When learning about Six Sigma, it may help to consider these charts, which detail how sigma level relates to defects per million opportunities (DPMO), and some realworld examples.
Sigma Performance Levels – One to Six Sigma  

Sigma Level  Defects Per Million Opportunities (DPMO) 
1  690,000 
2  308,537 
3  66,807 
4  6,210 
5  233 
6  3.4 
It’s one thing to see the numbers and it’s a whole other thing to see how it would apply to your daily life.
Realworld Performance Levels  

Situation/Example  In 1 Sigma World  In 3 Sigma World  In 6 Sigma World 
Pieces of your mail lost per year [1,600 opportunities per year]  1,106  107  Less than 1 
Number of empty coffee pots at work (who didn’t fill the coffee pot again?) [680 opportunities per year]  470  45  Less than 1 
Number of telephone disconnections [7,000 talk minutes]  4,839  467  0.02 
Erroneous business orders [250,000 opportunities per year]  172,924  16,694  0.9 


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Comments
why do you mean 3.4 ppm as six Sigma
3.4 ppm means that the opportunities or likelihood that the product/process will fail is 3.4 in a million. like for an instance when you bought a single lottery ticket. the chances of winning is only 3.4 out in a million or 0.00034%. in terms of defects the chances that the single product/process fails is only 3.4 out in a million. it is totally different of saying that producing one million parts you are expecting 3.4 defects which a lot of people misinterpret.
so, im actually going to use real odds from a colorado state lottery ticket. Id like to use Factual Data printed on the ticket just to use realtime data, instead of the defined values of level 6 Sigma Quality.
360,342 winning tickets of 1,800,000
odds of winning a top prize 1 in 600,000
odds of winning overall 1 in 5
so we can say that chances of winning a top prize jackpot is 2tickets per 1.2million sold.
now just quick rough estimate to calculate a value for ppm would be approx 1.8 ppm
so purchasing one colorado lottery scratch ticket, your odds of winning a top prize is 1.8 in 1million tickets, or 0.00018%.
now lets just say, if colorado lottery had a “Scratchology” department which is designated for the sole business of the scratch ticket games, and this was the only scratch game the colorado lottery offered at that time. now lets say they implimented and assembled a GreenBelt project team, gathered key metrics to calculate a DPMO score. lets say, they wanted to see what the DPMO is for the process of printing the scratch games, which consists of taking all the pre calculated data of how many tickets to be printed, and what information needs to be transferred on each individual ticket, with a covert and classified manner. now lets say they werent perfect, but were able to determine that the process has 3.1DMPO. A level 6Sigma quality production line.
now is it correct to say that 3.1 tickets failed to print out of one million tickets printed.
i hope my interpretation of the above information makes sense, and for the most part correct.
Because what stood out to me, is how you gave the lottery odds as a perfect way to differentiate ppm from DPMO. But seeming how they are alike, but not the same, though do have common denominatating value of units, can these numbers used in tandem, or is one apple, the other orange.
if the DPMO figure was factual data, and not the random value i came up with for my perfect example, would it be safe to say, that a computerized production line that printed Colorado Lottery Scratch Game Tickets, did so with a 3.1DPMO, and level 6 excellent, and not just of good quality. and then add that this game not only looks beautiful but it also has a 1:5 overall winning odds, then odds of winning a top prize jackpot is a 1.8 ppm. is it safe to compare the ppm to DPMO, cause id say the printing process is extremely flawed if the DPMO exceeded the ppm of odds of winning yop prizes. if theres the chance that the defectedtickets that didnt print could actuallt be the top prize winners, and change the odds of winning to odds of losing.
why can,t we say that seven sigma or eight sigma instead of six sigma? or explanation of Six?
We CAN use 7 Sigma, but by accident the “lore” developed as 6 Sigma
There is no fundamental law that dictates either
Air travel is approximately a 7 Sigma process
Six sigma 3.4 ppm mean 3.4 defect per million opportunity.
It means 99.99966 % data falls withing specification. so that 10099.99966=0.00034% = 3.4 ppm
no wrong !, 3.4 doesn’t mean 0.00034% defects or 99.99966% good
6 means 99.9997 % good
and there are six [6] sigma levels not 3.
calculated as
[ defect units / (no. of oppertunities * no. of units) ] * 1,000,000
But U know that in calculating of actual capability metrics the normal distribution is from 3s to +3s (centered process). In there are about 99,….% probability of execute products is six sigma level quality. So, from 3s to +3s there is 6 sigma. I must say that in banking exist 7 sigma level.
Thank you for your comments. I see that Six Sigma represents a measurement for the number of defects in a near perfect level of performance in anything we can do.
Defect counts are like random variables representing deviation from mean. They follow bell curve (normal distribution) and have a mean (mu) and a standard deviation (sigma) like any other random distribution.
By counting defects per million you can judge the quality maturity of your process in units of one two three or six times the standard deviation (sigma).
1 2 3 6 sigma = 68% 95% 99.7% and 99.9999998% (percentage of total area under normal bell curve)
http://en.wikipedia.org/wiki/Normal_distribution
How you calculate 1sigma=690,000, 2sigma= 390,538 etc.? Is there any formula?
So Beautifully n simply expressed – so easy for commoners – Gr8
Are you sure about those numbers? 67% failure rate at 1sigma?
the tails should be 16% each side at 1sigma
and 0.6ppm at 5sigma, not 233ppm
I get these:
N NORM.DIST(N,0,1,TRUE) 1this 2*1e6 x that
1 0.841344746 0.158655254 317310.5079
2 0.977249868 0.022750132 45500.2639
3 0.998650102 0.001349898 2699.796063
4 0.999968329 3.16712E05 63.34248367
5 0.999999713 2.86652E07 0.573303144
6 0.999999999 9.86588E10 0.001973175
I got the same answers as you did. So here you are taking the normal CDF (so we get the area except for the right tail above N sigma), and then we do 1normalcdf(N) to get one tail area, and then 2*tail_area to get the area of both tails, then multiply by one million to get ppm. I am having trouble working backward with a single formula to get all the ppm levels as shown in the article.