Page 257 - Six Sigma Demystified

Part 3 s i x s i g m a to o l s 237

Exponential Distributions

Example
Estimate the probability of an accident within the first 31 days of the last ac-
cident if the time between accidents averages 47 days (i.e., the accident rate is
13 percent, and the plant runs 365 days a year).

Minitab

Use Calc\Probability Distributions\Exponential|Cumulative Probability. Set Scale
= 47; Input Constant = 31.

Result (From Session Window)
Cumulative Distribution Function
Exponential with mean = 47.
x P(X ≤ x)
31 0.482929

Excel

Enter =EXPONDIST(31, 1/47, 1) into an Excel cell. The solution provided is
48.3 percent. Conversely, the probability of being accident-free for that period
is 1 – 0.4829, which equals 51.7 percent.

Normal Distribution

The average ( X ) of a sample can be calculated by summing the measurements
and dividing by the number of measurements (N). The standard deviation of
the N sample measurements can be calculated as

∑ N ( X − X ) 2
s = j =1 1
N

We calculate a z value to convert the given normal distribution into a stan-
dardized normal distribution, which has a mean of zero and a standard devia-
tion of 1:

( X − X)
z =
s