Homework2key - 1 Homework 2 (12 pts) due Sept. 9 (1.5 pts.)...

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Unformatted text preview: 1 Homework 2 (12 pts) due Sept. 9 (1.5 pts.) 1.54 (p.54-55). Suppose that 10% of all bits transmitted through a digital communication channel are erroneously received and that whether any particular bit is erroneously received is independent of whether any other bit is erroneously received. Consider sending a very large number of messages, each consisting of 20 bits. This is a binomial distribution with n = 20, = 0.1 a) What proportion of these messages will have at most 2 erroneously received bits? prop (x 2) = p(0) + p(1) + p(2) = ???? ≤ 2 ¡ = ? ¡ + ? 1 ¡ + ? 2 ¡ = 20! 0! 20! 0.1 (0.9) 20 + 20! 1! 19! 0.1 1 (0.9) 19 + 20! 2! 18! 0.1 2 (0.9) 18 = (0.9) 20 + 20 ¡ 0.1 ¡ 0.9 ¡ 19 + 190 ¡ (0.1 2 0.9 ¡ 18 = 0.1216 + 0.2702 + 0.2852 = 0.6769 b) What proportion of these messages will have at least 5 erroneously received bits? ???? ≥ 5 ¡ = 1 − ???? ≤ 4 ¡ = 1 − ¢? ¡ + ? 1 ¡ + ? 2 ¡ + ? 3 ¡ + ? 4 ¡£ = 1 − 0.1216 + 0.2702 + 0.2852 + 0.1901 + 0.0898 ¡ = 1 − 0.9568 = 0.0432 c) For what proportion of these messages will more than half the bits be erroneously received? prop(x > 10) = 1 – prop(x 10) = 1 – [p(0) + p(1) + + p(10)] = 1 – (0.9568 + 0.0319 + 0.0089 + 0.0020 + 0.0004 + 0.0001 +0.0000) = 1 – 1.000 = 0 (1 pt.) 1.56ab (p.55) Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter = 20. In the long run, in what proportion of time periods will the number of drives This Poisson distribution has a mass function of ? ¡ = ¤ − 20 20 ! a) be at most 10? prop (x 10) = p(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8) + p(9) + p(10) = 0.0000 + 0.0000 + 0.0000 + 0.0000 + 0.0000 +0.0001 + 0.0002 + 0.0005 + 0.0013 + 0.0029 + 0.0058 = 0.0108 2 b) exceed 20? prop(x > 20) = 1 – prop(x 20) = 1 – [prop(x 10) + p(11) + p(12) + p(13) + p(14) + p(15) + p(15) + p(16) + p(17) + p(18) + p(19) + p(20)] = 1 – [0.0108 + 0.0058 + 0.0106 + 0.0176 + 0.0271 + 0.0387 + 0.0516 + 0.0646 + 0.0760 + 0.0844 + 0.0888 + 0.0888] = 1 – 0.5591 = 0.4409 Note: If you add up all of the numbers greater than 20 in the table, you will get an answer of 0.428 because not all of the numbers are there....
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This note was uploaded on 03/26/2012 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue University.

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Homework2key - 1 Homework 2 (12 pts) due Sept. 9 (1.5 pts.)...

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