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Asst7modified_shiehn - York University MATH 2030 3.0(Elementary Probability Assignment 7 Solutions Nov 2011 3.3 No 3(a E[2X 3Y = 2E[X 3E[Y = 2 1 3

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York University MATH 2030 3.0 (Elementary Probability) Assignment 7 – Solutions, Nov 2011 § 3.3 No. 3 (a) E [2 X + 3 Y ] = 2 E [ X ] + 3 E [ Y ] = 2 × 1 + 3 × 1 = 5 (b) Var[2 X + 3 Y ] = Var[2 X ] + Var[3 Y ] by independence. This then equals 2 2 Var[ X ] + 3 2 Var[ Y ] = 4 × 2 + 9 × 2 = 26. (c) By independence E [ XY Z ] = E [ X ] E [ Y ] E [ Z ] = 1 × 1 × 1 = 1. (d) Var[ XY Z ] = E [( XY Z ) 2 ] - E [ XY Z ] 2 = E [ X 2 Y 2 Z 2 ] - 1 2 by part (c). By independence this equals E [ X 2 ] E [ Y 2 ] E [ Z 2 ] - 1. But E [ X 2 ] = Var[ X ] + E [ X ] 2 = 2 + 1 2 = 3 and the same is true for the other two variables. Therefore Var[ XY Z ] = 3 × 3 × 3 - 1 = 26 . § 3.3 No. 14 (a) Let X be the income of a family picked uniformly at random from the given area. So E [ X ] = 10 , 000. We are asked to estimate the percentage of families with incomes over 50,000, or in other words to estimate P ( X > 50 , 000). Since X 0 we can use Markov’s inequality, and get P ( X > 50 , 000) E [ X
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This note was uploaded on 01/23/2012 for the course MATH 2030 taught by Professor Salisbury during the Spring '08 term at York University.

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Asst7modified_shiehn - York University MATH 2030 3.0(Elementary Probability Assignment 7 Solutions Nov 2011 3.3 No 3(a E[2X 3Y = 2E[X 3E[Y = 2 1 3

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