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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 3.1.3 3.2.1 3.2.4 3.3.6 3.4.2 3.4.5 3.4.7 3.4.9 and 3.4.13 Problem 3.1.3 Solution In this problem, the CDF of W is F W ( w ) = w < 5 ( w + 5) / 8 5 ≤ w < 3 1 / 4 3 ≤ w < 3 1 / 4 + 3( w 3) / 8 3 ≤ w < 5 1 w ≥ 5 . (1) Each question can be answered directly from this CDF. (a) P [ W ≤ 4] = F W (4) = 1 / 4 + 3 / 8 = 5 / 8 . (2) (b) P [ 2 < W ≤ 2] = F W (2) F W ( 2) = 1 / 4 1 / 4 = 0 . (3) (c) P [ W > 0] = 1 P [ W ≤ 0] = 1 F W (0) = 3 / 4 (4) (d) By inspection of F W ( w ), we observe that P [ W ≤ a ] = F W ( a ) = 1 / 2 for a in the range 3 ≤ a ≤ 5. In this range, F W ( a ) = 1 / 4 + 3( a 3) / 8 = 1 / 2 (5) This implies a = 11 / 3. Problem 3.2.1 Solution f X ( x ) = cx ≤ x ≤ 2 otherwise (1) (a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1. Z 2 cx dx = 2 c = 1 (2) Therefore c = 1 / 2. (b) P [0 ≤ X ≤ 1] = R 1 x 2 dx = 1 / 4 1 (c) P [ 1 / 2 ≤ X ≤ 1 / 2] = R 1 / 2 x 2 dx = 1 / 16 (d) The CDF of X is found by integrating the PDF from 0 to x ....
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This note was uploaded on 08/25/2008 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.
 Spring '08
 Duman

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