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Few Six Sigma concepts cre ate moreconfusion (or wrath) than the so-

called 1.5 sigma shift. Over the years,

I ve come to discover that the majority of

those writing about the subject really dont

und erstand it. In this column, I ll try to set

the re c o rd straight by explaining the tru e

rationale behind this infamous s t at i s t i c a l

c o rre c t i o n .

The standard normal distribution pro-

vides the starting point for discussion. This

d i stribution has been extensive ly studied

and tabularized, and many natural systems

can be approximated with it. Although thedistribution extends to infinity in both direc-

tions, drawings of the distribution usually

only show the area from 3 standard devi-

ations to +3 standard

dev iations because

this ran ge incl ude s

99.73 percent of

the data. Until Six

Sigma became pop-

ular, all quality cal-

culations were based

on this distribution

without any adjustm ents. For ex ample,

assume surveys show that customers are

u n h ap py when hold times exceed 15 min-

utes and that actual hold times are per-

fect ly normally distributed, ave raging 12

minutes with a standard deviation of one

m i nu t e. The 15 minute specific ation is 3

standard deviations above the mean. Look-

ing up the area of a normal distribution

beyond +3 standard dev i at io ns, we pre dict

t hat 0.135 percent of customer hold times

will exceed the specificat i o n.

Six Sigma modifies this pro c e d u re by a d j u s t i n g the calculated mean by 1.5

sigma befo re estimating the perc e n t age

out of specifi c ation. For the hold-time

ex a m p l e, we would perfo rm the calcula-

tions with an ave rage hold time of 13.5

m i nutes instead of the calculated value of

12 minutes. This would give an estimate of

6.68 percent exceeding the specific at i o n .

There is a saying among enginee rs and

scient ist s: All models are wrong, but some

models are useful. The traditional norm a l

model, for example, is certainly wrong, but

its often still useful. The question, then, iswhether the 1.5 sigma adjustment cre at e s

a model that s more useful than the tra d i-

tional model. I believe it does. While all

models simplify re a l i t y, the tra d i t i o n a l

model ove rs i m p l i fies re a l i t y. It make s

things look mu ch better to us than they

look to our customers .

Consider the assumptions for the hold-

time ex a m pl e. The assumption of perfect

n o rmality for hold times is cert a i n ly

w ro n g. Th e re s also sampling erro r, t he

d e finition of hold time ( e. g. , cu st o m e r-

perceived hold time vs. clock time), tremen-dous va ri ability from customer to cus-

tomer and more. In a production exa m ple,

ove rs i m p l i fi c ations include estimat i n g

sigma based on short- te rm va ri at i o n ,m a k-

ing measurements on product that has

never been used, not considering shipping

and handling effects, failure to consider the

e nv i ronments to wh i ch product will be

ex p o s e d, and incomplete unders t a n d i n g

of the customers re q u i re m e n t s .

The 1.5 sigma shift is simply a cor-recti on that accounts for fa c t o rs not

i n cluded in our model of re a l i t y.

The table of process sigma values and

e q u ivalent parts per million shows the

e q u ivalent yields that you obtain wh e n

the 1.5 sigma shift is taken into account.

Use the table as fo l l ows:

n When data are from measure m e n t s ,

c a l c u l ate the ave rage and standard dev i a-

tion. Then determine how many standard

d ev i ations there are between the mean

and each specification. Find the PPM level

Six Sigma and Beyond T h o m a s Py z d e k

q d

In t e r n a l p e rf o r m a n c e m e t r ic s o f t e n u n d e re st i m a t e c u st o m e r p r o b l em s.

T h e 1.5 Sig m a Sh i f t

The Standard Normal Dist r ibut ion Curv e

Pro cess Sigma Levels and Equivalen t PPM Quality Levels*

Pro c ess Sig m a Lev el Pr o ce ss PPM Pro cess Sig m a Le v e l Pr o c ess PPM

* A ssu m e s t h a t i n t h e l o n g t e r m t h e p r o c e ss co u l d d r i ft b y 1 . 5

( P PM = e r r o r s o r d e f e c t s p e r m i l l io n o p p o r t u n i t i e s )

6 . 2 7 1 4 . 6 6 8 0 06 . 1 2 2 4 . 6 2 9 0 0

6 . 0 0 3 . 4 4 . 5 9 1 , 0 0 05 . 9 7 4 4 . 3 8 2 , 0 0 05 . 9 1 5 4 . 2 5 3 , 0 0 0

5 . 8 8 6 4 . 1 5 4 , 0 0 05 . 8 4 7 4 . 0 8 5 , 0 0 0

5 . 8 2 8 4 . 0 1 6 , 0 0 05 . 7 8 9 3 . 9 6 7 , 0 0 05 . 7 7 1 0 3 . 9 1 8 , 0 0 0

5 . 6 1 2 0 3 . 8 7 9 , 0 0 05 . 5 1 3 0 3 . 8 3 1 0 , 0 0 0

5 . 4 4 4 0 3 . 5 5 2 0 , 0 0 05 . 3 9 5 0 3 . 3 8 3 0 , 0 0 0

5 . 3 5 6 0 3 . 2 5 4 0 , 0 0 05 . 3 1 7 0 3 . 1 4 5 0 , 0 0 05 . 2 7 8 0 3 . 0 5 6 0 , 0 0 05 . 2 5 9 0 2 . 9 8 7 0 , 0 0 05 . 2 2 1 0 0 2 . 9 1 8 0 , 0 0 05 . 0 4 2 0 0 2 . 8 4 9 0 , 0 0 0

4 . 9 3 3 0 0 2 . 7 8 1 0 0 , 0 0 04 . 8 5 4 0 0 2 . 3 4 2 0 0 , 0 0 0

4 . 7 9 5 0 0 2 . 0 2 3 0 0 , 0 0 04 . 7 4 6 0 0 1 . 7 5 4 0 0 , 0 0 04 . 6 9 7 0 0 1 . 5 0 5 0 0 , 0 0 0

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