1 smu emis 7364 ntu to-570-n modeling process capability normal, lognormal & weibull models...
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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
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Normal Distribution:
• Mean or expected value of X
Mean = E(X) =
• Median value of X
X0.5 =
• Standard deviation
)(XVar
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Normal Distribution:
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
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x
0
z
x
Z
Normal Distribution:
f(x) f(z)
P (X<x’) = P (Z<z’)
X’ Z’
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Normal Distribution:
• 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
Normal Distribution - example solution
(b) P(X 91.7) =
= P(Z > - 0.83)= 1 - P(Z -0.83) = 1 - 0.2033= 0.7967
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1007.91P
x
100
x
0
z
f(x) f(z)
91.7 -0.83
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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
7.115
10
1001.87
x
P
100
x
f(x)
87.1 115.7
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Normal Distribution - example solution
(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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Normal Distribution - example
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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Normal Distribution - example solution
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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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
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Normal Distribution - example solution
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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19
20
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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.
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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 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