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HW14Answers(4.4) - Math331 Spring 2008 Instructor David...

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Math331, Spring 2008 Instructor: David Anderson Section 4.4 Homework Answers Homework: pgs. 173 - 174, #’s 2, 3, 7, 11, 12. 2. Suppose the person chooses to park illegally. Let X be the amount of money he will pay on a given day. We have E [ X ] = 25 * . 6 + 0 * . 4 = 15 . Therefore he can expect to pay 15 per day on average. It is better to park legally. 3. Let X be the winnings from a single ticket in dollars. Then, based on what we were given, p (1 , 200 , 000) = 1 / (2 * 10 6 ), p (800) = 500 / (2 * 10 6 ), p (30) = 4000 / (2 * 10 6 ) and p (0) = 1 - 1 / (2 * 10 6 ) - 500 / (2 * 10 6 ) - 4000 / (2 * 10 6 ). Thus, E [ X ] = 1 , 200 , 000 1 2 * 10 6 + 800 500 2 * 10 6 + 30 4000 2 * 10 6 + 0 = . 86 . 7. We are given that if X is the demand for a magazine at a specific newsstand, then R ( X ) = { 4 , 5 , 6 , 7 } . The magazine sells for $ a and costs $2 a/ 3 to the owner, and the owner can not return unsold magazines. So, how many magazines should be ordered? For each i R ( X ), let Z i represent the profit in a given week for the newsstand if the owner purchases i magazines. Then, we need E [ Z i ] for each i ∈ { 4 , 5 , 6 , 7 } . The maximum value will give us the number of magazines we should order. Therefore, E [ Z 4 ] = (4 a - 4 * (2 / 3) a )( p (4) + p (5) + p (6) + p (7) = (4 / 3) a * 1 = (4 / 3) a. E [ Z 5 ] = (4 a - 5 * (2 / 3) a ) p (4) + (5 a - 5 * 2 / 3)( p (5) + p (6) + p (7)) = (2 / 3) a 6 18 + (5 / 3) a * 12 18 = 4 3 a. E [ Z 6 ] = (4 a - 6 * (2 / 3) a ) p (4) + (5 a - 6 * (2 / 3) a ) p (5) + (6 a - 6 * 2 / 3)( p (6) + p (7)) = 0 6 18 + a 5 18 + (6 / 3)
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