ) = E X 3 [ g ( X 3 )] Solution: The probability that Elena does not die is equal to the probability that both Stefan...
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## This note was uploaded on 01/21/2012 for the course STAT 333 taught by Professor Chisholm during the Fall '08 term at Waterloo.

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STAT 333 midterm1sol - Stat 333 : Midterm 1 Solutions -...

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