Transcript qualityControl - Lyle School of Engineering

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Applications
Statistical Quality Control
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
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An Application of Probability & Statistics Statistical
Quality Control
Statistical Quality Control is an application of
probabilitistic and statistical techniques to quality
control
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Statistical Quality Control - Elements
Analysis
of process
capability
Process
improvement
Statistical
process
control
Acceptance
sampling
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Statistical Tolerancing - Convention
Normal Probability
Distribution
0.00135
LTL
-3
0.9973
Nominal

0.00135
UTL
+3
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Statistical Tolerancing - Concept
LTL
Nominal
UTL
x
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Caution
For a normal distribution, the natural tolerance
limits include 99.73% of the variable, or put
another way, only 0.27% of the process output will
fall outside the natural tolerance limits. Two points
should be remembered:
1. 0.27% outside the natural tolerances sounds
small, but this corresponds to 2700 nonconforming
parts per million.
2. If the distribution of process output is non
normal, then the percentage of output falling
outside   3 may differ considerably from 0.27%.
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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.
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Normal Distribution - Example Solution
Pmeeting spec   P0.2485  x  0.2515
 Px  0.2515  P0.2485  x 
 0.2515 - 0.2508 
 0.2485 - 0.2508 
 





0.0005
0.0005




 1.40   4.60
 0.91924  0.0000
 0.91924
f(x)
0.2500
0.2485
LSL
nominal
0.2508
0.2515
USL
x
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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
P0.2485  x  0.2515  Px  0.2515  P0.2485  x 
 0.2515 - 0.2500 
 0.2485 - 0.2500 
 
  

0.0005
0.0005




 3.00   3.00
 0.99865  0.00135
 0.9973
f(x)
0.2485
LSL
nominal
0.2500
x
0.2515
USL
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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 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!
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