me4504 02 basic reliability

8/9/2019 ME4504 02 Basic Reliability

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ME4504:Aircraft MaintenanceEngineering

Dr. Y.C. Tong

Location: A521, Chun Tze Yuen Buliding

Email: [email protected]

Tel: (852) 2766 4502

1
mailto:[email protected]:[email protected]


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Syllabus1.Maintenance Systems

2.Reliability and Rates of Failure

3.Redundancy

4.Failure Interactions

5.Risk Analysis and Error Reduction in Aircraft

Maintenance

6.Aircraft Maintenance Programme Management

2

ME4504

:AircraftMaintenanceMana

gement


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Reliability Analysis

3

Basic Reliability Mathematics


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What does Reliability has to

do with Safety?

To reduce injury rates and losses, one must reduce the number ofdefects and subsequent failures. To reduce failures and defects,reliability must be improved.

Safety is achieved by reliability management.Reliability management utilizes reliability methodsand tools to determine and reduce the probabilityof failure.

4

Safety and Reliability Go Hand in HandReliability is how Safety is quantified.

Safety: the state of being safe, freedom from the occurrence

or risk of injury, danger, or loss.

Reliability: the ability to perform a specific function under

given conditions for specified period of time without failure.


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What is Reliability? 5

The probability of performing aspecific function under givenconditions for specified period of timewithout failure.

Most of us will have some concept of what

reliability is from our everyday life. For example,we may express how unreliable our cars are, or

how reliable the MTR is (or not), or how reliable

the wireless network is, or how reliable aviation is.

One can deduce what it means to be Reliable isfrom these above cases. It could perhaps be: the

quality of a particular object, device or system in

consistently meeting a function time and time and

again. From mathematical perspective, reliability

is:


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Reliability 6When we say a particular device is reliable, it describes:

a characteristic of the device that it has demonstrated the ability toconsistently perform its intended function without failure.

Reliability is designed and planned for in the development stages

of product and in operation with maintenance procedures. It

cannot be added in after the design and development stage.

Quantitatively, reliability is a probability. Reliability is a

number between zero and one.

Reliability of 1 means the device, system or service performs

without failure 100% of the time.

Reliability of 0 means the device, system or service performs

without failure 0% of the time.

Understanding why these failures occur is key to improving the

design, and hence reliability.


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Terms and Definitions 7The probability of performing a specific function under given

conditions for specified period oftime without failure.

These are often said to be the 4 key elements of Reliability.

Function: The device whose reliability is inquestion must perform a specific function. If we

use a flat head screw driver as a wedge, andthen the screw driver breaks after a couple

knocks of the hammer, this would not be

classified as a defect or degradation failure. It

could be classified as a failure due to improper

use of the tool.

Condition: The device must perform its function within the designed

conditions. For example, if a watch is water resistance up to 20

meters, and a scuba driver took the watch below 50 meters of depth

of water and the watch became waterlogged, we should not

classified this event as a failure due to defect or degradation.

However, it may be classified as an accidental failure


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Failure8

Failure in general is defined as the performanceof a device (process) outside of a specified value,and inability to perform its function for a given timeperiod within specified conditions.

The non-performance can be due to: Defect: imperfection

Deficiency: lack of conformance to specs

Fault: Cause of failure

Malfunction: unsatisfactory performance

Break down

What constitute and how to classify failures must

be well defined prior to testing and the use of the

component or system under study.For example, if the function of a pump is to deliver at least 200

litres of fluid per minute and it is now delivering 150 litre per

minute, the pump failed to perform its intended function, it has

therefore failed by definition. Failure to perform its intended

function can have significant consequence on the system as a

whole.


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Terms and Definitions cont. Time: The moment a device is introduced into service, it is at the

mercy of the operating environment (and the maintenance

programme). Most devices are required to perform over a period

of time, and generally the reliability of the device will change over

time.

Random Variable: A random variable is a variable where its

value has a probability of occurrence associated with it. In rolling

a fair dice, the possible outcomes 1, 2, 3, 4, 5 and 6 are random

variables. Each has a probability of occurrence of 1/6.

In addition:

Reliability

Failure Probability

Probability density

Hazard function

9


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Reliability

The Reliability Function, R(t), is also

known by numerous other names.

