131A_1_hw6_solution

# 131A_1_hw6_solution - EE 131A Probabilities Instructor...

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EE 131A Problem Set #6 Probabilities Instructor: Vwani Roychowdhury 1. Problem 3 . 46. P [ X 1 > x ] = Q ( x - 20 4 ) P [ X 2 > x ] = Q ( x - 22 1 ) P [ X 1 > 20] = Q ( 20 - 20 4 ) = 0 . 5 P [ X 2 > 20] = Q ( 20 - 22 1 ) = 0 . 9722 P [ X 1 > 24] = Q ( 24 - 20 4 ) = 0 . 159 P [ X 2 > 24] = Q ( 24 - 22 1 ) = 0 . 023 Note that in the second case, the chip with the smaller mean (but larger variance) is selected. 2. Problem 3.47 λ = 1. X = Y 1 + Y 2 + Y 3 + Y 4 + Y 5 where Y i are exponential random variables, X is Erlang random variable with m = 5. P [ X < 6] = 1 - e - 6 4 X k =0 6 k k ! = 0 . 7949 P [ X < 8] = 1 - e - 8 4 X k =0 8 k k ! = 0 . 09963 3. Problem 3 f X ( x ) = e - ( x - η ) 2 2 σ 2 2 πσ 2 1

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(a) P [2 < X < 5] = Z 5 2 f X ( x ) dx = Z 5 - η σ 2 - η σ e - - x 2 2 2 π = - Q (2 / 3) + Q ( - 1 / 3) = Q (2 / 3) - Q (1 / 3) + 1 (b) P [ X > 0] = Z 0 f X ( x ) dx = Z - η σ e - x 2 2 2 π = 1 - Q (3 / 3) (c) P [6 < | X - 3 | ] = P [ X > 9] + P [ X < - 3] = Z 9 - η σ e - x 2
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## This note was uploaded on 03/27/2012 for the course EE EE 131A taught by Professor Roychowdhary during the Spring '05 term at UCLA.

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131A_1_hw6_solution - EE 131A Probabilities Instructor...

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