We must understand the sigma of a process is merely an equivalent unilateral figureofmerit. As such, it is always regarded (and reported) as a shortterm measure of capability. The metric can be directly computed when continuous data is at hand, or it can be statistically synthesized from discrete data. In the latter context, a reported sigma is only analogous to the Z statistic. In other instances, the sigma metric can be made to represent the normalized capability of several criticaltoquality characteristics (CTQs). To better facilitate our understanding, consider the following example.
For the first scenario, we will postulate a single random normal CTQ with a symmetrical bilateral performance specification. Respectively, we compute Z.usl.st = (USL – T) / S.st = +3.5 and Z.lsl.st = (LSL – T ) / S.st = 3.5. Of course, each Z must be transformed into a tailarea probability (by way of a table or Excel macros). Doing so reveals that p(Z.usl.st) = 1 – normsdist(Z.usl.st) = 1 – normsdist(+3.5) = 1 – .99976733 = .00023267. On the other end, we compute p(Z.lsl.st) = normsdist(Z.lsl.st) = normsdist(3.5) =.00023267.
Combining both probabilities exposes the total probability of defect. Doing so reveals 00023267 + .00023267 = .00046535. Given this, the shortterm firsttime yield was presented as Y.ft.st = 1 – .00046535 = .99953465. With this as a backdrop, the equivalent unilateral capability was given as Z.bench = Z.st = normsinv(Y.ft.st) = normsinv(.99953465) = 3.31071537. In rounded form, we accept the quantity 3.31 Sigma. Consequently, the equivalent longterm unilateral capability can be approximated as Z.lt = Z.st – 1.50 = 3.31 – 1.50 = 1.81. Translated, this is equivalent to DPMO = 35,148.
Let us now postulate a second application scenario. Suppose we examined a certain discrete CTQ and discovered the longterm defect rate to be DPMO = 4,200. Converting this quality metric to an equivalent longterm firsttime yield discloses the approximation Y.ft.lt = 1 (DPMO / 10^6) = 1 (4,200 / 10^6) = 1 – .00420 = .99580. Of course, such a longterm firsttime yield can be statistically converted to an “equivalent” longterm unilateral standard normal deviate. So doing provides the quantity Z.lt = normsinv(1 – Y.ft.lt) = normsinv(.99580) = 2.63626982, or 2.64 in rounded form. Since no shortterm information or data were available, the instantaneous (shortterm) capability was approximated as Z.bench = Z.st = Z.lt + 1.5 = 2.64 + 1.50 = 4.14. Thus, we assert the process has an inherent capability of 4.14 Sigma.
Pooling the two scenarios, we must convert the respective sigmas into firsttime yield values. For the first scenario, we have Y.ft.st.1 = .99953465. For the second scenario, we observe Y.ft.st.2 = .99998234. The shortterm rolledthroughput yield would then be computed as Y.rt.st = Y.ft.st.1 * Y.ft.st.2 = .99953465 * .99998234 = .99951700. The normalized yield would be given as Y.norm = Y.rt.st^(1/K) = .99951700^(1/2) = .99975847.
Converting Y.norm to an equivalent unilateral shortterm Z value reveals Z.norm.st = 3.49, meaning that the overall sigma of the process should be reported as 3.49. The normalized longterm unilateral capability was approximated as Z.norm.lt = Z.norm.st – 1.50 = 3.49 – 1.50 = 1.99. Interestingly, such a level of capability translates to a quality level of DPMO = 23,293. Thus, we are able to merge heterogeneous metrics for the purpose of quality reporting.


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Comments
Information given here is either wrong / print mistake…
Article: “How is the defects per Million opportunities (DPMO) to sigma level and Yield conversion chart determined?”
Para # 2 >> from Z.short term >> you got the DPMO
Para # 3 >> from same DPMO >> you calculated Z.short term
But, the numbers are not matching –>
Para # 2 >> it is shown as 3.5
Para # 3 >> it is shown as 3.31
How is that possible?