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HW 6 (Sect 5.2-5.3) Req Solutions

# HW 6 (Sect 5.2-5.3) Req Solutions - Problem Solutions for...

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Unformatted text preview: Problem Solutions for Required Homework #6 from Chapter 5, Sections 5.2 and 5.3 Required homework: Problems 11, from section 5.2, and 19 and 23 from section 5.3. (Also included is problem 17 section 5.3.) 11. a. Def. Let X be the r.v. of duration of a service call, in hours, with values x = 1, 2, 3, 4, and 5 hours. The phrase “at the same frequency” means that the outcomes of service call length are all equally likely. The probability distribution: x prob ( x ) 1 0.25 2 0.25 3 0.25 4 0.25 1.00 Notes: 1. Each probability is a number between 0 and 1, and altogether they sum to 1. Therefore, this is a valid probability distribution. 2. All the values are equally likely to occur. This is an example of a “uniform” probability distribution (the total probability is distributed evenly or “uniformly” over all the values.) b. The graph (asked here but won’t be asked on a quiz or a test): 0.10 0.20 0.30 f ( x ) x 1 2 3 4 ASW 5e Req HW #6 from Chap 5, Sect 5.2-5.3 1 c. prob ( x ) ≥ 0 for each value x = 1, 2, …, 4. Also, prob (1) + prob (2) + prob 3) + prob (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00 Therefore, this is a valid probability distribution. d. prob (3) = 0.25 e. Arrive at 3 PM; overtime if job goes past 5 PM, more than 2 hours after arrival. The values of x that are over 2 hours to qualify for overtime are x = 3 and x = 4. prob (overtime) = prob (3) + prob (4) = 0.25 + 0.25 = 0.50 Additional: The expected duration of a service call is the probability-weighted sum of the values, calculated as: x prob(x) E(x) 1 0.25 0.25 2 0.25 0.50 3 0.25 0.75 4 0.25 1.00 1.00 2.50 So the expected duration is 2.5 hours. That is, service call can be expected to take...
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HW 6 (Sect 5.2-5.3) Req Solutions - Problem Solutions for...

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