S-assignment-3-3502-S10

# S-assignment-3-3502-S10 - STAT 3502 Assignment 3 Total...

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STAT 3502 Assignment # 3 Total mark=25 Due: 19 July, 2010 prior to the start of class 1. If two random variables have the joint density f ( x,y ) = ( 6 5 ( x + y 2 ) 0 < x < 1 , 0 < y < 1 0 otherwise. a. [1] Find the probability that 0 . 2 < X < 0 . 5 and 0 . 4 < Y < 0 . 6. Sol: P (0 . 2 < X < 0 . 5 , 0 . 4 < Y < 0 . 6) = R 0 . 6 0 . 4 R 0 . 5 0 . 2 6 5 ( x + y 2 ) dxdy = 6 5 R 0 . 6 0 . 4 x 2 / 2 + xy 2 ] 0 . 5 0 . 2 = 6 5 R 0 . 6 0 . 4 0 . 105 - 0 . 3 y 2 = 6 5 [0 . 105 y + 0 . 3 y 3 3 ] 0 . 6 0 . 4 = 6 5 (0 . 0362) = 0 . 04344 b. [1] Find the joint cdf of the two random variables and use it to verify the value obtained for the probability in part a. Sol: R y 0 R x 0 6 5 ( t + w 2 ) dtdw = 6 5 R y 0 t 2 / 2 + tw 2 ] x 0 6 5 6 5 R y 0 x 2 / 2 + xw 2 dw = 6 5 ( x 2 y/ 2 + xy 3 / 3), so F ( x,y ) = 0 x 0 ory 0 3 5 x 2 y + 2 5 xy 3 0 < x < 1 , < y < 1 3 5 x 2 + 2 5 x 0 < x < y 1 3 5 y + 2 5 y 3 x < y < 1 1 x x 1 P (0 . 2 < X < 0 . 5 , 0 . 4 < Y < 0 . 6) = F (0 . 5 , 0 . 6) - F (0 . 5 , 0 . 4) - F (0 . 2 , 0 . 6)+ F (0 . 2 , 0 . 4) = (3 / 5)0 . 5 2 (0 . 6)+ (2 / 5)0 . 5(0 . 6) 3 - (3 / 5)(0 . 5) 2 0 . 6 - (2 / 5)0 . 5(0 . 4) 3 - (3 / 5)(0 . 2) 2 0 . 6 - (3 / 5)0 . 2(0 . 6) 3 + (3 / 5)(0 . 2) 2 0 . 4 + (2 / 5))0 . 2(0 . 4) 3 = 0 . 04344 c. [1] Find both marginal densities and use them to ﬁnd P ( X > 0 . 8) and P ( Y < 0 . 5). Sol:

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## This note was uploaded on 09/27/2010 for the course STAT 104157 taught by Professor Jhonbran during the Spring '09 term at California Coast University.

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S-assignment-3-3502-S10 - STAT 3502 Assignment 3 Total...

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