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Applied Reliability
------------------------------------------------------------------------------------------------------------
Copyright David C. Trindade, Ph. D.
STAT-TECH ®
Spring 2010
Applied
Reliability
Techniques for Reliability
Analysis
with
Applied Reliability Tools (ART)
(an EXCEL Add-In)
and
JMP® Software
AM216 Class 6 Notes
Santa Clara University
Applied Reliability
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2
Reference
Applied Reliability – Quality Control
Material Based on Chapter 9 in Text:
Applied Reliability, 2nd ed.
by Paul A. Tobias and David C. Trindade
Published 1995 by Chapman & Hall,
New York
ISBN No. 0-442-00469-9
Applied Reliability
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3
Table of Contents
Combinatorics
Multiplication Rule
Permutations (Spreadsheet Function)
Combinations (Spreadsheet Function)
Binomial DistributionBinomial Experiment
Probability Calculations
Parameter Estimation and Application
Confidence Intervals
Poisson DistributionApproximation to BinomialProbability Calculations
Parameter Estimation and Application
Confidence Intervals
Hypergeometric DistributionProbability Calculations
Application
Acceptance SamplingRisks
Operating Characteristic Curves
Sampling Plan
LTPD Plans
Adjusting Sampling Plans
Minimum Sample Size Plans
AOQ Curve and AOQL
Control Charts for ReliabilityCumulative Count Control Charts
Applied Reliability
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4
Multiplication Rule for Outcomes
If the first experiment can result in n1 possible
outcomes, and for each outcome, the second
experiment can result in n2 possible outcomes,
then there are a total of n1 n2 possible outcomes
for the two experiments together.
The multiplication rule is extendable to any
number of experiments in a sequence.
Example:
On a restaurant menu, there are five appetizers, seven entrees, and six desserts. How many different three course meals are possible?
Answer: 5x7x6 = 210.
Applied Reliability
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5
Combinatorics
Permutations
The number of ways to arrange n objects in
order:
If all n objects are used: n(n-1)(n-2)···1 = n!
If only r objects of n are used:
|------------ r terms -----------|
Examples:
Your child receives five birthday presents.
How many different ways can your child open
them?
Answer: 5! = 5x4x3x2x1 = 120.
Two musical notes are needed to complete the
song. In one octave, thirteen possible notes
exist. Without repeating the same note, how
many different ways can the song be ended?
Answer: 13x12 = 156.
!1 ( 2) 1
!
nn n n n r
n r
Applied Reliability
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6
Combinations
The number of ways of combining r objects
taken from n available in which order is not
important:
Note: 0! = 1
Examples:Eight players offer to play for the company golf
team in a tournament. How many different four
person teams can I make up?
Answer: 8!/(4!4!) = (8x7x6x5)/(4x3x2x1) = 70.
There are seven blade servers and five open
positions in the rack. How many ways can any
five be chosen for installing in any order on the
five rack positions? (Two will be left out.)
Answer:
7!/(5!2!) = (7x6x5x4x3)/(5x4x3x2x1) = 21.
Combinatorics
1 1 !
! !
n n n r n
r! r n r
n
r
10 n nn 1
Applied Reliability
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7
Class Project #1
Combinatorics:
Consider the three objects: A, B, C.
1. Taking all three objects, how many
permutations are there?
List them:
2. Taking two objects at a time, how
many permutations are there?
List them:
3. Taking two objects at a time, how
many combinations are there?
List them:
Applied Reliability
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8
Class Project #1
Combinatorics:
Consider the three objects: A, B, C.
1. Taking all three objects, how many
permutations are there?
3! = 3x2x1 = 6.
List them:
ABC ACB BAC
BCA CAB CBA
2. Taking two objects at a time, how
many permutations are there?
3!/(3-2)! = (3x2x1)/(1) = 6.
List them:
AB AC BA BC CA CB
3. Taking two objects at a time, how
many combinations are there?
3!/(2!1!) = 3.
List them:
AB AC BC
Applied Reliability
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9
Class Project #2
Combinatorics
1. How many different eleven letter words
can be formed from the letters in the word
MISSISSIPPI?
2. Let S denote a survival and F, a failure.
Given two failures among five items on stress,
how many different ways can the two failures
occur on the five objects? (Hint: One way is
SSSFF.) List all words.
3.How many different ways can r failures
occur among n units on stress?
This problem is analogous to asking how
many different words of n letters can be
formed from r F's and (n-r) S's.
Applied Reliability
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10
Class Project #2
Combinatorics
1. How many different eleven letter words
can be formed from the letters in the word
MISSISSIPPI?
2. Let S denote a survival and F, a failure.
Given two failures among five items on
stress, how many different ways can the
two failures occur on the five objects?