10

Reliability Probability R(t): The probability that a device

or system will perform its intended function for a given

interval of time under specified operating conditions.

R(t) is a probability, hence:

Equivalence:

R(t): Reliability distribution function S(t): Survival function or Probability

of Survival

Complementary distribution function

R(t) or S(t) or RX(x) or Rt(t) or P(Tt)

or P(Xx)

10 tR

Component / device / system

reliability is a function of time

due to inherent defect growth

and wear-out are functions of

time / usage.


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Failure ProbabilityFailure Probability F(t): The probability that a device or

system will not perform its intended function for agiven interval of time under specified operating

conditions.

11

Equivalence:

PoF(t): Probability of Failure

Cumulative Distribution function (CDF)

Failure function

F(t) or PoF(t) or Fx(x) or Ft(t) or P(Tt) or P(Xx)

Since a unit either fails or survives, the failure probability

F(t) given in terms of the Reliability function R(t) is:

F(t) is a probability, hence:

10 tF

tRtF 1

F(t)R(t)

Venn diagram

Population = 1


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12Probability Density Function (pdf)

For continuous probability distributions, the Probability

Density function f(t) is the instantaneous change of FailureProbability F(t) with respect to the random variable t, i.e.

the derivative of the Failure Probability.

Equivalence:

pdf: probability density function

Density function

Probability density

dt

tdFtf

Since F(t) is a monotonically increasing function,

ttf ofvaluesallfor0

0:)(

)()(

1

1

0

1

tdttf

dttftF

t

t

or


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Failure Rate / Hazard Rate / Risk

The hazard function h(t) is the ratio of the probability density

function f(t) to the survival function R(t).

The hazard function indicates how often a failure occurs per unit

time, and failure-rate values generally change over time.

tRtf

th

11

ln

lnln

1111

tR

RtR

tR

tdRdt

tR

dttdRdt

tR

tfdtth

tttt

1

exp1t dtthtR

An alternative definition of the reliability function in terms of the

hazard function is:

Equivalence:

FR(t): Failure rate

h(t) Hazard function

Risk

Here is the proof.


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Mutually ExclusiveWhen we say two events (say "A" and "B") are mutually

exclusive, it means it is impossible for them to happen together.Hence:

BA

ABPAPBAP |Probability of event A and event B equals

the probability of event A times the

probability of event B given event A has

occurred.

Conditional Probability

P(A and B) = 0, i.e. Probability of A

and B together equals 0.

The probability of A or B is the sum

of the individual probabilities, i.e.P(A or B) = P(A) + P(B)

B AAB


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Conditional Probability Example 1

From a deck of 52 cards, what is

the probability of drawing 2 Aces?

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Solution:

Event A is drawing an Ace first.

Event B is drawing an Ace second

Since there are 4 Aces in a deck of 52

cards, the chance of drawing an Ace in

the first go is 4 out of 52, P(A) = 4/52.

Given that one Ace has been drawn from

the deck, the probability of the 2nd card

drawn is an Ace is less, only 3 of the 51,P(B|A) = 3/51.

4 x As

52 Cards

So: P(AB) = P(A) x P(B|A)

= (4/52) x (3/51)

= 12/2652 = 1/221

So the chance of getting 2 Aces is 1 in 221, or about 0.5%.

3 x As

51 Cards


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16Conditional Probability Example 2From a deck of 52 cards, what is the probability of

drawing a 2 given Spade is being drawn?Solution:

Event A is drawing Spades ().

Event A B is drawing a 2 of Spades from a

deck of cards.

Since there are 13 Spades in a deck of 52

cards, the chance of drawing a Spade in the

first go is 13 out of 52, P(A) = 13/52

Drawing a 2 of Spade from a deck is no

different from drawing any 1 card, theprobability is, P(A B) = 1/52

2s

2

52 Cards

So: P(B|A) = P(AB) / P(A)

= (1/52) / (13/52)

= 1/13

The chance of getting 2 given Spade has been drawn is 1 in 13 ( ~ 7.69%).

2


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Failure Data

Life and failure data allow the likelihood of the failure occurring,

and probability of failure events from occurring.

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Data (life and failure) analysis is the process of collecting and

analyzing data and to categorize the cause of a failure. It is a

vital tools used in reliability analysis, and the subsequentdevelopment of new products and for the improvement of

existing products.

