121616949-math.232 - Z-∞ f x dx Z ∞ 20 f x dx ≈ 0015...

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218 Chapter 9 Applications of Integration 0 5 10 15 20 25 0.05 0.10 0.15 . Figure 9.18 Normal density function for the defective chips example. integrals exactly; computer software has been used to approximate the integral values in this discussion.) But this is misleading: Z 10 . 5 9 . 5 f ( x ) dx 0 . 126, which is larger, certainly, but still small, even for the “most likely” outcome. The most useful question, in most circumstances, is this: how likely is it that the number of defective chips is “far from” the mean? For example, how likely, or unlikely, is it that the number of defective chips is different by 3 or more from the expected value of 10? This is the probability that the number of defective chips is less than 7 or larger than 13, namely Z 7 -∞ f ( x ) dx + Z 13 f ( x ) dx 0 . 34 . So there is a 34% chance that this happens—not small at all. Hence the 15 defective chips does not appear to be cause for alarm. How about 20? Here we compute Z 0 -∞ f ( x ) dx + Z 20 f (
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Unformatted text preview: Z-∞ f ( x ) dx + Z ∞ 20 f ( x ) dx ≈ . 0015 . So there is only a 0 . 15% chance that the number of defective chips is more than 10 away from the mean; this would typically be interpreted as too suspicious to ignore—it shouldn’t happen if the process is running normally. The big question, of course, is what level of improbability should trigger concern? It depends to some degree on the application, and in particular on the consequences of getting it wrong in one direction or the other. If we’re wrong, do we lose a little money? A lot of money? Do people die? In general, the standard choices are 5% and 1%. So what we should do is find the number of defective chips that has only, let us say, a 1% chance of occurring under normal circumstances, and use that as the relevant number. In other words, we want to know when Z 10-r-∞ f ( x ) dx + Z ∞ 10+ r f ( x ) dx < . 01 ....
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