(Hint: One way is SSSFF.) List all words.
SSSFF SSFSF SSFFS SFFSS SFSFS
SFSSF FFSSS FSFSS FSSFS FSSSF
3.How many different ways can r failures
occur among n units on stress?
5
2
5!10
2!3!
11!34,650
4!4!2!
r
n
rnr
n
! !
!
Applied Reliability
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11
Permutations and Combinations in EXCEL
Applied Reliability
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12
Binomial Experiment
Necessary Conditions
1. Only two possible outcomes (success,
failure).
2. Fixed number of trials, n.
3. Constant probability of success, p, across
trials.
4. Trial outcomes are independent.
Applied Reliability
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13
Binomial DistributionDerivation
xnxn
x p1pxP
n
x
xnx p1p
Let p be the probability of failure, given
by the CDF at time t; then 1 - p is the
probability of survival. Consider n units
on test for time t. Designate a failure by F
and a survival by S. For the specific
sequence with x failures and (n-x) survivals:
F F ... F S S ... S
The probability of this sequence is:
Since there are
ways of x failures occurring among n units,
the probability of x failures among n units
on stress is given by the binomial distribution
probability mass function (PMF)
Applied Reliability
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14
Binomial DistributionProbability Function
20 units are placed on stress for 100 hours.
The CDF at 100 hours is 20%. The graph
below shows the binomial distribution
probabilities for x failures at 100 hours,
x = 0,1,2,…
0.0%
5.0%
10.0%
15.0%
20.0%
25.0%
0 1 2 3 4 5 6 7 8 9 10 11
x = Number of Failures
Pro
ba
bilit
y o
f x
Fa
ilu
res
n = 20, p = 0.2:
Applied Reliability
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15
Binomial DistributionCumulative Function
01)one least at(
1)()(
)()1()(
Note
)()()1()0()(
1
0
PP
xXPxXP
kPxXPxXP
kPxPPPxXP
n
xk
x
k
To get the cumulative probability of x or less
events, sum the individual binomial
probabilities.
Applied Reliability
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16
Binomial DistributionCumulative Function
20 units are placed on stress for 100
hours. The CDF at 100 hours is 20%. The
graph below shows the binomial
distribution cumulative probabilities of x or
less failures at 100 hours, x = 0,1,2,…
n = 20, p = 0.2:
Binomial CDF
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 1 2 3 4 5 6 7 8 9 10 11 12
x = Number of Failures
Pro
ba
bilit
y o
f x
or
les
s F
ailu
res
Applied Reliability
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17
Binomial DistributionProperties
Mean or expected number:
Expected number of successes (or failures) in
n trials with probability of success (or failure) p
per trial equals
Variance:
Standard Deviation:
np
2 1np p
1np p
Applied Reliability
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18
Binomial Distribution Calculations in EXCEL
=BINOMDIST(x, n, p, 0 or 1)
Fourth argument is:
0 or false = individual probability
1 or true = cumulative probability
Applied Reliability
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19
Class Project #3
Binomial Distribution:
A fair coin is tossed one hundred times.
1. What is the expected number of heads?
2. What is the standard deviation?
3. What is the probability of exactly fifty heads?
4. What is the probability of at most fifty heads?
5. What is the probability of 60 or more heads?
6. What is the probability of 40 or less heads?
7. What is the probability of greater than 40 but less
than 60 heads?
8. What is the probability of 100 heads?
Applied Reliability
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20
Class Project #3
Binomial Distribution:
A fair coin is tossed one hundred times.
1. What is the expected number of heads?
np = 100(0.5) = 50
2. What is the standard deviation?
3. What is the probability of exactly fifty heads?
4. What is the probability of at most fifty heads?
5. What is the probability of 60 or more heads?
6. What is the probability of 40 or less heads?
7. What is the probability of greater than 40 but
less than 60 heads?
8. What is the probability of 100 heads?
np p1 100(0.5)(0.5) 25 5
P X
50 100 50100 1 1
50 7.96%50 2 2
P X 60 2.844%
P X 50 53.98%
P X 40 2.844%
P X41 59 1 2(2.844) 94.31%
P X x
100 100 100
31100 1 1100 7.89 10
100 2 2
Applied Reliability
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21
Binomial Distribution PMF n = 100, p = 0.5
Applied Reliability
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22
Cumulative Binomial Distributionn = 100, p = 0.5
Applied Reliability
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23
Binomial DistributionParameter Estimation
Estimation of p:
In a binomial experiment, the probability of
failure p at a given time is estimated by the
proportion of failures observed out of n units.
Thus,
where x is the number of failures:
x = 0,1,2,...,n.