Reliability is assessed and improved by analyzing life and failure

data. Such data can be obtained from:

Field studies of system performance Controlled laboratory tests, sometimes called Life Tests.

Data can be Descriptive (like "high" or "fast") or Numerical

(numbers). Numerical Data can be Discrete or Continuous: Discrete data is counted

Continuous data is measured


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Probability Estimate 18 An estimate of the true probability of an event can be

obtained from a number of trials (experiments) of the event.The larger the number of trials the more likely it is that we are getting very close to the

true probability.

When the number of trials is small, the estimate may be

quite off, it may be over-pessimistic or over-optimistic, but it

also may be very close to the true value.

The mathematical relation between the true probability of an

event and an estimate of that probability obtained from N

trials of the event is a limit function, i.e., the true probability

is obtained as N approaches infinity. If in N trials, weobtainednf outcomes which yielded the event tti , then the

probability estimate of tti is defined as:

1

lim

N

ttnttF

if

N

i


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Example Crankshaft Fatigue Failures19

10 crankshafts are randomly selected and

tested in accelerated stress test. Table

shows the cycles to failure obtained from the

test. The probability distribution estimate of

the cycle to failuret is calculated by:

The result is showed in the table and the

figure below.

Consider the case of the fatigue failure of

the crankshaft.

1

N

ttnttF

if

i


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Exercise20

1: Compute and plot the probability

distribution estimate of the data set shown inthe table.

Solution1:

1lim

N

ttnttF

if

Ni

2: Give reasons why you think the denominator of the probability

distribution estimate equation isN+1, and not justN.

Answer:ForanysampleofNdatapoints,amean(average)canbecalculated.

Themeanisequivalenttooneextradatapointofthesample,hence(N+1)


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The mean, E(X), is the average value of the random variable.

The median, xmedian, is the value of the random variable that

corresponds to F(x) = 0.5 (50%).

The mode, xmode, is the value of the random variable that ismost frequent, i.e. max(f(x)).

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For symmetrical probability distributions, the mean, median and mode

are the same. For non-symmetrical distribution functions, the mean,

median and mode are not the same.

Mean, Median and Mode


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Exercise22

Calculate the mean, median and mode for

the data showed in the table.


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23

THE BATHTUB CURVEA bathtub curve describes the failure rate (hazard function)

over the life cycle of a population of a product.


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The Bathtub CurveThe pattern of the death rate (death/day or death/year) of people over

a life-time has a shape of a bathtub hence the name Bathtubcurve. The Bathtub curve has 3 distinctive modes of failure:

An infant morality

Deaths by accidents

Aging

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It was discovered that the failure

rate of electronic equipment over

the life-cycle also follows the

pattern of a bathtub curve. The

Bathtub curve has 3 distinctivephases or modes of failure:

Early failures

Random failures

Wear-out failures


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Early-Failure / Infant Morality Phase

Phase 1: An infant mortality early life phase characterized

by a decreasing failure rate (Phase 1). Failureoccurrence during this period is not random in time

but rather the cause of sub-standard components

with gross defects and that escaped the quality

controls of the manufacturing process. In software,

it is caused by undetected programming errors.

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Impact of Infant Morality

Infant Mortality, from a customer satisfaction viewpoint, is

unacceptable. They damages customer confidence and companyreputation.

They are caused by design flaws, defects built into a product.

However, even the best design intent can fail to cover all possible

interactions of components in operation.


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Early-Life Failure Prevention

Stress test shall be applied: Stress testing shouldbe started at the earliest development phases and

be used to evaluate design weaknesses and

uncover specific assembly and materials problems.

Stress testing shall be applied with increasing

stress levels until ultimate stresses or failures are

brought about. The failures should be investigated

and design improvements should be made toeliminate design and material defects that would

otherwise show up with product failures in the field.

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For some industries, a stress test after manufacturing of products can be

valuable to burn-in the product to weed out manufacturing defects in a

product that escaped product quality control.

To avoid infant mortalities, the product manufacturer

must have a reliability system and process, and

feedback loop set in place to determine andeliminate al potential defects. This includes also

appropriate specifications and standards, adequate

design tolerance and sufficient component de-

rating.


