1041110967 - Name / Khaled Abdulaziz ID/ 1041110967...

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Question 1: a) i) + - 2 0 ) 1 )( 1 ( dx x x = - 2 0 ) 1 ² ( dx x = [x 3 /3 –x ] 2 0 = 2/3 – 0 = 2/3 This function does not satisfy condition 1 and 2 .therefore, it is not legitimate probability distribution. ii) - + 1 1 ) 1 ( 5 . 1 dx x x = 1.5 - + 1 1 ) ² ( dx x x = 1.5[ 1/3 x 3 +1/2 x² ] 1 1 - = 1.5[2/3] = 1 Even though this function satisfy condition 2 .but it does not satisfy condition 1 .since x is between -1 and 1, the function give us negative value f(x) < 0. Therefore, it is not legitimate probability distribution. b) First we find the values of c. ∫ ∫ - + c dxdy x xy 0 1 1 ) 5 ( = 1 ∫ ∫ - + c dxdy xy y x 0 1 1 ) 5 ² ( = 1 => dy x x y c - + 0 1 1 3 ²] 2 / 5 3 / 1 [ =1 c ydy 0 3 / 2 =1 1/3[c²] =1 b c²=3 b c = ± 3 We need only positive c = 3 Z=(X-1)² , z only depends on x whereas x {-1 <x<1} The inverse of z is: x – 1 = ± Z x = 1 ± Z As x varies from -1 to 1 it never exceeds 1 and therefore x = 1- z .or if we draw graph Z=(X-1)² and then we will see when x is from -1 to 1 this correspond to the left branch of
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This note was uploaded on 08/31/2009 for the course FET eem taught by Professor Alan during the Spring '09 term at Multimedia University, Cyberjaya.

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1041110967 - Name / Khaled Abdulaziz ID/ 1041110967...

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