Consider the case where a particular system is comprised of three subordinate elements. In this case, we will say the first-time yield (quality confidence) of each element is known to be .85, .90, and .95, respectively. The cross product would then be given as .85 x .90 x .95 = .612, or about 61.2 percent. 

This value can also be referred to as rolled throughput yield. As such, it is the statistical probability that all three of the system related elements will “yield,” thereby providing a quality system. In other words, it is the statistical confidence that the system will deliver its quality entitlement. From yet another perspective, it can be viewed as the likelihood of a risk-free system.

Leveling the system yield of .612 reveals the normalized yield to be .612^(1/3) = .8490, or about 84.9 percent. In this sense, the normalized yield is “kind of like” an average. From this angle, the normalized yield can be interpreted as the “typical” yield per element, given only the knowledge of rolled throughput yield and the number of elements. As conformation of this assertion, please recognize that .8490^3 = .612.

Notice that the arithmetic average provides the quantity (.85 + .90 + .95) / 3 = .850. This method of normalization is improper because .850^3 = .6141. Obviously, this value (.614) is not the rolled-throughput of the system (as we recall the previously established value of .612). Thus, we conclude the simple arithmetic average should not be used to normalize a set of yield values.

Given the aforementioned discussion, we are now free to “sigma-tize” the normalized yield. In this case, the inverse normal recognizes the probability of .8490 to be the equivalent of Z = 1.032. Of interest, this particular Z value must be recognized as a long-term measure of capability (owing to the discrete nature of its origins). By convention, we provide the short-term capability as = + 1.5 = 1.032 +1.500 = 2.532, where 1.5 is the shift correction.

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