20061ee131A_1_131A5SOL

# 20061ee131A_1_131A5SOL - EE 131A Homework #5 Solution...

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EE 131A Homework #5 Winter 2006 Solution K. Yao 1 The average number of calls in 15 minutes is (1 / 4) × 2 = 1 / 2 = λ . a. P(no calls in 15 minutes) = e - 1 / 2 = 0 . 6065 . b. P(no more than 1 call in 15 minutes) = P(no call) + P(one call) = e - 1 / 2 + (1 / 2) e - 1 / 2 = 0 . 9098. 2. (37) a. p 0 = λ 0 0! e - λ = e - 15 = 3 . 06 × 10 - 7 , λ = 15 . b. P [ N > 10] = 1 - P [ N 10] = 1 - 10 k =0 (15) k k ! e - 15 = 0 . 8815 . 3. (42) a. P [ X x ] = 1 - e - λx = r 100 , 1 - r 100 = e - λx π ( r ) = x = - 1 λ ln(1 - r 100 ) = 1 λ ln( 100 100 - r ) . π (90) 2 . 3 λ , π (95) 3 λ , π (99) 4 . 6 λ . b. P [ X x ] = 1 - Q ( x σ ) . Using Tables 3.3 and 3.4: 1 - Q ( x σ ) = 0 . 90 x σ = 1 . 28 π (90) = 1 . 28 σ 1 - Q ( x σ ) = 0 . 95 x σ 1 . 5 π (95) 1 . 5 σ 1 - Q ( x σ ) = 0 . 99 x σ 2 . 33 π (99) 2 . 33 σ 4. (44) P [ X < m ] = P [ X m ] = Φ( m - m σ ) = Φ(0) = 1 2 . P

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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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20061ee131A_1_131A5SOL - EE 131A Homework #5 Solution...

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