Note this is the same statistic for estimating
the CDF F(t) at time t, that is,
ˆ /p x n
tFp ˆˆ
Confidence Intervals on p:
Instead of just a point estimate for p, we
may specify a confidence interval that
captures the true (population) p with a
certain level of confidence.
Applied Reliability
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24
Binomial Distribution
95% Confidence Intervals for Proportions Clopper and Pearson Chart
Applied Reliability
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25
Binomial Distribution
Exact Confidence Limits Using
EXCEL:
Exact confidence limits for the population parameter
p may be found using the Inverse Beta function in
EXCEL.
Lower limit: 0 for x = 0; =BETAINV(a/2, x, n-x+1) for
x between 1 and n.
Upper limit: =BETAINV(1-a/2, x+1, n-x) for x
between 1 and n-1; 1 for x = n.
Example: n =100, x = 2,
The estimate of p = 2/100 = 0.02 = 2%.
Thus, a 90% 2-sided confidence interval on the
population p is:
LCL: = BETAINV(.05, 2, 99) = 0.36%
UCL: = BETAINV(.95, 3, 98) = 6.16%
Applied Reliability
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26
Class Project #4
Binomial Distribution:
Fifty devices are stressed for 168 hrs. The
probability of a device failing by 168 hrs is 0.05
or 5%.
Determine:
A. The probability all devices survive 168 hours.
B. The expected number of failures.
C. The probability of at least one failure.
D. At the end of the test there are 10 failures.
Provide a 95% confidence interval for the
population failure proportion. Does the interval
include the assumed 5% failure probability?
Applied Reliability
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27
Class Project #4
Binomial Distribution:
Fifty devices are stressed for 168 hrs. The
probability of a device failing 168 hrs is 0.05 or 5%.
Determine:
A. The probability all devices survive 168 hours.
B. The expected number of failures.
np = 50x0.05 = 2.5
C. The probability of at least one failure.
D. At the end of the test there are 10 failures.
Provide a 95% confidence interval for the population
failure proportion. Does the interval include the
assumed 5% failure probability?
From the Clopper Pearson chart, we get the interval
(10% to 34%),
which does not include the 5% failure probability.
50
( 0) 1 .05 0.077P X
( 1) 1 ( 0) 1 0.077 0.923P X P X
Applied Reliability
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28
The Poisson Distribution
Approximation to Binomial
Distribution
For n large and p small, the Poisson
distribution provides an excellent
approximation to the binomial distribution.
Let l = n x p, the expected number. The
probability of exactly x occurrences is:
The mean or expected number is l. The
variance is also l. The standard deviation is
Approximation is good for large n and small p,
such that l= n x p is less than, roughly, 7.
!
xeP x
x
ll
.l
Applied Reliability
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29
The Poisson Distribution
Comparison of Binomial and Poisson Probabilitiess
n = 90, p = 0.01, l = np = 0.9
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0 1 2
Number of Failures
Pro
bab
ilit
y F
un
cti
on
Binomial
Poisson
210)2( PPPXP
9381.01654.03679.04047.02 XP
Approximation to Binomial:
There are ninety units placed on stress for 100
hours. If the probability of failure is 0.01 for an
individual unit, what is the probability of 2 or less
units failing?
We want
Binomial Solution:
Poisson Solution:
9337.01647.03659.04066.02 XP
Applied Reliability
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30
Poisson Distribution
Applications of Poisson Distribution
The Poisson is also a distribution on its own
to model situations where the probability of
a single event over a period of time or
space is constant.
For example, for calculations involving
density, such as the number of defects per
wafer, the Poisson distribution (with l = the
mean number of defects for a given area)
can be used to model the defect distribution.
The Poisson has been applied to the
number of phone calls in a given period, the
number of repairs in time, the number of
bugs in software, the number of raisins in a
box of cereal, the number of people in a
queue, the number of flaws per yard of
insulated wire, the number of misprints per
page, and so on.
Applied Reliability
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31
Poisson Distribution Calculations in EXCEL
=POISSON(x, l, 0 or 1)
Applied Reliability
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32
Class Project #5
Poisson Distribution:
The average defect density is 0.01 defects per square
inch. The wafer area is 200 square inches. Ten wafers
are randomly selected from a lot of fifty.
What is:
1. The expected number of defects per wafer?
2. The expected number of defects for the ten wafers?
3. The probability any wafer is defect free?
4. The probability any wafer has at least one defect?
5. The probability any wafer has exactly five defects?
6. The probability all ten wafers are defect free?
7. The probability exactly two of the ten wafers are
defect free?
Applied Reliability
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33
Class Project #5
Poisson Distribution:
The average defect density is 0.01 defects per square
inch. The wafer area is 200 square inches. Ten wafers
are randomly selected from a lot of fifty.
