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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 financial
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
Minitab’s 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 data’s interpretation hinged on the value of P(X ≥ 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) = e−1.75 (1.75)
k!
the calculation of P(X ≥ 8) using the CDF command and the fact that P(X ≥ 8) =
1 − P(X ≤ 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