# constant

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- This topic has 2 replies, 2 voices, and was last updated 17 years, 8 months ago by Monlui.

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- June 12, 2002 at 4:59 pm #29628
Hi my name is Raul.

Some body can help me to know

where does d2, D4, D3, A2 constants come from?

Thanks0June 12, 2002 at 6:32 pm #76349

GabrielParticipant@Gabriel**Include @Gabriel in your post and this person will**

be notified via email.Imagine that you have a population that is normally distributed. Both R (range) and sigma are parameters of population’s spread. Whe you make a Xbar-R control chart, you get the ranges from each subgroup and you average them to get Rbar. Knowing Rbar, which is the most probable value of the population sigma? Rbar/d2. This is conceptually where does d2 come from. I don’t know the formula to get there.

In a stable process, 99,7% of the points of the Xbar chart will fall within the average +/- 3 sigmas of the Xbar distribution. It can be proven that the Xbar distrbution always has the same average than the individuals (X) distribution. Given the data collected in a control chart the best estimation of the population average (or the average of Xbar) is the average of all points that is mathematically the same that the average of all Xbar calculated, i.e Xbarbar. On the other hand, the sigma of the Xbar distribution is the sigma of the individuals population divided by sqr(n) (sqr=square root, n=subgroup size). Then the limits within we expect 99.7% of the Xbar points will fall can be estimated as Xbarbar+/-3x(sigma/sqr(n)). As said in the first paragraph, sigma can be estimated as Rbar/d2. Then the limits can be calculated as Xbarbar+/-3x((Rbar/d2)/sqr(n)). If we call A2=3/(d2xsqr(n)) the limits can be written as Xbarbar+/-A2xRbar, the famous formula for the upper and lower control limits of the subgroups average.

Similarly the values of D3 and D4 are obtained for the limits that containt 99.7% of the subgroup ranges, but I don’t remember the step by step to get there as I do for A2.

Hope it is not as complicated to read as it was to write.

Gabriel0June 12, 2002 at 7:30 pm #76351Gabriel: Thank you for your information,now I understand more about it.

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