1. The expected number of defects per wafer?
np = 200x0.01 = 2.0
2. The expected number of defects for the ten wafers?
10x2 = 20
3. The probability any wafer is defect free?
4. The probability any wafer has at least one defect?
5. The probability any wafer has exactly five defects?
6. The probability all ten wafers are defect free?
7. The probability exactly two of the ten wafers are
defect free?
Binomial:
Poisson: l = np = 10(0.135) = 1.35
21 (0) 1 0.865P e
036.0!5
2 25
e
020102 ee
2 8
10 2 2
2 0.258(2) 1P e e
2 1.351.35(2) 0.236
2
eP
20 0.135P e
Applied Reliability
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34
Poisson Distribution
Estimation of l
The estimate of the population
parameter l is just the average
observed number of occurrences
over the collection of time periods,
samples, objects, lines, etc.
Confidence Intervals for l
Tables for the 90%, 95%, and 99%
confidence levels for various observed
counts are shown next. Note the
confidence limits depend only on the
confidence level and the observed
count. Conversion to a CI for l may be
done using these tables.
Exact confidence limits can also be
calculated directly using functions in
EXCEL.
Applied Reliability
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35
Poisson Distribution
Table of Confidence Limits Based
on Observed Count
Confidence
Level
alphaObserved
Count
X lower limit upper limit lower limit upper limit lower limit upper limit
0 0.00 3.00 0.00 3.69 0.00 5.30
1 0.05 4.74 0.03 5.57 0.01 7.43
2 0.36 6.30 0.24 7.22 0.10 9.27
3 0.82 7.75 0.62 8.77 0.34 10.98
4 1.37 9.15 1.09 10.24 0.67 12.59
5 1.97 10.51 1.62 11.67 1.08 14.15
6 2.61 11.84 2.20 13.06 1.54 15.66
7 3.29 13.15 2.81 14.42 2.04 17.13
8 3.98 14.43 3.45 15.76 2.57 18.58
9 4.70 15.71 4.12 17.08 3.13 20.00
10 5.43 16.96 4.80 18.39 3.72 21.40
11 6.17 18.21 5.49 19.68 4.32 22.78
12 6.92 19.44 6.20 20.96 4.94 24.14
13 7.69 20.67 6.92 22.23 5.58 25.50
14 8.46 21.89 7.65 23.49 6.23 26.84
15 9.25 23.10 8.40 24.74 6.89 28.16
16 10.04 24.30 9.15 25.98 7.57 29.48
17 10.83 25.50 9.90 27.22 8.25 30.79
18 11.63 26.69 10.67 28.45 8.94 32.09
19 12.44 27.88 11.44 29.67 9.64 33.38
20 13.25 29.06 12.22 30.89 10.35 34.67
21 14.07 30.24 13.00 32.10 11.07 35.95
22 14.89 31.41 13.79 33.31 11.79 37.22
23 15.72 32.59 14.58 34.51 12.52 38.48
24 16.55 33.75 15.38 35.71 13.26 39.74
25 17.38 34.92 16.18 36.90 14.00 41.00
26 18.22 36.08 16.98 38.10 14.74 42.25
27 19.06 37.23 17.79 39.28 15.49 43.50
28 19.90 38.39 18.61 40.47 16.25 44.74
29 20.75 39.54 19.42 41.65 17.00 45.98
30 21.59 40.69 20.24 42.83 17.77 47.21
90%
0.1
99%95%
0.05 0.01
Applied Reliability
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36
Poisson DistributionExact Confidence Limits Using
EXCEL:
Exact confidence limits for the population parameter
l may be found using the inverse chi-squared
distribution (CHIINV(probability; degrees of freedom))
in EXCEL.
Lower limit: 0 for x = 0; =0.5* CHIINV(1-a/2, 2x) for x
> 0.
Upper limit: =0.5* CHIINV(a/2, 2*(x+1)) for x > 0.
Example: Observe 2 bugs in 100 lines of code,
The estimate of l: x = 2 bugs/100 LOC
Thus, a 90% 2-sided confidence interval on the
population l = bugs/100 lines of code is:
LCL: = 0.5* CHIINV(.95, 4) = 0.36
UCL: = 0.5* CHIINV(.05, 6) = 6.30
Applied Reliability
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37
Poisson Confidence Limits
Distinguish between Poisson confidence limits on:
• the total count (as in the previous table)
• the average count per unit of measure (as a % in ART).
EXAMPLE:
Suppose we open three boxes of cereal and find a totalof 9 raisins. The average estimate lis 3 raisins per box. The 90% confidence interval on:
• the total count for three boxes is 4.70 to 15.71
• the average per box l is 1.57 to 5.24.
(CI width is 3.67.)
