hw07soln

# hw07soln - ECE 310 Spring 2005 HW 7 solutions Problem E6.7...

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ECE 310, Spring 2005, HW 7 solutions Problem E6.7 According to what was derived in pages 182-184 of the textbook, we have, substituting n = 4, k = 2, a = 3 and b = 3: P ( K = 2) = 4 2 Γ(6)Γ(5)Γ(5) Γ(3)Γ(3)Γ(10) ) = ( 4! 2!2! ) 5!4!4! 2!2!9! = 2 7 Problem E6.8 a) Assuming the lifetime is exponentially distributed, F ( x ) = (1 - e - αx ) U ( x ): F (5000) = 1 2 1 2 = (1 - e - 5000 α ) α = ln 2 5000 Therefore, the mean chip lifetime is 1 α = 5000 ln 2 b) P ( { X < 1000 } ∪ { X > 10000 } ) = P ( X < 1000) + P ( X > 10000) = P ( X 6 1000) + (1 - P ( X 6 10000)) = F (1000) + 1 - F (10000) = 1 - e - ln 2 5 + 1 - (1 - e - 2 ln 2 ) = 1 - ( 1 2 ) 1 / 5 + ( 1 2 ) 2 Problem E6.10 Assuming Exponential distribution, and denoting by T the random variable corresponding to the duration of a call, we have: F T (1) = P ( T 1) = 1 - P ( T > 1) = 0 . 5 But, F T (1) = 1 - e - λ , so we have: 1 - e - λ = 0 . 5 λ = ln(2) Then, P ( T > 0 . 5) = 1 - F T (0 . 5) = 1 - (1 - e - 0 . 5 ln(2) ) = 2 - 0 . 5 0 . 707 Problem E6.11 Assuming that the waiting time is exponentially distributed, F ( t ) = (1 - e - αt ) U ( t ): P ( T > 1) = 1 - P ( T 6 1) = 1 - F (1) = e - α = 0 . 1 α = ln 10 So, F ( t ) = [1 - ( 1 10 ) t ] U ( t ) 1

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Problem E6.16 a) Assuming the waiting time is exponentially distributed, if the mean value is 1 ns , then λ = 1 (for time measured in
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