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Unformatted text preview: Solutions: Assignment 4 October 4, 2005 3.28 a. E [ X ] = Σ x p ( x ) = 0(0 . 08) + 1(0 . 15) + 2(0 . 45) + 3(0 . 27) + 4(0 . 05) = 2 . 06 b. V ( X ) = X i ( x i μ ) 2 p ( x i ) = (0 2 . 06) 2 (0 . 08) + (1 2 . 06) 2 (0 . 15) + (2 2 . 06) 2 (0 . 45) ··· = 0 . 9364 c. σ x = p V ar ( x ) = √ . 9364 = 0 . 9677 d. V ( X ) = E [ X 2 ] ( E [ X ]) 2 E [ X 2 ] = X x x 2 p ( x )) = 0 2 (0 . 08) + 1 2 (0 . 15) + 2 2 (0 . 45) + 3 2 (0 . 27) + 4 2 (0 . 05) = 5 . 18 V ( X ) = 5 . 18 (2 . 06) 2 = 0 . 9364 1 3.35 Note: the deductible is the part of the damages that the INSURED has to pay. If you have a $100.00 deductible, you must pay the first $100.00 of any damages. X =damage incurred Y =what the insurance company pays; Y =damage  deductible or 0 if damage  deductible < 0. X 1000 5000 10000 p(X) 0.8 0.1 0.08 0.02 Y 500 4500 9500 Profit is $100.00, so charge $100.00 + (what the insurance company ex pects to pay) = 100 + E [ Y ]....
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This homework help was uploaded on 01/31/2008 for the course STAT 220 taught by Professor Shalizi during the Fall '05 term at Carnegie Mellon.
 Fall '05
 Shalizi
 Statistics

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