Suppose we open ten boxes and find a total of 30 raisins, The average number of raisins per box is 3, as before. The 90% confidence interval on:
• the total count for ten boxes is 21.59 to 40.69
• the average per box l is 2.159 to 4.069.
(CI width for l is now 1.91 since the sample is larger.)
The width of the confidence interval depends on the total count.
Applied Reliability
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38
Class Project #6
Poisson Distribution:
1000 randomly selected lines of code are
inspected. Twenty three bugs are found.
Assuming a Poisson distribution:
A. What's the estimate of the expected
number of bugs per 1000 lines of code
(KLOC) in the population?
B. Provide a 90% confidence interval on
the expected number of bugs per KLOC in
the population.
Applied Reliability
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39
Class Project #6
Poisson Distribution:
1000 randomly selected lines of code are
inspected. Twenty three bugs are found.
Assuming a Poisson distribution:
A. What's the estimate of the expected
number of bugs per 1000 lines of code
(KLOC) in the population?
Answer: 23
B. Provide a 90% confidence interval on
the expected number of bugs per KLOC in
the population.
Answer:
From Table of CL for Poisson, CI is 15.72 to
32.59
Using EXCEL:
LCL: = 0.5* CHIINV(0.95, 46) = 15.72
UCL: = 0.5* CHIINV(0.05, 48) = 32.59
Applied Reliability
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40
Poisson Distribution
No Limit on Defect Count
The Poisson distribution has no limit (like nfor the binomial distribution) on the number of defects per unit of measure that can be counted.
Defective Unit versus Defects
A unit is defective if it has one or more defects (a defect is defined as a nonconformance to specifications). Thus, defective units are treated separately from the number of defects per unit or area.
A defective unit is handled by the binomial or hypergeometric distributions. For calculations involving density (e.g., defects per wafer, failures per period, etc.), the Poisson distribution is often used.
Applied Reliability
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41
Hypergeometric Distribution
Application:
Fixed number of trials.
Two possible outcomes: Success or failure.
Sampling from finite population. Hence,
probability of success with each trial changes
and events are not independent.
Example: What’s the probability of drawing two
aces in a row from a well shuffled deck of
cards?
Example: 20 lines of code in 100 are defective
(contain a bug). Randomly select 10 lines.
What’s the probability of finding 0,1,2,…
defective lines?
Applied Reliability
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42
Hypergeometric Distribution
Formula:
The probability of getting exactly x rejects in a
sample of size n drawn from a finite lot of size N
that contains a total of m rejects is
where x = 0,1,2,3,...
For n<0.1N, binomial distribution is a good
approximation for P(X) where p = m/N.
N
n
mN
xn
m
xxP
Applied Reliability
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43
Class Project #7
Hypergeometric Distribution:
Your dresser drawer contains twenty
socks, half red and half black. It’s early
morning and still dark. You don’t want to
turn on the light and disturb your spouse.
So in darkness you randomly select two
socks.
1. What is the probability that you
choose a matched pair of black socks?
2. What is the probability that you
choose a matched pair of either color?
Applied Reliability
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44
Class Project #7
Hypergeometric Distribution:
1. What is the probability that you
choose a matched pair of black socks?
2. What is the probability that you
choose a matched pair of either color?
P(matched pair, either color) = 2xP(2) =
0.474
10 10
2 0
20
2
45 12 0.237
190
m N m
x n x
N
n
xP
Applied Reliability
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45
Hypergeometric Distribution Calculations in EXCEL
= hypgeomdist( x; n; m; N)
Applied Reliability
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46
Fisher’s Exact Test
In two groups we observe failure proportions 1/6 and 7/9. The small sample sizes are small, but we want to test whether Group 1 is significantly better than Group 2. The results are displayed in a 2x2 contingency table with margin totals:
Is the population from which the first group was drawn significantly better than the second group population or is the difference likely to occur by chance alone?
For small sample sizes, Fisher’s Exact Test, based on the hypergeometric distribution, calculates the probability of observing differences equal to or greater than the results observed.
Group 1 Group 2 Totals
Pass 5 2 7
Fail 1 7 8
Totals 6 9 15
Applied Reliability
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47
Fisher’s Exact Test
For both groups, note the total sample size is N= 15 and the total failures is m = 8. Assume a randomly drawn sample size of n = 6 for Group 1. The probability of getting exactly 1 failure out of the 8 possible is given by the hypergeometric distribution
Group 1 Group 2 Totals
Pass 5 2 7
Fail 1 7 8
Totals 6 9 15
P
8 15 8
1 6 1
15
6
1 0.03357
Applied Reliability
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48
Fisher’s Exact Test(Continued)
An even more extreme event (zero failures in Group 1) with the same marginal totals is shown below.
