hw4 solutions

Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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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 ) = 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 = V ar ( x ) = 0 . 9364 = 0 . 9677 d. V ( X ) = E [ X 2 ] - ( E [ X ]) 2 E [ X 2 ] = 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

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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 0 1000 5000 10000 p(X) 0.8 0.1 0.08 0.02 Y 0 500 4500 9500 Profit is \$100.00, so charge \$100.00 + (what the insurance company ex- pects to pay) = 100 + E [ Y ]. E [ Y ] = 0(0 . 8) + 500(0 . 1) + 4500(0 . 08) + 9500(0 . 02) = 600 Charge 600 + 100 = 700. 3.38 X =number of lots ordered E [ X ] = 1(0 . 2) + 2(0 . 4) + 3(0 . 3) + 4(0 . 1) = 2 . 3 V ( X ) = E [ X 2 ] - ( E [ X ]) 2 E [ X 2 ] = 1(0 . 2) + 4(0 . 4) + 9(0 . 3) + 16(0 . 1) = 6 . 1 V ( X ) = 6 . 1 - 2 . 3 2 V ( X ) = 0 . 81 2
Now, the number of pounds left is a function of X . Number of pounds left = 100 - 5 X , since there are 5 pounds per lot.

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