solutions04

# solutions04 - Math 302.102 Fall 2010 Solutions to...

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Unformatted text preview: Math 302.102 Fall 2010 Solutions to Assignment #4 1. (a) We find P { < X < 2 } = Z 2 f ( x )d x = Z 2 1 x- 2 d x =- x- 1 2 1 = 1- 1 2 = 1 2 . (b) We find P { < X < 2 } = Z 2 f ( x )d x = Z 2 7 e- 7 x d x =- e- 7 x 2 = 1- e- 14 . (c) We find P { < X < 2 } = Z 2 f ( x )d x = Z 1 e e 2- 5 x 2 e- x d x =- e e 2- 5 · e- x x 2 + 2 x + 2 1 = 2 e- 5 e 2- 5 . (d) We find P { < X < 2 } = Z 2 f ( x )d x = Z 2 xe- x d x = [- xe- x- e- x ] 2 = 1- 3 e- 2 . 2. (a) Notice that we must necessarily have 0 < a < 4. Since P { X ≤ a } = Z a 1 8 x d x = x 2 16 a = a 2 16 , we find that in order for P { X ≤ a } = 1 / 2 we must have a 2 16 = 1 2 implying that a 2 = 8. The restriction that a > 0 implies that the unique value of a such that P { X ≤ a } = 1 / 2 is a = √ 8. (b) As in (a) , we must necessarily have 0 < a < 4. Since P { X ≥ a } = Z 4 a 1 8 x d x = x 2 16 4 a = 1- a 2 16 , we find that in order for P { X ≤ a } = 1 / 2 we must have 1- a 2 16 = 1 4 implying that a 2 = 12. The restriction that a > 0 implies that the unique value of a such that P { X ≥ a } = 1 / 4 is a = √ 12....
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## This note was uploaded on 04/07/2011 for the course MATH 302 taught by Professor Israel during the Spring '08 term at UBC.

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solutions04 - Math 302.102 Fall 2010 Solutions to...

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