This probability is
So the one-sided probability of observing the results or stronger is the sum
0.03357 + 0.001399 = 0.034965 3.5%
At ~96.5% confidence, the test results are significantly different.
P
8 15 8
0 6 0
15
6
0 0.001399
Group 1 Group 2 Totals
Pass 6 1 7
Fail 0 8 8
Totals 6 9 15
Applied Reliability
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49
Class Project #8Fisher’s Exact Test
In an experiment to compare different
treatments, the old method produced four
rejects out of twenty. In the second
experiment, the new procedure resulted in
zero rejects out of fifteen. How statistically
significant is the improvement between the
new and old?
Applied Reliability
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50
Class Project #8Fisher’s Exact Test
In an experiment to compare different
treatments, the old method produced four
rejects out of twenty. In the second
experiment, the new procedure resulted in
zero rejects out of fifteen. How statistically
significant is the improvement between the
new and old?
New Old Totals
Pass 15 16 31
Fail 0 4 4
Totals 15 20 35
0925.0035
15
435
015
4
0
P
=HYPGEOMDIST(0,4,35,15)=0.092532
Applied Reliability
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51
Acceptance Sampling
< Y % > Y %
Accept Correct Type II error
Reject Type I error Correct
Decision
on Lot
Population Value (% Defective)
Matrix of Possible Choices
Risks
Accept or reject a lot based on the
results of looking at a sample of size n
randomly drawn from a lot of size N.
Several possible risks involved in
deciding whether or not a lot (the
population) has a maximum allowed
percent defective level, say Y%:
-Reject a good lot with percent
defective below Y% and commit a Type
1, a, or producer's error.
-Accept a bad lot with percent
defective above Y% and commit a Type
2, b, or consumer's error.
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52
Operating Characteristic(O.C.) Curve
Plot of PA versus p:
n = 50 c = 3
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20%
Lot Percent Defective
Pro
bab
ilit
y o
f A
ccep
tan
ce
OC Curve
For a given sampling plan (sample size n and
acceptance number c), the probability of
acceptance will depend on the incoming lot
percent defective p. A plot of the probability of
acceptance PA versus the lot percent defective
p is called an O.C. curve.
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53
Calculations for an O.C. Curve
Use binomial distribution with n, varying p, to
calculate probability of acceptance, that is, P( X < c ):
Example:
Sketch the O.C. curve for a sample size n = 100 and
acceptance number c = 2.
The probability of acceptance is:
With n = 100, now try different p values, using ART.
For p = 0.01, the probability of acceptance P is
0.3660+0.3697+0.1849 = 0.9206.
For p = 0.05, the probability of acceptance P is
0.0059+0.0312+0.0812=0.1183.
For p = 0.10, the probability of acceptance P is
0.0000+0.0003+0.0016=0.0019.
1
22
2 0 1 2
0 1
1 1
12 1
2
n
n
n
P X P P P
P p
P np p
n nP p p
Method:
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54
OC Curves
Fixed c, Varying Sample Sizes n:
Fixed n, Varying Acceptance Numbers c:
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0% 5% 10% 15% 20% 25%
Lot Percent Defective
Pro
ba
bilit
y o
f A
cc
ep
tan
ce
c = 3
n = 500
n = 100
n = 300
n = 50
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0% 5% 10% 15% 20% 25%
Lot Percent Defective
Pro
ba
bilit
y o
f A
cc
ep
tan
ce
n = 50
c = 0 c = 2c = 1 c = 3
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55
O.C. Curve Calculations Using ART
Applied Reliability
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56
Sampling Plan
A sampling plan is a specified sample size n and
acceptance number c. A sampling plan generates a
specific O.C. curve.
The sampling plan is determined by specifying four
values, equivalent to two points on the O.C. curve:
1. The p value for an incoming lot for which the
probability of acceptance is very high (e.g., 95% or
producer’s risk a = 5%).
2. The p value for an incoming lot for which the
probability of rejection is very high ( e.g., 90% or
consumer’s risk b = 10%).
These four numbers (two p values and associated
risks) uniquely determine the sampling plan.
We assume the lot size N is large relative to n, or
equivalently, that we are sampling from a process, rather
than from lots. This assumption allows using simpler
binomial distribution for calculations rather than the more
complex hypergeometric distribution.
Several procedures exist to generate sampling
plans: published tables, nomograph, computer
programs, iterative calculations, etc.
Applied Reliability
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Sampling Plan Nomograph
57
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58
Generating a Sampling Plan Using ART
2%, 5%; 8%, 10%AQL RQLa b
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59
LTPD Sampling Plans
What if our primary concern was viewing sampling plans that would assure rejection with high probability, say 90%, of a defect level that was the maximum value a consumer could tolerate.
