l berkley davis copyright 2009 mer301: engineering reliability lecture 5 1 mer301: engineering...

L Berkley DavisCopyright 2009

MER301: Engineering ReliabilityLecture 5

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MER301: Engineering Reliability

LECTURE 5:

Chapter 3:Probability Plotting,The Rules of Counting, Binomial Distribution, Poisson Distribution



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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

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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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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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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 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


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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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 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


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)


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



l berkley davis copyright 2009 mer301: engineering reliability lecture 5 1 mer301: engineering...

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engineering reliabilitylecture

n objects

number of arrangements

specific order ismer301

order isthis

order of selection

rulethe total number

specific order equals