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

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  = 3.14159… and e = 2.7183... f(x) x

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

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.

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

Standard Normal Distribution Table of Probabilities http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html Enter table with and find the value of  Excel z  z f(z)

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

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

Normal Distribution - example solution (a) P(X < 105.3) = = P(Z < 0.53) = F(0.53) = 0.7019 100 x z f(x) f(z) 105.3 0.53

Normal Distribution - example solution (b) P(X  91.7) = = P(Z > - 0.83) = 1 - P(Z  -0.83) = 1 - 0.2033 = 0.7967 100 x z f(x) f(z) 91.7 -0.83

Normal Distribution - example solution (c) P(87.1 < X  115.7) = = P(-1.29 < Z < 1.57) = F(1.57) - F(-1.29) = 0.9418 - 0.0985 = 0.8433 100 x f(x) 87.1 115.7

Normal Distribution - example solution (d) P(X  x) = 0.05 P(Z  z) = 0.05 implies that z = 1.64 P(X  x) = therefore x - 100 = 16.4 x = 116.4 100 x f(x) 116.4

Normal Distribution - example The diameter of a metal shaft used in a disk-drive unit is normally distributed with mean 0.2508 inches and standard deviation 0.0005 inches. The specifications on the shaft have been established as 0.2500  0.0015 inches. We wish to determine what fraction of the shafts produced conform to specifications. 0.2508 0.2515 USL 0.2485 LSL 0.2500

Normal Distribution - example solution

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 0.2500. Then we have

Normal Distribution - example solution

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 0.9973, 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.

Normal Distribution - example The probability that any specific unit of product is non-defective is 0.9973 x 0.9973 x . . . x 0.9973 = (0.9973)100 = 0.7631 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!

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

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(,)

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

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?

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.

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

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

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)

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

Weibull Distribution - example The random variable X can modeled by a Weibull distribution with  = ½ and  = 1000. The spec time limit is set at x = 4000. What is the proportion of items not meeting spec?

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.