Such plans are called LTPD (lot tolerance percent defective) plans. The next table is based on a 10% consumer’s risk at various rejectable quality levels.
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60
LTPD TableConsumer’s Risk b = 10%
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61
Adjusting Sampling Plans
Reducing the Sample Size:
Suppose the qualification sampling plan is
to accept on 3 or less rejects out of 300
units. Because of the cost or lack of
availability of 300 units, we wish to reduce
the sample size while holding the
consumer’s risk constant. What is the
sample size if 2 or less rejects are allowed,
or 1 or less, or even zero?
Applied Reliability
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Adjusting Sampling Plans
Procedure:
Refer to figures following.
1. Find the graph corresponding to the present acceptance number, say 3.
2. Find the intersection of the sample size on the horizontal x-axis with the diagonal line in the graph labeled with the probability of acceptance value “10,” corresponding to a 10% consumer’s risk.
3. Read, on the vertical y-axis, the percent defective value associated with the 10% consumer’s risk.
4. Find the graph with a lower acceptance number and reverse the above procedure.
5. Using the percent defective value previously found for the consumer’s risk on the y-axis of the new graph, find its intersection with the labeled “10” line.
6. Dropping vertically at this intersection point to the x - axis, read off the new sample size for this lower acceptance number.
62
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Graphs for Adjusting Sampling Plans
63
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Graphs for Adjusting Sampling Plans
64
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65
Class Project #9
Adjusting a Qualification Sampling Plan:
The qualification plan calls for allowing
two rejects out of two hundred units.
The engineer needs to reduce the
sample size of the study. If he wants to
keep the same defect level at which the
consumer’s risk is 10% for lot
acceptance, what is the necessary
sample size, allowing only one reject.
What is the LTPD?
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66
Class Project #9
Adjusting a Qualification Sampling Plan:
The qualification plan calls for allowing
two rejects out of two hundred units.
The engineer needs to reduce the
sample size of the study. If he wants to
keep the same defect level at which the
consumer’s risk is 10% for lot
acceptance, what is the necessary
sample size, allowing only one reject.
What is the LTPD?
From Figure 9.15 (c = 3), 10% consumer’s
risk is at 2.2% defective.
From Figure 9.13 (c = 1) , at 2.2% defective,
the 10% consumer’s risk corresponds to
sample size 175 units.
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67
Minimum Sample Size Plans
The smallest sample size for a plan having a specified rejectable quality level occurs when the acceptance number is zero.
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68
Minimum Sample Size PlansZero Failures
Protection Against Consumer's Risk:
For zero failures and rejection
probability (1 - b), the minimum
sample size to reject a lot with fraction
defective value p is given by:
ln
ln 1n
p
b
Plot of Minimum Sample Size Versus Fraction Defective
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69
Using ART “Minimum Size Sampling Plans”
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70
Class Project #10
Sampling Plans:
We're willing to accept at most a 10%
chance of permitting a failure fallout as
high as 1.0%.
A. If we allow up to three rejects, what
sample size do we need ? See Table
9.3 or ART Sampling Plans.
B. It turns out, we have only 500 pieces
available. What should our acceptance
number be now?
C. What is the minimum sample size we
need, that is, allowing zero failures?
Applied Reliability
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71
Class Project #10
Sampling Plans:
We're willing to accept at most a 10% chance
of permitting a failure fallout as high as 1.0%.
A. Allowing up to three rejects, what sample
size do we need?
Table 9.3: 668
B. It turns out, we have only 500 pieces
available. What should our acceptance
number be now?
Table 9.3: c = 1 for 390 units
c = 2 for 533 units
C. What is the minimum sample size we need,
that is, allowing zero failures?.
Table 9.3: 231
ART (Min. SS): 230
Formula:
ln0.1229
ln(1 .01)n
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72
Sampling PlansThe Meaning of AQL and LTPD
AQL: Acceptable quality level
The incoming lot percent defective with
typically a 95% chance of acceptance. If
a lot is rejected (that is, the number of
rejects is above the acceptance number),
we state with 95% confidence that the lot
defect level is above the AQL.
LTPD: Lot tolerance percent defective
The incoming lot percent defective with a
10% chance of acceptance. If a lot is
accepted (that is, the number of rejects is
equal to or below the acceptance number),
we can state with 90% confidence that the
lot defect level is below the LTPD.
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73
Sampling PlansOutgoing Quality
AOQ: Average outgoing quality
Average level of defects shipped. Found
by multiplying lot percent defective by
probability of acceptance, that, is pxP.
Generates AOQ curve. Assumes rejected
lots are screened to make lot perfect.
AOQL: Average outgoing quality limit
Maximum value of AOQ curve. With 100%
inspection of rejected lots, AOQ to customer
is never any worse than AOQL.
