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L Berkley DavisCopyright 2009
MER301: Engineering ReliabilityLecture 5
1
MER301: Engineering Reliability
LECTURE 5:
Chapter 3:Probability Plotting,The Rules of Counting, Binomial Distribution, Poisson Distribution
L Berkley DavisCopyright 2009
MER301: Engineering ReliabilityLecture 5
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Summary of Topics Probability Plotting
Rules of Counting
Binomial Distribution
Poisson Distribution
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Probability Plotting Graphical method of testing a data set to
see if it conforms to some specific distribution
For an ordered data set x(j) the Cumulative Distribution Function is plotted x(j)vs (j-0.5)/n where j=1,…,n
Probability plotting often used in failure analysis to predict future failures….
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Normal Probability Plot
180 190 200 210 220
95% Confidence Interval for Mu
190 200 210
95% Confidence Interval for Median
Variable: C1
A-Squared:P-Value:
MeanStDevVarianceSkewnessKurtosisN
Minimum1st QuartileMedian3rd QuartileMaximum
185.662
9.652
184.315
0.2570.636
195.700 14.032
196.90.511343-6.2E-01
10
176.000184.500191.500207.250220.000
205.738
25.617
208.081
Anderson-Darling Normality Test
95% Confidence Interval for Mu
95% Confidence Interval for Sigma
95% Confidence Interval for Median
Descriptive Statistics
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Normal Probability Plot
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Normal Probability Plot
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Rules of Counting
Fundamental Rule of Counting
Permutation Rules
Combinations
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Rules of Counting
Fundamental Counting Rule The total number of ways a sequence of k events can occur
with ni denoting the number of ways the ith event (i= 1 to k) can occur is
For buying a computer, there are choices of three hard drives n1, two levels of RAM n2, two video cards n3, and three monitors n4 so the total number of options is
knnnn 21
3632234321 nnnnn
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Rules of Counting-con’t Permutation Rules
A permutation is an arrangement of n distinct objects in a specific order. The number of arrangements of n distinct objects in a specific order equals
If the n objects are taken k at a time, then the number of arrangements in a specific order is
A case where the order of selection is not important is called a Combination. The number of ways of selecting k objects from n objects without regard to order is
This is the permutation divided by k!, the number of ways the k objects can be arranged…
)!(
!
kn
n
!n
)!(!
!
knk
n
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Rules of Counting-con’t
Combinations A case where the order of selection is not important is called a Combination.
The number of ways of selecting k objects from n objects without regard to order is
This is the permutation divided by k!, the number of ways the k objects can be arranged…
For the case where the objects are taken n at a time but some are identical, the number of possible arrangements in a specific order is
)!(!
!
knk
n
nrrr
rrr
n
j
j
21
21 !!!
!
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Rules of Counting-con’t
A Combination for n objects taken k at a time is equal to the Permutation of the n objects (n!/(n-k)! )divided by the number of ways the k objects can be arranged (k!)
For five objects taken three at a time
ARRANGEMENTS OF A,B,C,D,E TAKEN THREE AT A TIME
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDEACB ADB AEB ADC AEC AED BDC BEC BED CEDBCA BDA BEA CDA CEA DEA CDB CEB BED DECBAC BAD BAE CAD CAE DAE CBD CBE DBE DCECAB DAB EAB DAC EAC EAD DBC EBC EBD ECDCBA DBA EBA DCA ECA EDA DCB ECB EDB EDC
10606
1
)!35(
!5
!3
1
)!35(!3
!5
)!(!
!
knk
n
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Binomial Distribution
The Binomial distribution describes the results of n independent identical success-failure trials Constant chance of a success or failure
outcome(called probability p) Knowing the outcome of any one
repetition does not change chance of any other repetition
Must be able to count the number of successes and failures
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Binomial Distribution
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Binomial Distribution Combinations and Mean/Variance
From counting combinations, the number of combinations of n distinct objects selected x at a time is given by
Mean and Variance of the Binomial Distribution
)!(!
!
xnx
n
x
n
123)2()1(! nnnn
)1(
)(
ppnV
pnXE
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Binomial Distributions for n=10 and p=0.1,0.5,0.9
X n p p(X=x) x*p(X=x)0 10 0.9 1E-10 01 10 0.9 9E-09 9E-092 10 0.9 3.64E-07 7.29E-073 10 0.9 8.75E-06 2.62E-054 10 0.9 0.000138 0.0005515 10 0.9 0.001488 0.007446 10 0.9 0.01116 0.0669627 10 0.9 0.057396 0.4017698 10 0.9 0.19371 1.5496829 10 0.9 0.38742 3.48678410 10 0.9 0.348678 3.486784
n*p Mean9 9
Variance=0.9
X n p p(X=x) x*p(X=x)0 10 0.1 0.34867844 01 10 0.1 0.38742049 0.3874204892 10 0.1 0.19371024 0.3874204893 10 0.1 0.05739563 0.1721868844 10 0.1 0.01116026 0.0446410445 10 0.1 0.00148803 0.0074401746 10 0.1 0.00013778 0.0008266867 10 0.1 8.748E-06 6.1236E-058 10 0.1 3.645E-07 2.916E-069 10 0.1 9E-09 8.1E-0810 10 0.1 1E-10 0.000000001
n*p Mean1 1
Variance=0.9
X n p p(X=x) x*p(X=x)0 10 0.5 0.0009766 01 10 0.5 0.0097656 0.009765632 10 0.5 0.0439453 0.087890633 10 0.5 0.1171875 0.35156254 10 0.5 0.2050781 0.82031255 10 0.5 0.2460938 1.230468756 10 0.5 0.2050781 1.230468757 10 0.5 0.1171875 0.82031258 10 0.5 0.0439453 0.35156259 10 0.5 0.0097656 0.0878906310 10 0.5 0.0009766 0.00976563
n*p Mean5 5
Variance=2.5
L Berkley DavisCopyright 2009
Binomial Distributions for n=10 and p=0.1,0.5,0.9
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Excel formula for Binomial=binomdist(x,n,p,f(X=x)=false)
=binomdist(x,n,p,cumulative=true)
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Example 5.1 An electronics manufacturer claims that 10%
or less of power supply units fail during warranty. To test this claim, an independent laboratory purchases 20 units and conducts accelerated life testing to measure failure during the warranty period. Let p denote the probability that a power supply unit
fails during the testing period. The laboratory data resulting from testing will be
compared to the claim that p0.10. Let X denote the number among the 20 sampled that
fail. What is the expected value and variance of X if the manufacturer’s claim is true?
