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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.
 Spring '08
 ISRAEL
 Math, Probability

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