Applied Reliability
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74
Sampling Plans
AOQ Curve and AOQL
EXCEL Worksheet
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75
AOQ Curves Using ART
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76
Statistical Process Control
Control Charts for Reliability
If p is the historical process average,
the upper control limit is given by
Three-Sigma Control Chart for Binomial Proportions
1
1UCL
p pp z
na
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77
Cumulative Count Control Chart
Control Based on Good Units
CCC chart concentrates on number of
good units produced instead of number of
defective items.
Accumulated number of good units is
plotted in time and compared to control
limits such that a cumulative count of good
units will fall inside the limits when a
process is in control.
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78
Cumulative Count Control Chart
Example
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79
Cumulative Count Control ChartPoints Outside Control Limits
Interpretation
If the first failure occurs before the
cumulative count exceeds the lower
limit, the process is not capable of
meeting the PPM level specified for the
control limits.
If the first failure occurs after the
cumulative count exceeds the upper
limit, the process is demonstrating even
better performance than the PPM level
specified for the control limits.
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80
Cumulative Count Control Charts for Low PPM
p
.Median
69310
pln
/lnnLCL
a
1
21
pln
/lnnUCL
a
1
2
Centerline Set at Median:
Lower Control Limit:
Upper Control Limit:
Centerline and 90% LCL and UCL can be
obtained from the following table by reading in
the row for a given PPM under the columns
labeled 0.5, 0.95, and 0.05, respectively.
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81
Minimum Sample Sizes for Zero Rejects at Various Probabilities
Probability of Zero Rejects
PPM 0.999 0.995 0.99 0.975 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.025 0.01 0.005 0.001
1 1001 5013 10051 25318 51294 105361 223144 356675 510826 693147 916291 1203973 1609438 2302584 2995731 3688878 4605168 5298315 6907752
6 sigma:
3.4 295 1475 2956 7447 15087 30989 65631 104905 150243 203867 269497 354110 473364 677230 881097 1084963 1354460 1558326 2031690
5 201 1003 2011 5064 10259 21073 44629 71335 102165 138630 183258 240794 321887 460516 599145 737775 921032 1059661 1381548
10 101 502 1006 2532 5130 10536 22315 35668 51083 69315 91629 120397 160943 230258 299572 368887 460515 529830 690773
20 51 251 503 1266 2565 5268 11158 17834 25542 34658 45815 60199 80472 115129 149786 184443 230257 264914 345385
50 21 101 202 507 1026 2108 4463 7134 10217 13863 18326 24079 32188 46051 59914 73776 92102 105964 138152
75 14 67 134 338 684 1405 2976 4756 6811 9242 12217 16053 21459 30700 39942 49184 61400 70642 92100
100 11 51 101 254 513 1054 2232 3567 5109 6932 9163 12040 16094 23025 29956 36887 46050 52981 69075
200 6 26 51 127 257 527 1116 1784 2554 3466 4581 6020 8047 11512 14978 18443 23024 26489 34536
5 sigma:
233 5 22 44 109 221 453 958 1531 2193 2975 3933 5167 6907 9882 12856 15831 19763 22737 29644
500 3 11 21 51 103 211 447 714 1022 1386 1833 2408 3219 4605 5990 7376 9209 10594 13813
1000 1 6 11 26 52 106 224 357 511 693 916 1204 1609 2302 2995 3688 4603 5296 6905
5000 1 3 6 11 22 45 72 102 139 183 241 322 460 598 736 919 1058 1379
4 sigma:
6210 1 2 5 9 17 36 58 83 112 148 194 259 370 481 593 740 851 1109
10000 1 3 6 11 23 36 51 69 92 120 161 230 299 368 459 528 688
3 sigma:
66807 1 2 4 6 8 11 14 18 24 34 44 54 67 77 100
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82
Class Project #11
Cumulative Count Control Chart:
A process is assumed to run at 500 PPM.
A cumulative count control chart is
desired for monitoring purposes.
Find the centerline and upper and lower
non-defective cumulative count values for
a 95% control limit band.
Applied Reliability
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83
Class Project #11
Cumulative Count Control Chart:
A process is assumed to run at 500 PPM.
A cumulative count control chart is desired for
monitoring purposes.
Find the centerline and upper and lower non-
defective cumulative count values for a 95%
control limit band.
From Table: CL = 1386, LCL = 51, UCL = 7376
Formulas:
Compare to the tabulated values for “Minimum
Sample Sizes for Zero Rejects at Various
Probabilities.”
Matches!
6
0.69311386
500 10
ln(1 0.05 / 2)51
ln(1 0.0005)
ln(0.05 / 2)7376
ln(1 0.0005)
CL
LCL
UCL
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84
Using ART for “Cumulative Count Control Charts”