Yield to Sigma Relationship

Using the process shown an example of yield to sigma relationships are calculated:

Yield


The probability of having a defect in Process 1:

P(1) = 1 – TPY

Simply says that 100% minus the % of not having a defect from either scrap or rework in ONLY Process 1 is the probably of having a defect.

P(1) = 1 – 0.80 = .20 = 20%

Z-table for 0.20 is approximately 0.84 sigma. This is considered a long term sigma value.

For an entire series of processes use the RTY in place of TPY above. The probability of having a defect in the entire series is:

P(All) = 1 – RTY = 1 – 0.475 = 0.525 = 52.5%

Z-table for 0.525 is less than 0 sigma, long term. Over half will be defective. A negative sigma value means that most of the process is performing outside your customer's specification range (LSL and USL).



Using Excel to calculate DPMO and Sigma

There are a couple commonly used Microsoft Excel functions that convert Defects per Million Opportunities (DPMO) to a process sigma (z-score) and vice versa.

To convert from DPMO to process sigma:

Process Sigma = NORMSINV * ( 1 - ( DPMO / 1,000,000)) + 1.5

To convert from process sigma to DPMO:

DPMO = 1,000,000 * ( 1 - ( NORMSDIST * (process sigma - 1.5 ))

CAUTION:

The 1.5 shift is subjective, but some experts use this as conversion from long to short term performance estimates (and vice versa). You can use another number for your estimation.

NOTE:

A Six Sigma process refers to the process short-term performance or how it is performing currently. When referring to DPMO of the process, we are referring to long-term or projected performance behavior.

A six sigma level of performance has 3.4 defects per million opportunities (3.4 DPMO). A current six sigma process now will have a estimated shift of 1.5 sigma (lower) in the future and will perform at a 4.5 sigma level, which produces 3.4 DPMO.




Example of z-distribution

What is the probability that Z is greater than or equal to -1.96 and smaller than or equal to -1.4?

Consider that Z is a standard normal random variable.

ANSWER: 0.0558

Use the Z-table to help solve.





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