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278 Chapter 4 Special Distributions Figure 4.7.6 Frequency 25 m = $117.00 y = $159.10 20 15 10 5 0 $100 $200 Warranty cost ($200) $300 $400 $500 $600 Simulated repair costs The histogram in Figure 4.7.6 shows the distribution of repair costs incurred in one hundred simulated two-year periods, one being the sequence of events detailed in Figure 4.7.5. There is much that it tells us. First of all (and not surprisingly), the warranty costs more than either the median repair bill (= $117.00) or the mean repair bill (= $159.10). The customer, in other words, will tend to lose money on the optional protection, and the company will tend to make money. On the other hand, a full 33% of the simulated two-year breakdown scenarios led to repair bills in excess of $200, including 6% that were more than twice the cost of the warranty. At the other extreme, 24% of the samples produced no maintenance problems whatsoever; for those customers, the $200 spent up front is totally wasted! So, should you buy the warranty? Yes, if you feel the need to have a ï¬nancial cushion to offset the (small) probability of experiencing exceptionally bad luck; no, if you can afford to absorb an occasional big loss. Appendix 4.A.1 Minitab Applications Examples at the end of Chapter 3 and earlier in this chapter illustrated the use of Minitabâs PDF, CDF, and INVCDF commands on the binomial, exponential, and normal distributions. Altogether, those same commands can be applied to more than twenty of the probability distributions most frequently encountered, including the Poisson, geometric, negative binomial, and gamma pdfs featured in Chapter 4. Recall the leukemia cluster study described in Case Study 4.2.1. The dataâs interpretation hinged on the value of P(X â¥ 8), where X was a Poisson random variable k , k = 0, 1, 2, . . . . The printout in Figure 4.A.1.1 shows with pdf, p X (k) = eâ1.75 (1.75) k! the calculation of P(X â¥ 8) using the CDF command and the fact that P(X â¥ 8) = 1 â P(X â¤ 7). Figure 4.A.1.1 MTB > cdf 7; SUBC > poisson 1.75. Cumulative Distribution Function Poisson with mean = 1.75 x P(X <= x) 7 0.999532 MTB > let k1 = 1 - 0.999532 MTB > print k1 Data Display k1 0.000468000