1. Modeling Process Capability Normal, Lognormal & Weibull Models
SMU
EMIS 7364
NTU
TO-570-N
Modeling Process Capability Normal, Lognormal & Weibull Models
Updated: 1/14/04
Statistical Quality Control
Dr. Jerrell T. Stracener, SAE Fellow

2. Normal Distribution:
A random variable X is said to have a normal (or
Gaussian) distribution with parameters  and ,
where -  <  <  and  > 0, with probability
density function
-  < x < 
where  = … and e =
f(x) x

3. Normal Distribution:
Mean or expected value of X
Mean = E(X) = 
Median value of X
X0.5 = 
Standard deviation

4. Normal Distribution:
Standard Normal Distribution
If X ~ N(, ) and if , then Z ~ N(0, 1).
A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.

5. P (X<x’) = P (Z<z’) x z
Normal Distribution:
f(x)
f(z)
P (X<x’) = P (Z<z’) X’ Z’

6. Standard Normal Distribution Table of Probabilities
Enter table with
and find the
value of 
Excel z  z
f(z)

7. Properties of the Normal Model - the effects of  and 

8. Normal Distribution - example
The following example illustrates every possible
case of application of the normal distribution.
Let X ~ N(100, 10)
Find:
(a) P(X < 105.3)
(b) P(X  91.7)
(c) P(87.1 < X  115.7)
(d) the value of x for which P(X  x) = 0.05

9. Normal Distribution - example solution
(a) P(X < 105.3) =
= P(Z < 0.53)
= F(0.53) =
100 x z
f(x)
f(z)
105.3
0.53

10. Normal Distribution - example solution
(b) P(X  91.7) =
= P(Z > )
= 1 - P(Z  -0.83) = =
100 x z
f(x)
f(z)
91.7
-0.83

11. Normal Distribution - example solution
(c) P(87.1 < X  115.7) =
= P(-1.29 < Z < 1.57)
= F(1.57) - F(-1.29) = =
100 x
f(x)
87.1
115.7

12. Normal Distribution - example solution
(d) P(X  x) = 0.05
P(Z  z) = 0.05 implies that z = P(X  x) =
therefore
x = 16.4
x = 116.4
100 x
f(x)
116.4

13. Normal Distribution - example
The diameter of a metal shaft used in a disk-drive unit
is normally distributed with mean inches and
standard deviation inches. The specifications
on the shaft have been established as 
inches. We wish to determine what fraction of
the shafts produced conform to specifications.
0.2508
0.2515
USL
0.2485
LSL
0.2500

14. Normal Distribution - example solution

15. Normal Distribution - example solution
Thus, we would expect the process yield to be
approximately 91.92%; that is, about 91.92% of
the shafts produced conform to specifications. Note
that almost all of the nonconforming shafts are too
large, because the process mean is located very
near to the upper specification limit. Suppose we
can recenter the manufacturing process, perhaps
by adjusting the machine, so that the process mean
is exactly equal to the nominal value of
Then we have

16. Normal Distribution - example solution

21. Normal Distribution - example
Using a normal probability distribution as a model
for a quality characteristic with the specification
limits at three standard deviations on either side of
the mean. Now it turns out that in this situation the
probability of producing a product within these
specifications is , which corresponds to 2700
parts per million (ppm) defective. This is referred to
as three-sigma quality performance, and it actually
sounds pretty good. However, suppose we have a
product that consists of an assembly of 100
components or parts and all 100 parts must be
non-defective for the product to function
satisfactorily.

22. Normal Distribution - example
The probability that any specific unit of product is
non-defective is
x x x = (0.9973)100 =
That is, about 23.7% of the products produced under
three sigma quality will be defective. This is not an
acceptable situation, because many high technology
products are made up of thousands of components.
An automobile has about 200,000 components and
an airplane has several million!

23. The Lognormal Model:
Definition - A random variable X is said to have the
Lognormal Distribution with parameters  and ,
where  > 0 and  > 0, if the probability density
function of X is:
, for x > 0
, for x  0

24. Properties of the Lognormal Distribution
Probability Distribution Function
where (z) is the cumulative probability distribution
function of N(0,1)
Rule: If T ~ LN(,), then Y = lnT ~ N(,)

25. Properties of the Lognormal Model:
Mean or Expected Value
Variance
Median

26. Lognormal Model - example
A theoretical justification based on a certain material
failure mechanism underlies the assumption that ductile
strength X of a material has a lognormal distribution.
Suppose the parameters are  = 5 and  = 0.1
(a) Compute E(X) and Var(X)
(b) Compute P(X > 120)
(c) Compute P(110  X  130)
(d) What is the value of median ductile strength?
(e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120?
(f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

27. The Weibull Probability Distribution Function
Definition - A random variable X is said to have the
Weibull Probability Distribution with parameters 
and , where  > 0 and  > 0, if the probability
density function of X is:
, for x  0
, elsewhere
Where,  is the Shape Parameter,  is the Scale
Parameter. Note: If  = 1, the Weibull reduces to the
Exponential Distribution.

28. The Weibull Distribution: Probability Density Function
f(x) x
x is in multiples of 

29. Weibull Distribution - Probability Distribution Function
, for x  0
Where x is in multiples of q
5.0
3.44
2.5
1.0
0.5

30. Properties of the Weibull Distribution
100pth Percentile
and, in particular
Mean or Expected Value
Note: See the Gamma Function Table to obtain
values of (a)

31. Properties of the Weibull Distribution
Variance of X
and the standard deviation of X is 

32. Weibull Distribution - example
The random variable X can modeled by a Weibull distribution with  = ½ and  = The spec time limit is set at x = What is the proportion of items not meeting spec?

33. Weibull Distribution - example
The fraction of items not meeting spec is
That is, all but about 13.53% of the items will not meet spec.