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Unformatted text preview: Math 182 Homework 6.7 Problem 4 Suppose that f ( x ) = braceleftbigg kx 2 (1 x ) if 0 x 1 otherwise (a) Find k so that f ( x ) is a probability density function. We need k to be such that integraltext 1 f ( x ) dx = 1 Therefore, k integraldisplay 1 x 2 (1 x ) dx = k integraldisplay 1 ( x 2 x 3 ) dx = k bracketleftbigg x 3 3 x 4 4 bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x =1 x =0 = k parenleftbigg 1 3 1 4 parenrightbigg = k 12 and therefore k 12 = 1 so k = 12. (b) For this value of k , find P ( X 1 2 ) . This is just 12 integraldisplay 1 1 2 x 2 (1 x ) dx = 12 bracketleftbigg x 3 3 x 4 4 bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x =1 x = 1 2 = 12 bracketleftbigg 1 12 parenleftbigg 1 24 1 64 parenrightbiggbracketrightbigg 12 11 192 = 11 16 = 0 . 6875 (c) Find the mean. = integraldisplay 1 xf ( x ) dx = 12 integraldisplay 1 x 3 (1 x ) dx = 12 integraldisplay 1 ( x 3 x 4 ) dx = bracketleftbigg 3 x 4 12 x 5 5 bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x =1 x =0 = 3 12 5 = 15 12 5 = 3...
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This note was uploaded on 09/19/2009 for the course MATH 283 taught by Professor Mortensen during the Spring '08 term at Nevada.
 Spring '08
 MORTENSEN
 Probability

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