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Steady State / Useful Life Period

Phase 2: A useful life period where devices, products have a

relatively constant failure rate caused by randomlyoccurring defects, accidents and overstresses. Failures

are caused by unexpected and sudden over stress

conditions. The aviation safety management system,

and most reliability analyses pertaining to electronic

systems are concerned with lowering the failurefrequency during this period.

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Does Phase 2 Really Exist?

Some reliability specialists pointed out that real products

don't exhibit constant failure rates.Whether this is true or not depends on how one defines the

Steady State phase. Failure in the Steady State phase isn't

usually related to defects or wear-out, but rather related to

accidents in the field. Similar to how a persons life can be

loss (randomly) via accidents before reaching old age.


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Steady-State Period

The Steady State period is characterized by:

Constant failure rate, Steady-state period, where failure rate ismuch lower than in early-life failure period.

Failures are caused by random environmental shocks and/or

operation accidents, where times between failures can be

characterized by the MTTF and MTBF, and hence the

Exponential distribution. It is possible that there are still "infant" mortalities occurring

well beyond the typical burn-in period and well into the steady

state period. It can only be differentiated via failure root cause

analysis.

Steady State

(constant FR)

InfantMortality

(Early Life

Failures)

FailureRateh(t)

Operating Time

Infant morality

running well into

steady state period


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Prevent Steady State Period Failure

Random failures, being random, cannot be directly prevented

by replacement or maintenance of the device or system. Infact, intervention can increase the failure rate via human errorand infant morality failures.

Minimize and prevent human errors: Human error has beendocumented as a primary contributor to more than 70%* of

commercial airplane hull-loss accidents. Human error istypically associated with flight operations, maintenancepractices and air traffic management. Reducing human errorand improving human performance will help towards lowerfailure rates, improving safety and efficiency.

Implement best standards and practices, and Safetymanagement policies: Dedication to better understanding howhumans can most safely and efficiently be integrated with thetechnology. Then translate this understanding into design,training, policies, or procedures to help humans perform better.

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* (OHare, Wiggins, Batt, & Morrison, 1994; Wiegmann and Shappell, 1999; Yacavone, 1993)


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Wear-Out Phase

Phase 3: A wear out period where the failure rate increases due

to critical parts wearing out (Phase 3). This correspondsto a normal wear and tear, and degradation period. As

parts wear out, it takes less stress to cause failure and

the overall system failure rate increases, accordingly

failures do not occur randomly in time.

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Wear-out phase is characterized by:

Failure rate increasing rapidly with age.

Wear-out phase is most applicable to mechanical systems

and other systems that are subjected to some kind of stress

cycles (e.g. thermal and vibrational).

Some electronic hardware do not exhibit wear out failure

during its intended service life.

Weibull Failure Model can be used


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32Wear-out Failure Prevention

For many mechanical assemblies, the wear-out time are oftendesigned to be less than the designed operational life of the whole

product. In such cases, preventive replacement of assemblies are

used to extend the operational life of the product - replacement

and overhaul are normal routines. For example, relays, generators,switching devices, engine parts and hydraulic components in aircraft are replaced on a

periodic basis, usually before they fail, to enable the aircraft to fly for many years ofsafe operation. Tires and brake components are replaced several times over the

period of time that the automobile is in use.

Wear-out, such as fatigue, corrosion and damage growth, can be predicted by

the appropriate sciences with reliability considerations.

Replacements, overhauls and preventive actions restore the

equipment to an operational condition of low failure probability.

Good preventive maintenance is capable of eliminating risk due to

wear-out failure, enabling reliable operation for very long time is

possible.

Everything Eventually Wears Out


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Composite Failure Rate

ththththT 321

3

1

321

0

3

0

2

0

1

0 3

0 2

0 1

0

111

111

1

expexpexp

exp

exp

i

i

t

t

t

t

t

t

t

t

t

t

t

t

t

t T

tRtRtRtR

dtthdtthdtth

dtthdtthdtth

dtthtR

Based on the Bathtub Curve, the

total failure rate is given by:

Then, the total reliability is given by:

This proves that the total reliability of the device / system at time t is

simply the product of the reliability of all three phases at time t.

where1 = Phase 1, Early-life failure,

Infant morality.

