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1 of 3 Beautiful Homework No. 4 Engineering 323 Problem 3-13 Bridges Problem Statement 3-13: A mail-order computer business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the probability mass function of X is as given in the accompanying table. x 0 1 2 3 4 5 6 P(x) .10 .15 .20 .25 .20 .06 .04 Probability Mass Function 0.3 P(x) values 0.25 0.2 0.15 p(x) 0.1 0.05 0 0 1 2 3 4 5 6 7 x values Getting Started: The probability mass function (pmf) of a discrete random variable X, is defined for every number x by p(x) = P(X = x) = P(all s ε S : X(s) = x) For every possible value x of X, P(X = x) specifies the probability of observing that value when the experiment is performed. Now using the table we get, Beautiful Homework No. 4 2 of 3 Engineering 323 Problem 3-13 Bridges P(X=0) = p(0) = .10 P(X=1) = p(1) = .15 P(X=2) = p(2) = .20 P(X=3) = p(3) = .25 P(X=4) = p(4) = .20 P(X=5) =p(5) = .06 P(X=6) = p(6) = .04 To calculate the probability of each of the following events, we need to find out what probabilities are in the set given then add them up. a. {At most 3 lines are in use} This set includes the probability of 0, 1, 2, and 3. Therefore, P(X ≤ 3) = p(0) +p(1) + p(2) + p(3) = .1 + .15 + .2 + .25 = .7 b. {Fewer than 3 lines are in use} This question is similar to part (a), except 3 is not included. P(X < 3) = p(0) + p(1) + p(2) = .1 + .15 + .2 = .45 c. {At least 3 lines are in use} This is the compliment to part (b) or we can add the probabilities of 3, 4, 5, and 6. P(X ≥3) = p(3) + p(4) + p(5) + p(6) = .25 + .2 + .06 + .04 = .55 Or P(X ≥ 3) = 1 – P(X < 2 ) = 1 - .45 = .55 d. {Between 2 and 5 lines, inclusive, are in use} Again this is just adding up the probabilities we want, in this case 2, 3, 4, and 5. Beautiful Homework No. 4 3 of 3 Engineering 323 Problem 3-13 Bridges P(2 ≤ X ≤ 5) = p(2) + p(3) + p(4) + p(5) = .2 + .25 + .2 + .06 = .71 e. {Between 2 and 4 lines, inclusive, are not in use} This problem has to be handled a bit differently. Because our random variable is defined in terms of lines in use, we need to interpret our given set in those terms. Notice if 2 lines are not in use then 4 are in use. If 3 lines are not in use then 3 lines are in use. If 4 lines are not in use, then 2 lines are in use. Looking at it this way our new set becomes {Between 4 and 2 lines, inclusive, are in use}. Now that we have our problem in terms of lines in use like our random variable we can calculate like before in part (d). P(4 ≥ X ≥ 2) = p(4) + p(3) + p(2) = .2 + .25 + .2 = .65 f. {At least 4 lines are not in use} Again we need to define are set in terms of lines in use. This works out to be {At most 2 lines are in use}. P(X ≤ 2) = p(0) + p(1) + p(2) = .1 + .15 + .2 = .45