# Does Operating at Six Sigma Mean 6.84 DP/2 Mill?

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- This topic has 4 replies, 5 voices, and was last updated 1 year, 5 months ago by Joel Mason.

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- February 12, 2019 at 7:16 pm #236314

jodean22Participant@jodean22**Include @jodean22 in your post and this person will**

be notified via email.Hello, I am new to Six Sigma and trying to work my way through the math, if operating at 6σ means 3.42 DPMO it stands to reason that 2,000,000 opportunities would yield only 6.84 defects. However, when I do the math I doesn’t quite compute… 3.42 defects / 1,000,000 opportunities = .000342% defect rate (or 99.99966% effective). However .000342% * 2,000,000 = is 684 defects?

What am I missing?

Thank you,

0February 13, 2019 at 7:37 am #236361Hi Jodean,

This is a great question – Lots of new SixSigma users will have questions with the math and once you see this, its actually pretty straight forward:

First, lets swap out the numbers to easier values. What if we had 500 defects per 1000 opportunities. We would get a defect rate of 500/1000 = 0.5 , i.e. the defect rate per single opportunity is 0.5. We then multiply this by 100 to get that as a percentage rate of 50%.

(watch out because if you format a cell in Excel as a % it will do the multiply by 100 automatically, this often leads to errors!).

So to get the number of expected defects for 2000 opportunities we should multiply by the defect rate per single opportunity, e.g. 2000*0.5 = 1000 expected defects. We should not use the 50% (as this has already been multiplied by 100) and would wrongly give 2000*50 = 100,000 expected defects (I think I used to work at this factory).

If we swap back in the original numbers; first we should convert the % rate to the rate per opportunity, so have 0.000342/100 to get 0.00000342 . Then do the multiplication to get 0.00000342*2,000,000 = 6.84 expected defects.

3February 13, 2019 at 10:51 am #236363

Mike CarnellParticipant@Mike-Carnell**Include @Mike-Carnell in your post and this person will**

be notified via email.@jodean22 Make sure you understand that 3.4 dpmo is a shifted value for Six Sigma.

1February 18, 2019 at 5:38 am #236432

99SnowyParticipant@99Snowy**Include @99Snowy in your post and this person will**

be notified via email.3.42 defects/1,000,000 oppotunities = 0.000342%, = 0.000000342 DPO

Therefore for 2,000,000 oppotunities x 0.00000342 (DPO) = 6.84 defects1February 18, 2019 at 11:07 am #236433

Joel MasonParticipant@joelmason35**Include @joelmason35 in your post and this person will**

be notified via email.I’d like to add more detail to what @MikeCarnell offered, because I suspect his point is at the heart of your question. Looking at a standard normal probability distribution (mean = 0, standard deviation = 1), 3.4 x 10-6 of the observations lie to the right and left of 4.5 standard units away from the mean in both directions. That means the area under the standard normal curve on the right tail and the left tail beyond 4.5 sigma from the mean is 3.4 x 10-6. That’s what Mike means by saying it is a shifted value. Companies like Motorola that were early adopters of what we call Six Sigma understood that most of their processes in fact did not have stable central tendencies. They knew that their processes “wondered around” a little. Walter Shewhart knew that as well back in the 1920’s when he was working for Western Electric and developed the first SPC charts that we still use today. Without modeling that “wondering mean”, what do you do? Generally speaking, you give yourself room for it to happen. That’s why most DPMO tables that you will find online and in literature allow for a 1.5 sigma shift in the mean. So 6-sigma DPMO allows for the central tendency to shift right or left by 1.5 standard units.

Over the years, academics and industry professionals have figured out elegant ways of modeling processes that have central tendencies that move. As you progress in your mastery of the six sigma body of knowledge, I hope you allow yourself to learn more about those approaches. Good luck.

Joel

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