2 = Phase 2, Constant failure

rate, Steady State.

3 = Phase 3, Wear-out.


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The bathtub curve is a representation of the reliabilityperformance, or the failure rate, of a population of

components or non-repaired items over the life-cycle.

The bathtub curve is introduced and mostly used as a

visual model to illustrate the three key periods of product

failure.o Early-life / Infant morality failure

o Useful-life / Steady state failure

o Aging / Wear-out failure

The bathtub curve is generally not calibrated to depict a

graph of the expected behavior for a particular productfamily. It is rare to have enough short-term and long-term failureinformation to actually model a population of products with a calibrated

bathtub curve.

34Bathtub Curve Summary


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Bathtub Curve - Electronics


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Bathtub Curve - Mechanical


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Bathtub Curve - Software


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Basic Reliability Mathematics MTTF and MTBF

Binomial Distribution Poisson Distribution

Uniform Distribution

Exponential Distribution

Weibull Distribution

Normal Distribution

Lognormal Distribution


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MTTF and MTBF

39


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Mean Time To Failure (MTTF) is commonly used to characterize the

expected operation time to failure for non-repairable device orsystems that exhibit constant failure rate h(t).

wherenf is the number of failures, MTTF must account for failed andnon-failed devices not just the failed devices.

MTTF

f

N

i

i

n

TTF

MTTF

Example:

The table showed the data of 10 devices that

operated in the fields gathered at 1,000,000

cycles.

For this example, the TTF is 8,859,923 cycles

and there were 4 failures. The MTTF is

8,859,923 / 4 = 2,214,981 cycles.

Strangely, no one specimen has been tested to over 2,000,000 cycles and all failures so far occurred

under 1,000,000 cycles. How could the mean time to failure be over 2,000,000 cycles?


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MTTF can be misleadingIf we take a careful look at the MTTF formulation, the MTTF would

only approach the true mean time to failure where nf N.

41

f

N

i

i

n

TTF

MTTF

If nf


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MTBFMean Time between Failure (MTBF) is commonly used to

characterize the expected operation time to failure for repairabledevice or systems that exhibit constant failure rate h(t). Once the

device failed, it is repaired and put back into service.

42

f

N

i

ii

n

uptimeofStartdowntimeofStart

MTBF

Strictly speaking, MTBF and

MTTF are not the same.

MTBF deals with a group of

same devices that are

repeatedly repaired and put

back into service. Whereas,MTTF works with new

devices only.

If MTBF were carried out

correctly, it will suffer the

same pitfall and misuse of

MTTF to a lesser degree

since majority TBF are linked

to failures.

Example:

In field operation, the TBF for 3 identical repairable pumps were

collected and shown in the table. There are 7 failures up to this

stage and 2 still running (Total 9 of devices). The MTBF of these

pumps are:

ID Device 1 Status Device 2 Status Device 3 Status

1 1765 Repaired 3082 Repaired 356 Repaired

2 1750 Repaired 875 Repaired 2789 Repaired

3 1084 censored 2373 In repair 1338 Censored

2202

7

15411MTBF


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MTTF and MTBF Limitations

If nf


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Probability Distributions

44

A probability distribution describes the probability of eachpossible outcome, set of outcomes, or the probability that an

outcome is in a particular interval. Distributions can be

expressed with a table, equation, or graph.


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Discrete and ContinuousProbability:

Discrete probability Continuous probability

45

A discrete probability distribution is one in which

the data can only take on certain values, such

as integers. For a discrete distribution, the

values in the distribution has associatedprobabilities - for example, "the probability that

a three will result from a roll of a dice is 1/6."

A continuous probability distribution is one in

which data can take on any value within a

specified range. A continuous probabilitydistribution therefore has an infinite number of

possible values, and the probability associated

with any particular value of a continuous

distribution is 1/ = zero. Time and distance

are examples of continuous variables.


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Discrete DistributionSome data are discrete by nature. For example, people.

The number of students in a class is discrete, as you can't

have half a student. Another example is the results of rolling

2 dice is discrete, as we can only obtain the values 2, 3, 4,

5, 6, 7, 8, 9, 10, 11 and 12, and nothing else in between

these numbers, and nothing more.

