1 SMU EMIS 7364 NTU TO-570-N Modeling Process Capability Normal, Lognormal & Weibull Models Updated:...

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SMUEMIS 7364

NTUTO-570-N

Modeling Process Capability Normal, Lognormal & Weibull Models

Updated: 1/14/04

Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow

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Normal Distribution:

A random variable X is said to have a normal (orGaussian) distribution with parameters and ,where - < < and > 0, with probability density function

- < x <

where = 3.14159… and e = 2.7183...

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x2

1

e2

1)x(f

f(x)

x

3


• Mean or expected value of X

Mean = E(X) =

• Median value of X

X0.5 =

• Standard deviation

)(XVar

4


Standard Normal Distribution

If X ~ N(, ) and if , then Z ~ N(0, 1).

A normal distribution with = 0 and = 1, is calledthe standard normal distribution.

X

Z

5

x

0

z

x

Z


f(x) f(z)

P (X<x’) = P (Z<z’)

X’ Z’

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• Standard Normal Distribution Table of Probabilities

http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html

Enter table with

and find thevalue of

• Excelz

0

z

f(z)

x

Z

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Properties of the Normal Model - the effects of and

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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

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Normal Distribution - example solution

(a) P(X < 105.3) =

= P(Z < 0.53)= (0.53)= 0.7019

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1003.105P

x

100

x

0

z

f(x) f(z)

105.3 0.53

10


(b) P(X 91.7) =

= P(Z > - 0.83)= 1 - P(Z -0.83) = 1 - 0.2033= 0.7967

10

1007.91P

x

100

x

0

z

f(x) f(z)

91.7 -0.83

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(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

7.115

10

1001.87

x

P

100

x

f(x)

87.1 115.7

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(d) P(X x) = 0.05P(Z z) = 0.05 implies that z = 1.64P(X x) =

therefore

x - 100 = 16.4x = 116.4

10

1001

10

100P

10

100P

xxZ

xx

64.110

100

x

100

x

f(x)

116.4

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The diameter of a metal shaft used in a disk-drive unitis normally distributed with mean 0.2508 inches andstandard deviation 0.0005 inches. The specificationson the shaft have been established as 0.2500 0.0015 inches. We wish to determine what fraction ofthe shafts produced conform to specifications.

0.2508 0.2515 USL

0.2485 LSL

0.2500

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0.2515x0.2485P

x0.2485P0.2515xP

0005.0

0.2508-0.2485

0.0005

0.2508-0.2515

60.41.40

0000.091924.0

91924.0

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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

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0.2515x0.2485P

x0.2485P0.2515xP

0005.0

0.2500-0.2485

0.0005

0.2500-0.2515

00.33.00

00135.099865.0

9973.0

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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.

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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 underthree sigma quality will be defective. This is not anacceptable situation, because many high technology products are made up of thousands of components.An automobile has about 200,000 components andan airplane has several million!

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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

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xln2

1

e2x

1 )x(f

0

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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(,)

xln

)x(F

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Properties of the Lognormal Model:

• Mean or Expected Value

2

2

1

e)X(E

1ee)X(Var222

• Variance

• Median

ex 5.0

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Lognormal Model - example

A theoretical justification based on a certain materialfailure 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?

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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.

x1ex )x(f

0

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The Weibull Distribution: Probability Density Function

f(x)

x

x is in multiples of

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Weibull Distribution - Probability Distribution Function

, for x 0 β

θ

x

e-1 xXPF(x)

0.000

0.200

0.400

0.600

0.800

1.000

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

x

F(x)

Where x is in multiples of

5.0

3.442.5

1.0

0.5

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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)

1

p)-ln(1- θ t p

θ t0.632

1

1 )X(E

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Properties of the Weibull Distribution

Variance of X

and the standard deviation of X is

1

11

2 222

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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?

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Weibull Distribution - example

The fraction of items not meeting spec is

4000XP

That is, all but about 13.53% of the items will not meet spec.

1353.0

e

e

)4000(F1

)4000(P1

2

1000

40001/2

X

1 SMU EMIS 7364 NTU TO-570-N Modeling Process Capability Normal, Lognormal & Weibull Models Updated:...

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