Sigma Limits and Defects per Million Opportunities

The relationship between the sigma level (SL) of a process and the defects per million opportunities (DPMO) is calculated using the cumulative distribution function (F(z)) of the normal distribution where F(z) is the probability of observing a value less than z. The F(z) values below were calculated using the NORMSDIST(z) function in Excel 2000.

Our calculations show the SL ranging from 0 to 7 in steps of 0.25 in the first column. The second and third columns calculate F(SL + 1.5) and F(1.5 - SL). The 1.5 accounts for the 1.5 shift assumed by Six Sigma. These values from the cumulative distribution function are used in further calculations. The fourth column (probability good) gives the probability of an observation that is not a defect. The values in this column are simply the difference between the second and third columns:

probability good = F(SL + 1.5) - F(1.5 - SL)

The fifth column calculates the probability of a defect as 1 - (probability good). The last column converts the probability of a defect to DPMO by multiplying by 1,000,000.

Our defect counts for SL < 3.5 are slightly larger than the counts shown by Mikel J. Harry because, for these cases, it is necessary to consider both tails of the distribution.¹ When only one tail of the distribution is considered, the DPMO values are calculated as one million times the second column (1,000,000 x F(1.5 - SL)).

When DPMO calculations are carried out to the nearest defect, the one-tail approximation differs from the correct two-tail value for all SL < 3.5. The most extreme example is that the one-tail zero sigma DPMO value is 933,193 instead of one million. There is no difference between the one-tail approximation and the correct two-tail calculation when the SL is > 1 and the calculations are carried out to only two significant figures. We recommend using the single tail approximation and using only two significant figures when the SL is greater than 1. For smaller SL values, it is necessary to consider both tails of the distribution (see Table 1).

The third or fourth column of the table can be used to convert the observed probability of a defect to a SL. Robert J. Gnibus gives a one-tail approximation that usually works adequately; however, it will give incorrect answers when the probability of a defect is large.²

Gnibus' second example concerns processing speeds of mortgage customers, where "all the defects (loans in a monthly sample taking more than five days to process) are counted, and it is determined that there are 600 loans in the 1,000 applications processed last month that don't meet this new customer requirement."

For this example, the probability of a defect is 600 / 1,000 = 0.6; the rule for our example (Table 1) is to pick the largest SL whose probability of a defect value is larger than the observed probability of a defect. Table 1 shows the SL is 1.25 because the SL has the probability of a defect = 0.602. Gnibus' one-tail approximation uses the NORMSINV function in Excel to calculate the z value for the corresponding "probability good." The one-tail approximation (with a 1.5 shift) is:

SL = 1.5 + NORMSINV (probability good)

For the example:

SL = 1.5 + NORMSINV (0.4) = 1.5 + (-0.253) = 1.247

Gnibus rounded this to a SL of 1.2, while the correct value is closer to 1.3. Again, the one-tail approximation for SL gives a value close to the correct answer because the SL > 1.

Gnibus' third example shows his and Harry's one-tail approximations both agree when SL = 1.0; both give 690,000 DPMO where the correct DPMO value is 700,000. When the fraction of defects is larger, the one-tail approximation can give meaningless answers. The one-tail approximation will give negative SL values when the probability of a defect is greater than 0.933.

This can be clearly seen in an additional example. Pretend you want to test the computation skills of 1,000 students and give them slide rules to assist them with their calculations. Slide rules are outdated computing aids, so most of the students will be unsatisfied. If you find the SL number for the number of satisfied students, you will be faced with two cases: zero students who are satisfied and 66 students who are satisfied. The SLs are zero and 0.25, respectively, when you use two-sided limits. Using the one-sided approximation gives - for the first SL and zero for the second.

Note I am pointing out technical errors in a basic table used by Harry to sell Six Sigma to management. This means that it is worthwhile to question other aspects of Six Sigma.

REFERENCES

1. Mikel J. Harry, "Six Sigma: A Breakthrough Strategy for Profitability," Quality Progress, May 1998.

2. Robert J. Gnibus, "Six Sigma's Missing Link," Quality Progress, November 2000.