46


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Discrete Probability

Probability is the measure of the likeliness that an event

will occur.

47

A coin has two sides: Head

and Tail. In a game of

tossing a coin, they form the

complete set of all possibleoutcomes. The probability of

tossing a head in a fair toss is

0.5 (or 50%), and the

probability of tossing a tail in

a fair toss is also 0.5.

A typical dice has 6 faces

marked 1, 2, 3, 4, 5 and 6,

which forms the complete set of

all possible outcomes. It isimpossible to obtain for

example, 1.2, 5.4, 5.85, 8 etc

In a fair roll of a dice, the

probability of obtain any one of

the outcome is 1/6.

P b bili Di ib i f


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Probability Distribution of

Discrete Random VariableThe two important probability distribution of discrete

random variable are:

Binomial Distribution

Poisson Distribution

48


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Given N trial, the probability of n successes out of N trials is given

by:

Where:

Binomial DistributionA binomial distribution has the following 4 characteristics:

Only two exclusive outcomes are possible in each trial. Oneoutcome is called a success and the other a failure.

The process consists of a sequence of N trials.

The probability of a success denoted by p, does not change

from trial to trial. The probability of failure is 1-p and is also

fixed from trial to trial. The trials are independent: the outcome of previous trials do

not influence future trials.

49

nNn ppnNNnxP

1|

!!!

nNn

N

n

N

This is the number of ways inwhich n successes can be

obtained from Ntrials.

E l 1


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Example 150

In a game, a coin is tossed 3 times. The probability of

obtaining a head or a tail on each toss is 0.5. What isthe probability that exactly two tails are obtained in this

game?

Toss

1 2 3

H H H

T H H

H T H

H H T

T T H

T H T

H T T

T T T

Solution: Note that for this example,

success means obtaining a tail. Thenumber of all possible outcomes of

binomial distribution is 2n = 23 = 8.

The number of outcomes with 2

success is 3!/(2! (3-2)!) = 3.

The probability that exactly two tails

are obtained is given by:

375.05.025.03

5.015.02

33|2

232

xP


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Probability Distribution of Example 1

To compute the entire probability distribution, in additional to the

probabilities of exactly two tails, the probabilities of exactly zero,one and three tails needs to be obtained.

51

125.0125.011

5.015.00

33|0

030

xP

375.025.05.03

5.015.01

33|1

131

xP

125.01125.01

5.015.03

33|3

333

xP

For exactly 1 tail out of 3 tosses

For exactly 0 tails out of 3 tosses

For exactly 3 tails out of 3 tosses

E l 2


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Example 252

A pump is demanded on average 3 times per

week. The probability of pump failure per

demand is 0.05. If the pump fails, it isreplaced with a new one immediately. What

is the probability that the pump will fail exactly

twice in a week?

Demand

1 2 3

F F F

S F F

F S F

F F SS S F

S F S

F S S

S S S

Solution: Note that for this example, pump

failure = success. The number of all

possible outcomes of binomial distribution

is 2n = 23 = 8.

The number of outcomes with 2 success is

3!/(2! (3-2)!) = 3. The probability that the

pump will fail exactly twice in a week isgiven by:

007125.095.00025.03

05.0105.02

33|2

232

xP


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Example 3 53What is the probability that the pumps of example 2 will fail at most

once per week?Solution:

Failing at most once per week include those of 1 failure and 0 failure.

For 0 failure: For 1 failure:

The total probability is:

135.095.005.03

05.0105.01

33|1

2

131

xP

857.095.011

05.0105.03

33|0

3

030

xP

992.0135.0857.0

05.0105.01

305.0105.0

0

3

1|

131030

0

n

nNn ppnNNnxP

Q lit C t l P bl


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Quality Control ProblemA (very poor) manufacturer makes a

product with a 20% defect rate (p=0.2).

If we select 5 randomly chosen items atthe end of the assembly line, what is the

probability of having 1 defective item in

our sample?

Solution: Note that the number of all

possible outcomes of binomialdistribution is 2n = 25 = 32.

