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# Hw5Sol - SIEO 3600(IEOR Majors Introduction to Probability...

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SIEO 3600 (IEOR Majors) Solution to Assignment #5 Introduction to Probability and Statistics February 24, 2010 Solution to Assignment #5 1. X is a continuous random variable with probability density function f X ( x ) = 1 / 2 , if 2 < x < 4; 0 , otherwise . (a) Compute P ( X < 3). (b) Compute R 0 xf ( x ) dx and R 0 x 2 f ( x ) dx . (c) Show that the cumulative distribution function F X ( x ) = 0 , if x 2; ( x - 2) / 2 , if 2 < x < 4; 1 , if x 4 . (d) Let ¯ F X ( x ) 1 - F X ( x ), show that ¯ F X ( x ) = 1 , if x 2; (4 - x ) / 2 , if 2 < x < 4; 0 , if x 4 . (e) Graph both F X ( x ) and ¯ F X ( x ). (f) Show that R 0 xf ( x ) dx = R 0 ¯ F ( x ) dx . Solution: (a) P [ X < 3] = R 3 2 1 2 dx = 1 2 . (b) The mean is E [ X ] = R 4 2 x 2 dx = 3, The second moment is E [ X 2 ] = R 4 2 x 2 2 dx = 28 3 . (c) If x < 2 we never make it to the interval [2 , 4] where f lives and thus the integral or, the probability (the total area is always 1!) under the graph is zero. If x [2 , 4] we have F ( x ) = R x -∞ f ( t ) dt = R x 2 1 2 dt = x - 2 2 . Finally, if x > 2 we have F ( x ) = R x -∞ f ( t ) dt = R 4 2 1 2 dt = 1. (d) To get ¯ F ( x ), just figure out 1 - F ( x ) in each interval. (e) Omitted. (f) Finally, we have R 0 ¯ F ( t ) dt = R 2 0 1 dt + R 4 2 4 - t 2 dt + R 4 0 dt = 2+2 * 2 - 1 2 (16 - 4) = 3 = E ( X ) given in (b). This gives us an alternative way to compute

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Hw5Sol - SIEO 3600(IEOR Majors Introduction to Probability...

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