This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE 310, Spring 2005, HW 7 solutions Problem E6.7 According to what was derived in pages 182184 of the textbook, we have, substituting n = 4, k = 2, a = 3 and b = 3: P ( K = 2) = 4 2 (6)(5)(5) (3)(3)(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 x ) U ( x ): F (5000) = 1 2 1 2 = (1 e 5000 ) = ln 2 5000 Therefore, the mean chip lifetime is 1 = 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 , so we have: 1 e = 0 . 5 = ln(2) Then, P ( T > . 5) = 1 F T (0 . 5) = 1 (1 e . 5ln(2) ) = 2 . 5...
View
Full
Document
This homework help was uploaded on 09/26/2007 for the course ECE 3100 taught by Professor Haas during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HAAS

Click to edit the document details