The number of outcomes with 1 success

is 5!/(1! (5-1)!) = 5. The probability of

having exactly 1 defective item in our

sample is:

54Sample ID

1 2 3 4 5

F F F F F

S F F F F

F S F F F

F F S F F

F F F S F

F F F F S

S S F F F

S F S F F

S F F S F

S F F F S

F S S F F

F S F S F

F S F F S

F F S S F

F F S F S

F F F S S

S S S F F

S S F S F

F S F F S

F S S F F

:

S S S S S

4096.08.02.05

2.012.01

55|1

4

151

xP

and many, many more possibilities


55/84

Aircraft Li-Battery Problem 55A particular aircrafts operation and safety is heavily dependent on the

reliable function of a Lithium battery. The probability of failure of theLi-battery is found to be 0.001 per flight. If the aircraft operates 380

flights per year, what is the probability of no failures over 1 year?

Solution: Note that the number of all possible outcomes of binomial

distribution is 2n = 2380 = 2.4626x10114. The number of outcomes with

0 failures is 380!/(0! (380-0)!) = 1. The probability of having exactly 0failure is:

684.0999.001

001.01001.00

380380|0

380

03800

xP

Hence, the probability of having 1 or more

battery failure over a year is:

316.0684.01

380|01380|0

xPxPThis high level of probability of failure is

unacceptable. Maintenance program must be

put in place to bring the PoF down to an

acceptable level.

Poisson Distribution


56/84

Poisson DistributionThe Poisson Distribution is a subset of the Binomial Distribution. It is alimiting case of a Binomial distribution when the number of trials, N, getsvery large and p, the probability of success, is small.

A proof is given by Mathematic as:

56

(htt

p://mathworld.wolfram.c

om/PoissonDistribution.html)

nNnNN

ppn

NNnxPnxPo

1lim|lim

P i Di t ib ti
http://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.htmlhttp://mathworld.wolfram.com/PoissonDistribution.html


57/84

The Poisson distribution, is a one parameter probabilitydistribution,

applies when the following conditions are met:

Discrete outcomes (n = 0, 1, 2, 3, 4, .)

Describes the distribution of discrete and infrequent eventsover an interval

The number of occurrences of the event in each interval canrange from zero to some very large countable number.

Each event is independent of the other event.

Expected number of occurrences of the event in a giveninterval of time or space is assumed to be known andconstant throughout the experiment.

Poisson Distribution 57

spaceortimeofinterval

occurrenceofno.mean

en

nPn

P!


58/84

Example 1 58The number of flaws in a fibre optic cable follows a Poisson distribution. The

average number of flaws in 50m of cable is 1.2.

(i) What is the probability of exactly three flaws in 150m of cable?

Solution: The mean number of flaws in 150m of cable is 3.6. So the

probability of exactly three flaws in 150m of cable is:

(ii) What is the probability of at least two flaws in 100m of cable?

Solution: Mean number of flaws in 100m of cable is 2.4, then

2125.0!3

6.33 6.3

3

ePP

6916.03084.01

!1

4.2

!0

4.21

1012

4.21

4.20

ee

xPxPxP PPP

2.150

6.3150

4.2100


59/84

Example 1 cont. 59(iii) What is the probability of exactly one flaw in the first 50m of

cable and exactly one flaw in the second 50m of cable?Solution: Exactly one flaw in a 50m section of cable is:

Then, the probability of exactly one flaw in any 50m of cable is also

Therefore, the probability of exactly one flaw in the first 50m andexactly one flaw in the second 50m is:

3614.0!1

2.11 2.1

1

ePP

3614.0!1

2.11 2.1

1

ePP

1306.03614.03614.011 PP PP

2.150

2.150

Ai ft B tt P bl


60/84

Aircraft Battery Problem 60The operation and safety of advanced aircraft today rests heavily on

the reliable function of batteries. The probability of failure of a

particular type of battery is found to be 0.001 per flight. If the aircraft

operates 380 flights per year, what is the probability of no failures over

1 year?

Solution: In 380 flights, the expected average number of failure is

0.001 x 380 = 0.380. The probability of having no failure (n=0) over

one year is:

6839.06839.01

!0

38.00 38.0

0

enP

Therefore, the probability that failure will occur

with one year is:

3161.0

6839.0101

nP

1001.0 10001

380.0380

f


61/84

Aircraft Battery Problem 61The Poisson distribution of the probability of failures over one year

is computed and showed. The probability of 1 battery failure over one year is 26%

The probability of 2 battery failures over one year is 5%

:

The probability of 6 battery failure over one year is approx. 10-6.

