s-assignment2

# S-assignment2 - Solution for assignment 2 1 a = 2 P(X 2 = 1 P(X 1 = 0.594 b = 4 P(X = 0 = e4 = 0.018 c P(X = 0|10 calls before = P(X = 0 = 0.018

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Solution for assignment # 2 1. a) λ = 2 , P ( X > 2) = 1 - P ( X 6 1) = 0 . 594 b) λ = 4 , P ( X = 0) = e - 4 = 0 . 018 c) P ( X = 0 | 10 calls before) = P ( X = 0) = 0 . 018, because the # of events in any interval are independent from that of any other mutually exclusive interval d) If we wait k-hours, then W =# of calls in the next k-hours satisﬁes W P (4 k ) so P ( W > 1) = 1 - P ( W < 1) = 1 - P ( W = 0) = 1 - e - 4 k > 0 . 90 k > - ln(0 . 10) / 4 = 0 . 58, which is almost 35 minutes. 2. a) P ( X 6 1 2 ) = R 1 2 0 2 e - 2 x dx = - e - 2 x ± 1 2 0 = 0 . 632 b) P ( X > 1) = R 1 2 e - 2 x dx = - e - 2 x ± 1 = 0 . 135 c) P ( 1 2 6 X 6 1) = - e - 2 + e - 1 = 0 . 233 d) F ( x ) = R x 0 2 e - 2 y dy = 1 - e - 2 x 3. a) R 4 0 cxdx = cx 2 2 ± 4 0 = c (8) = 1 c = 1 8 b) R a 0 x 8 dx = x 2 16 ± a 0 = a 2 16 = 0 . 5 a = 8 = 2 . 828 c) R b 0 x 8 dx = b 2 16 = 0 . 95 b = 3 . 899 4. X 1 N (3 , (0 . 1) 2 ) , X 2 N (3 . 04 , (0 . 02) 2 ) a) P (2 . 9 < X 1 < 3 . 1) = P ( 2 . 9 - 3 0 . 1 < Z < 3 . 1 - 3 0 . 1

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## This note was uploaded on 09/27/2010 for the course STAT 104157 taught by Professor Jhonbran during the Spring '09 term at California Coast University.

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S-assignment2 - Solution for assignment 2 1 a = 2 P(X 2 = 1 P(X 1 = 0.594 b = 4 P(X = 0 = e4 = 0.018 c P(X = 0|10 calls before = P(X = 0 = 0.018

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