# CHAPTER 20

# 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 = 0.25*x* 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 = 0.1*x* + 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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