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solution06

# solution06 - Solutions to Homework 06 1 a From the figure...

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Solutions to Homework 06 1. a. From the figure, f X ( x ) = ( 0 | x | > a c 1 - | x | a | x | ≤ a Now, 1 = Z -∞ f X ( x ) dx = Z a - a c 1 - | x | a ! dx = 2 Z a 0 c 1 - x a dx = 2 c " x - x 2 2 a # a 0 = ac c = 1 a b. F X ( x ) = 0 for x < - a and F X ( x ) = 1 for x > a . For - a x 0, | x | = - x . Then F X ( x ) = 1 a Z x - a 1 + x 0 a ! dx 0 = 1 2 + 1 a x + x 2 2 a ! For 0 x a , | x | = x . Then F X ( x ) = Z 0 - a f X ( x 0 ) dx 0 + Z x 0 f X ( x 0 ) dx 0 = 1 2 + 1 a Z x - a 1 + x 0 a ! dx 0 = 1 2 + 1 a x - x 2 2 a ! c. 1 4 = P [ | X | < b ] = F X ( b ) - F X ( - b ) = 2 a b - b 2 2 a ! b = a 1 - s 3 4 1

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2. F X ( x ) = n X k =0 n k ! p k (1 - p ) n - k u ( x - k ) f X ( x ) = n X k =0 n k ! p k (1 - p ) n - k δ ( x - k ) a. Following the example done in class (refer to page 99, Example 3.10 of textbook) F X ( x | A ) = ( 0 x < 4 F X ( x ) - F X (4) 1 - F X (4) for x > 4 f X ( x | A ) = ( 0 x < 4 f X ( x ) 1 - F X (4) for x > 4 b. P [ X = x | A ] = P [ { X = x } ∩ A ] P [ A ] = P [ { X = x } ∩ { X > 4 } ] P [ { X > 4 } ] = P [ { X = x } ] 1 - F X (4) for x > 4 3. Since the packet arrival rate is 5 packets per second, in half a second, the average number of packets arriving is λ = 0 . 5 * 5 = 2 . 5 packets. a. P [0 packets in 0.5 second] = λ 0 0! e - λ = e - 2 . 5 = 0 . 0821 b. P [1 or less packets in 0.5 second] = P [0 packets in 0.5 second] + P [1 packet in 0.5 second] = λ 0 0! e - λ + λ 1 1! e - λ = e - 2 . 5 + 2 . 5 e - 2 . 5 = 0 . 2873 c. P [1 or more packets in 0.5 second] = 1 - P [0 packets in 0.5 second] = 1 - λ 0 0! e - λ = 1 - e - 2 . 5 = 0 . 9179 2
4. Let X be the random variable denoting the thickness of the shim. Thus we need P [ { X > 4 . 9 } or { X > 5 . 1 } ]. P [ { X < 4 . 9 } or { X > 5 . 1 } ] = 1 - P [4 . 9 X 5 . 1] = 1 - P [4 . 9 < X 5 . 1] = 1 - F X (5 . 1) + F X (4 . 9) = 1 - Φ 5 . 1 - 5 . 0 0 . 05 + Φ 4 . 9 - 5 . 0 0 . 05 = 1 - [1
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