20065ee131A_1_EE131AF06HW4Solution

20065ee131A_1_EE131AF06HW4Solution - Solution of Assignment...

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Unformatted text preview: Solution of Assignment 4, EE 131A Due: Monday November 13, 2006 Problem 1. If the distribution function of a discrete random variable X is given by F X ( x ) = if x < , 1 2 if x < 1 , 3 5 if 1 x < 2 , 4 5 if 2 x < 3 , 9 10 if 3 x < 3 . 5 , 1 if x 3 . 5 , calculate the probability mass function of X . Solution: The discontinuity points of F X ( x ) is the set S X , the range of X . Therefore, S X = { , 1 , 2 , 3 , 3 . 5 } ; and p X (0) = 1 2- 0 = 1 2 , p X (1) = 3 5- 1 2 = 1 10 , p X (2) = 4 5- 3 5 = 1 5 , p X (3) = 9 10- 4 5 = 1 10 , p X (3 . 5) = 1- 9 10 = 1 10 . Problem 2. The distribution function of a random variable X is defined as follows: F X ( x ) = if x < , . 2 + 0 . 5 x 2 if x < 1 , 1 if x 1 . (a) What type of random variable is X ? (b) Find the following probabilities: P [ X <- . 5] P [ X < 0] P [ X 0] P [0 . 3 X < 1] P [0 . 3 X 1] P [ X > . 5] P [ X = 1] P [ X 5] P [ X < 5] Solution: (a) X is a random variable of mixed type since it is continuous except for discontinuity at 0 and at 1. (b) P [ X <- . 5] = F X (- . 5- ) = 0 P [ X < 0] = F X (0- ) = 0 P [ X 0] = F X (0) = 0 . 2 P [0 . 3 X < 1] = F X (1- )- F X (0 . 3- ) = 0 . 7- ( . 2 + 0 . 5(0 . 3) 2 ) = . 455 P [0 . 3 X 1] = F X (1)- F X (0 . 3- ) = 1- ( . 2 + 0 . 5(0 . 3) 2 ) = 0 . 755 P [ X > . 5] = 1- F X (0 . 5) = 1- ( . 2 + 0 . 5(0 . 5) 2 ) = 0 . 675 P [ X = 1] = F X (1)- F X (1- ) = 1- . 7 = 0 . 3 P [ X 5] = 1- F X (5- ) = 1- 1 = 0 P [ X < 5] = F X (5- ) = 1 Problem 3. Three balls are to be randomly selected without replacement from an urn containing balls numbering 1 through 20. The random variable X is defined as the largest selected number.is defined as the largest selected number....
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20065ee131A_1_EE131AF06HW4Solution - Solution of Assignment...

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