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# Asst4-solved - Statistics 427 Homework#4 Solutions to...

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Unformatted text preview: Statistics 427: Homework #4 Solutions to Assigned Problems 3.30 [6 pts (3,3)] (a) We can just use the definition of the expected value of Y : E [ Y ] = 3 X y =0 yp Y ( y ) = 0(0 . 60) + 1(0 . 25) + 2(0 . 10) + 3(0 . 05) = 0 . 6 . (b) Now we are dealing with the function H ( Y ) = 100 Y 2 . The expected value of the surcharge is therefore E [ H ( Y )] = 3 X y =0 H ( y ) p Y ( y ) = 3 X y =0 100 y 2 p Y ( y ) = 0(0 . 60) + 100(0 . 25) + 400(0 . 10) + 900(0 . 05) = \$110 3.31 [8 pts (3,1,4)] We can calculate the variance via Var [ Y ] = E Y 2- (E [ Y ]) 2 . First we find E [ Y ] = 55 X y =45 yp Y ( y ) = 48 . 84 . Next we find E Y 2 = 55 X y =45 y 2 p Y ( y ) = 2389 . 84 , which means Var [ Y ] = 2389 . 84- (48 . 84) 2 = 2389 . 84- 2385 . 346 = 4 . 494. The standard deviation is just the square root of the variance: σ Y = p Var [ Y ] = √ 4 . 494 = 2 . 12. To find the probability that Y is within one standard deviation of its mean value, we calculate P ( μ Y- σ Y ≤ Y ≤ μ Y + σ Y ) =...
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Asst4-solved - Statistics 427 Homework#4 Solutions to...

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