This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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: 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 ( Y Y Y Y + Y ) =...
View
Full
Document
 Spring '08
 Staff
 Statistics

Click to edit the document details