Quiz3Aans - - = 1.6 1.44 = 0.16 . SD ( X ) = 16 . = 0.4 ....

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STAT 408 Spring 2009 Version A Name ANSWERS . Section __________ Quiz 3 (10 points) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. Let X be a continuous random variable with the probability density function f ( x ) = 4 3 2 3 3 2 x x - for 0 x 2, f ( x ) = 0 otherwise. a) (2) Find P ( X < 1 ). P ( X < 1 ) = ( 29 - = - 1 0 3 2 1 0 3 2 2 4 3 4 3 2 3 dx x x dx x x = 16 5 = - = - 4 4 1 3 3 4 2 3 0 1 4 1 3 2 4 3 4 3 x x = 0.3125 . b) (2) Find E ( X ). E ( X ) = ( 29 - dx x f x = ( 29 - = - 2 0 4 3 2 0 3 2 2 4 3 4 3 2 3 dx x x dx x x x = 8 . 4 6 5 4 32 3 4 4 16 2 3 0 2 5 1 4 2 4 3 5 4 - = - = - x x = 1.2 .
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1. (continued) c) (3) Find SD ( X ). Var ( X ) = ( 29 ( 29 [ ] 2 2 X E - - dx x f x = [ ] 2 2 0 3 2 2 2 . 1 4 3 2 3 - - dx x x x = ( 29 [ ] 2 2 0 5 4 2 . 1 2 4 3 - - dx x x = [ ] 2 6 5 2 . 1 0 2 6 1 5 2 4 3 - - x x = 44 . 1 6 4 64 3 5 4 32 2 3 -
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Unformatted text preview: - = 1.6 1.44 = 0.16 . SD ( X ) = 16 . = 0.4 . 2. (3) Consider f ( x ) = ( 29 &lt; &lt;-otherwise 2 1 2 3 x x x = &lt; &lt;-otherwise 2 2 3 2 3 2 x x x Is f ( x ) a valid probability density function? Justify your answer. No credit will be given without proper justification. Circle one: f ( x ) is a valid p.d.f. f ( x ) is NOT a valid p.d.f. Even though ( 29 -x x f d = -2 2 2 3 2 3 x x x d = 2 2 3 4 3 2 1 -x x = 1, f ( x ) is NOT a valid p.d.f. since f ( x ) &lt; 0 for 0 &lt; x &lt; 1....
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Quiz3Aans - - = 1.6 1.44 = 0.16 . SD ( X ) = 16 . = 0.4 ....

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