Week 9 - Chapter 4 Continuous Random Variables and...

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Chapter 4: Continuous Random Variables and Probability Distributions 134 12. a. P(X < 0) = F(0) = .5 b. P(-1 X 1) = F(1) – F(-1) = 6875 . 16 11 = c. P(X > .5) = 1 – P(X .5) = 1 – F(.5) = 1 - .6836 = .3164 d. F(x) = F (x) = - + 3 4 32 3 2 1 3 x x dx d = ( 29 2 2 4 09375 . 3 3 4 32 3 0 x x - = - + e. ( 29 5 . ~ = m F by definition. F(0) = .5 from a above, which is as desired. 13. a. 3 3 1 ) 1 )( 3 ( 0 1 3 3 1 1 1 1 4 = = - - = - - = = k k k k dx x k x b. cdf: F(x)= 3 3 1 1 4 1 1 1 3 3 3 3 ) ( x x dy y dy y f y x x x - = + - = - - = = - - - . So ( 29 1 1 , 1 , 0 3 - = - x x x x F c. P(x > 2) = 1 – F(2) = 1 – ( 29 8 1 8 1 1 = - or .125; ( 29 ( 29 088 . 875 . 963 . 1 1 ) 2 ( ) 3 ( ) 3 2 ( 8 1 27 1 = - = - - - = - = < < F F x P d. 2 3 2 3 0 2 2 3 3 3 ) ( 1 1 3 1 4 = + = - - = = = x x dx x dx x x x E 3 3 0 1 3 3 ) ( 3 1 1 2 1 4 2 2 = + = - - = = = x x dx x dx x x x E 4 3 4 9 3 2 3 3 )] ( [ ) ( ) ( 2 2 2 = - = - = - = x E x E x V or .75 866 . ) ( 4 3 = = = x V s e. ) 366 . 2 ( ) 366 . 2 ( ) 866 . 5 . 1 866 . 5 . 1 ( F x P x P = < = + < < - 9245 . ) 366 . 2 ( 1 3 = - = -
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