HW 2 Solution

# HW 2 Solution - STAT 350 Assignment 2 Solutions (55 Pts)...

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STAT 350 Assignment 2 Solutions ( 55 Pts ) 54. (a) (1) Let x = number of bits erroneously transmitted. Then, x is binomial with n = 20, = .10, so Proportion(x 2) = .122 + .270 + .285 = .677 (from Table II). (b) (1) Proportion(x 5) = .032 + .009 + .002 + .000 + ….+ .000 = .043. (c) (1) More than half' means 11 or more, so Proportion(x 11) = .000 + … + .000 = .000. 57. (a) (1) Proportion(x = 1) = .164 (Table III, = .2) (b) (1) Proportion(x 2) = 1 - Proportion(x 1) = 1 - (.819 + .164) = 1 - .983 = .017. (c) (1) Let y denote the number of missing pulses on 10 disks. Then, Proportion(y 2) = 1 - Proportion(y 1) = 1 - (.135 + .271) = .594 (Table III, with = 2). 58. (2) Although x has a binomial distribution with n = 1000 and = 1/200 = .005, its distribution can be approximated by a Poisson distribution with = n = 1000/200 = 5. Therefore, using Table III, Proportion(x 8) = .065 + .036 + .018 + .008 + .003 + .001 = .131. Note: because the Table entries are rounded to 3 places, you would get a slightly different answer (of .135) if you worked the problem by first adding the proportions for x 7, then subtracting from 1. Proportion( 5 x 10) = .175 + .146 + .104 + .065 + .036 + .018 = .544. 1. (a) (1) The sum of the n = 11 data points is 514.90, so x = 514.90/11 = 46.81. (b) (2) The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: 4.4 16.4 22.2 30.0 33.1 36.6 40.4 66.7 73.7 81.5 109.9. The sixth value, 36.6 is the middle, or median, value. The mean differs from the median because the largest sample observations are much further from the median than are the smallest

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## This note was uploaded on 09/21/2011 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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HW 2 Solution - STAT 350 Assignment 2 Solutions (55 Pts)...

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