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Unformatted text preview: X i represent an error for the i th bit. Then we have E ( X i ) = 0 . 15 and V ar ( X i ) = 0 . 1275. The total number of error bits is given as S 100 = X 1 + + X 100 , with the normalized form: Z 100 = S 100 10 = S 10015 3 . 57 Therefore we have the probability of having 20 or less bit errors to be: P ( S 100 20) = P ( Z 100 2015 3 . 57 ) 1Q (1 . 4) = 0 . 92 4. Chebyshev bound . Problem 7.17 , page 404 of ALG The mean of a die toss is: = 3 . 5. The variance of a die toss is: 2 = 2 . 92. We can then represent the probability of having the sample mean between 60 / 20 = 3 and 80 / 20 = 4 by Chebyshev inequality as: P (  M n3 . 5  < . 5) 1 2 n 2 12 . 91 20 . 5 2 = 0 . 418 2...
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This note was uploaded on 02/03/2011 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.
 Fall '08
 LORENZELLI

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