11ProbHw7sol

11ProbHw7sol - IEOR 3658 Probability Prof. Mariana...

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IEOR 3658 Assignment #7 Solutions Probability November 13, 2011 Prof. Mariana Olvera-Cravioto Page 1 of 3 Assignment #7 Solutions 1. (a) Since b < 0 and shape of the density is triangular, bx + c has to have a x-intercept of a and an area of one. The height of this triangle is the y -intercept: c . While the base of the triangle is a . We have two equations: ac 2 = 1 ba + c = 0 This solves to c = 2 a and b = - 2 a 2 . (b) The CDF is obtained via integration. P ( X x ) = 0 x < 0 2 x a - x 2 a 2 0 x a 1 a < x 2. (a) 1 = Z 1 0 K (1 - y ) 2 ydy = K Z 1 0 ( y - 2 y 2 + y 3 ) dy = K ± y 2 2 - 2 y 3 3 + y 4 4 ²³ ³ ³ ³ y =1 y =0 = K ± 1 12 ² so K = 12. (b) Integrating gives that the mean is Z 1 0 y (12)(1 - y ) 2 y dy = 2 5 . The second moment is Z 1 0 y 2 (12)(1 - y ) 2 y dy = 1 5 . Hence the variance is 1 5 - 4 25 = 1 25 . 3. Suppose the height of the triangle is h , then consider the CDF of X as follows: for any 0 x h , b B h x
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2 IEOR 3658, Assignment #7 Solutions 1 - F ( x ) = P ( X > x
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11ProbHw7sol - IEOR 3658 Probability Prof. Mariana...

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