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CHAPTER 20 SOLUTIONS

20.1   SECTION 20.1

20.1 The first seven values of the cumulative distribution function for a Poisson(3) variable are 0.0498, 0.1991, 0.4232, 0.6472, 0.8153, 0.9161, and 0.9665. With 0.0498 ≤ 0.1247 < 0.1991, the first simulated value is x = 1. With 0.9161 ≤ 0.9321 < 0.9665, the second simulated value is x = 6. With 0.6472 ≤ 0.6873 < 0.8153, the third simulated value is x = 4.

20.2 The cumulative distribution function is

For u = 0.2, solve 0.2 = 0.25x for x = 0.8. The function is constant at 0.5 from 2 to 4, so the second simulated value is x = 4. For the third value, solve 0.7 = 0.1x + 0.1 for x = 6.

20.2   SECTION 20.2

20.3 Because 0.372 < 0.4, the first value is from the Pareto distribution. Solve for x = 100 [(1 − 0.693)−1/3 − 1] = 48.24. Because 0.702 ≥ 0.4, the second value is from the inverse Weibull distribution. Solve 0.284 = e−(200/x)2 for x = 200[− ln(0.284)]−1/2 = 178.26.

20.4 For the first year, the number who remain employees is bin(200, 0.90) and the inversion method produces a simulated value of 175. The number alive but no longer employed is bin(25, 0.08/0.10 = 0.80) and the simulated value is 22. The remaning 3 employees die during the year. For year 2, the number who remain employed is bin(175, 0.90) and the simulated value ...

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