Find the probability that less than 5 of the 20 power supplies will fail if the manufacturer’s claim is true
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Normal Approximation to the Binomial Distribution For values of np>5 and n(1-p)>5,
probabilities from the binomial distribution can be approximated
as follows:
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Poisson Distribution
The Poisson Distribution gives the predicted probability of a specific number of events occurring in an interval of known size, when the mean number of events in such intervals is known Number of goals scored in a game Errors in transmission of data
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Poisson Distribution Assumptions for Poisson Distribution
Random discrete events that occur in an interval that can be divided into subintervals
Probability of a single occurrence of the event is directly proportional to the size of a subinterval
If the sampling subinterval is sufficiently small, the probability of two or more occurrences of the event is negligible
Occurrences of the event in nonoverlapping subintervals are independent
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Poisson Distribution
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Poisson Examples:lambda =1,4.5,9
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lambda=1lambda x f(X=x) Cumulative
1 0 0.3678794 0.367879441 1 0.3678794 0.735758881 2 0.1839397 0.91969861 3 0.0613132 0.981011841 4 0.0153283 0.996340151 5 0.0030657 0.999405821 6 0.0005109 0.999916761 7 7.299E-05 0.999989751 8 9.124E-06 0.999998871 9 1.014E-06 0.999999891 10 1.014E-07 0.999999991 11 9.216E-09 11 12 7.68E-10 11 13 5.908E-11 11 14 4.22E-12 11 15 2.813E-13 1
lambda=4.5lambda x f(X=x) Cumulative
4.5 0 0.011109 0.0111094.5 1 0.04999 0.06109954.5 2 0.112479 0.17357814.5 3 0.168718 0.3422964.5 4 0.189808 0.53210364.5 5 0.170827 0.70293044.5 6 0.12812 0.83105064.5 7 0.082363 0.91341354.5 8 0.046329 0.95974274.5 9 0.023165 0.98290734.5 10 0.010424 0.99333134.5 11 0.004264 0.99759574.5 12 0.001599 0.99919494.5 13 0.000554 0.99974844.5 14 0.000178 0.99992634.5 15 5.34E-05 0.9999797
Poisson Distribution
lambda=9.0lambda x f(X=x) Cumulative
9 0 0.00012341 0.000123419 1 0.001110688 0.0012340989 2 0.004998097 0.0062321959 3 0.014994291 0.0212264869 4 0.033737155 0.0549636419 5 0.060726879 0.1156905219 6 0.091090319 0.206780849 7 0.117116124 0.3238969649 8 0.13175564 0.4556526049 9 0.13175564 0.5874082449 10 0.118580076 0.705988329 11 0.097020062 0.8030083839 12 0.072765047 0.8757734299 13 0.050375802 0.9261492319 14 0.032384444 0.9585336759 15 0.019430666 0.9779643419 16 0.01092975 0.9888940919 17 0.005786338 0.9946804299 18 0.002893169 0.9975735989 19 0.001370449 0.9989440469 20 0.000616702 0.999560748
Excel Formula for Poisson=poisson(x,lambda,f(X=x)=false)
=poisson(x,lambda,cumulative=true)
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Poisson Examples:lambda =1,4.5,9
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Poisson Process
Observing discrete events in a continuous “interval” of time, length or space The number of white blood cells in a drop
of blood The number of times excessive pollutant
levels are emitted from a gas turbine power plant during a three-month period
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Example 5.2 Yeast is added when mash is prepared for fermentation in the
beer making process. The yeast is cultured in vats and the exact amount of yeast added to the mash is of critical importance. The yeast cell count in culture fluid averages 6000 yeast cells per cubic millimeter of fluid in the culture vats. It is however necessary to know the concentration in a specific vat to know how much fluid to add to the mash. The distribution of yeast cells is known to follow a Poisson Distribution. To establish the yeast cell concentration, a 0.001 cubic
millimeter drop is taken and the number of yeast cells X is counted
What is the probability of getting two or less yeast cells in a single sample?
What is the probability of getting more than two yeast cells in a single sample?
How many yeast cells are expected if the sample is from a typical culture vat?
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Normal Approximation of the Poisson Distribution
Poisson developed for case of n approaching infinity For mean >5 then:
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Summary of Topics Probability Plotting
Rules of Counting
Binomial Distribution
Poisson Distribution
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