Page 258 - Six Sigma Demystified

238 Six SigMa DemystifieD

Appendix 1 provides the cumulative percentage points of the standard nor-
mal distribution. The z value allows us to estimate the probability of being less
than or greater than any given x value, such as a customer requirement. For
example, if we calculate that the average cycle time of an order fulfillment
process is 4.46 days and the standard deviation of the cycle time is 2.97 days,
we can calculate a z value for an upper specification of 10 days as follows:

(10 − . 4 46 )
z = = . 1 86
. 2 97
From Appendix 1 we find that a z value of 1.86 corresponds to a cumulative
probability of 0.9686, implying that 96.86 percent of the process will be within
the customer requirement. Conversely, 1 – 0.9686 (or 3.14 percent) will exceed
the requirement of 10 days, assuming that the process is modeled adequately
by the normal distribution. This assumption is conveniently verified for a set of
data using the tests discussed below (see “Goodness-of-Fit Tests”).

Normal Distributions

Example
Estimate the probability of not exceeding a customer limit of 10 days for order
fulfillment if the process average is 4.46 days and the standard deviation is 2.97
days.

Minitab

Use Calc\Probability Distributions\Normal|Cumulative Probability. Set Mean =

4.46; Standard Deviation = 2.97; Input Constant = 10.
Result (From Session Window)
Cumulative Distribution Function
Normal with mean = 4.46 and standard deviation = 2.97.
x P(X ≤ x)
10 0.968932
Excel

Enter =NORMDIST(10, 4.46, 2.97, 1) into an Excel cell. The solution pro-
vided is 96.9 percent. Conversely, the probability of being accident-free for that
period is 1 – 0.969, which equals 3.1 percent.