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Unformatted text preview: STAT 230 Test 5 July 12, 2006 3:30 4:10 pm 1. Three towels are hung outside on a clothes line to dry on a windy day. Since there are only four pegs available, the towels are hung in a line. Pegs 1 and 4 are used to secure the outer sides of towels 1 and 3, while peg 2 secures inner side towel 1 and one side of towel 2, and peg 3 secures the other side of towel 2 and the inner side of towel 3. Each peg has a probability . 3 of coming loose in the wind, independently of the other pegs, and a towel will fall to the ground if both of the pegs securing it come loose. Let X i = 1 if towel i falls, and 0 otherwise, for i = 1 , 2 , 3. Let X = X 1 + X 2 + X 3 be the total number of towels that fall. [2] (a) Find P ( X i = 1) for each i . Soln: P ( X i = 1) = P (Peg i and Peg i+1 come loose) = 0 . 3 2 = 0 . 09 [2] (b) Find E ( X i ) for each i , and hence find E ( X ). Soln: E ( X i ) = 0 P ( X i = 0) + 1 P ( X i = 1) = 0 . 09 Hence E ( X ) = E ( 3 i =1 X i ) = n i =1 E ( X i ) = 0 . 27 [2] (c) Find V ar ( X i ) for each i . Soln: V ar ( X i ) = E ( X 2 i ) E ( X i ) 2 . Since X i = 0 or X i = 1, X 2 i = X i , and so var ( X i ) = E ( X i ) E ( X 2 i ) = 0 . 09 . 09 2 = 0 . 0819 [2] (d) Find Cov ( X 1 ,X 2 ), Cov ( X 2 ,X 3 ) and Cov ( X 1 ,X 3 ). Soln: Cov ( X 1 ,X 2 ) = E ( X 1 X 2 ) E ( X 1 ) E ( X 2 ) = E ( X 1 X 2 ) . 09 2 . Now E ( X 1 X 2 ) = P (towels 1 and 2 fall) = P (Pegs 1, 2 and 3 come loose) = 0 . 3 3 So Cov ( X 1 ,X 2 ) = 0 . 3 3 . 09 2 = 0 . 0189. By symmetry, Cov ( X 2 ,X 3 ) = Cov ( X 1 ,X 2 ) = . 0189. For Cov ( X 1 ,X 3 ), E ( X 1 X 3 ) = P (towels 1 and 3 fall) = P (All 4 pegs come loose) = 0 . 3 4 So Cov ( X 1 ,X...
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