Despite the probability reduces rapidly, in aviation, one criticalfailure is more than enough to achieve catastrophic failure and loss.


62/84

ExampleIn 2012, the accident rate recorded of the aviation industry is 4.2 accidents per 1 million

departures. Assuming the occurrence of accidents follows the Poisson distribution, what isthe probability of more than 10 accidents occurring in the next 3 million departures?

62

Solution: The rate of accidents in 3 million departures is:

The probability of more than 10 accidents occurring during the next 3 million departures

is

7124.0

!10

3.12...

...!2

3.12

!1

3.12

!0

3.12110

10...

....321110

3.1210

3.122

3.121

3.120

e

eeenP

nP

nPnPnPnP

6.122.4 million3million1

10

3million 3million

!110110

i

i

ei

nPnP


63/84

Continuous Distribution

Some data is continuous by nature. Continuous Data can takeany value within a range. As an examples, a person's height

could be any value (within the range of human heights).

Another example is time in a race: you could even measure it

to fractions of a second.

63

UNIFORM DISTRIBUTION

EXPONENTIAL DISTRIBUTION WEIBULL DISTRIBUTION

NORMAL DISTRIBUTION

LOGNORMAL DISTRIBUTION

C ti P b bilit


64/84

Continuous ProbabilityContinuous Distribution: A continuous distribution describes

the probability of continuous values. The probability of any singlevalue is zero. For example,P(X=0.25)=0.

So only the probability of intervals is of interest, for exampleP(X0.3) or P(0.20 X 0.30).

A continuous distribution is normally described in terms of

probability density, f(t). The following slides explore several

commonly used continuous distribution functions in aviation.

64

0lim20

20

2

2

20

20

0

22

x

xx

x

x

xx

dxxfxXxP

dxxfxXxP

x

x

Uniform Random Variable: U(a b)


65/84

Uniform Random Variable: U(a,b)The Uniform Distribution is also called the Rectangular

Distribution. The Uniform Distribution has equal pdf for all values

of the Random variable between the range a and b:

The density function is given by:

elseeverywhere

for

0

1 bxaxf ab

The uniform distribution implies that the

values between a and b are equally

likely to occur. Keep in mind that a

constant f(t) is not the same as constant

failure rate h(t).

elseeverywhere

for

0

1 bxaxh xb

Uniform density


66/84

Uniform density

The 1st moment of the distribution (mean or average) is:

The 2nd

moment of the distribution (variance) is:

baxdxxfXEb

a 21

21212 abdxXExxfXV

b

a

The cumulative distribution functionis given by:

b

bxa

ax

xFbaUabax

for x

for

for

1

0

,

The uniform distribution is a symmetrical distribution,

hence the mean equals the median (50% point).

f


67/84

Uniform distribution (Continued)Let X be a uniform random variable and its distributed according

to U(4,10).

Compute the following:

1. probability P(5.1X5.9)

2. probability P(X>5.9)

3. Find the 75

th

percentile of X.4. The mean, median and mode

5. h(x=7)

Solutions:

Since pdf = 1/(b-a) = 1/(10-4) = 1/6

1. P(5.1X5.9) = pdf x (5.9 5.1) = 0.8 / 6 = 0.13332. P(X>5.9) = pdf x (10 5.9) = 4.1 / 6 = 0.6833

3. 75th percentile = P(X>x) = 0.75 = (x 4) / 6 -> x = 8.5

4. Mean = Median = Mode = (10 + 4) / 2 = 7

5. h(x = 7) = 1 / (b - x) = 1 / (10 7) = 0.3333



68/84

68Exponential Distribution

Probability density:

Cumulative probability:

Failure rate:

When a population of item is subject to failures that occurs in

random intervals and the expected number of failures is the samefor long periods of time, then the probability distribution of failures

is said to fit an Exponential Distribution. The pdf, CDF and hazard

function in terms of the variable xare given as:

0and0where

exp

x

xxf

xxF

exp1

x

xx

exp

exp

* The risk of failure of an old item is the same as that of a new item.

E ti l Di t ib ti


69/84


The 1st moment of the distribution (mean or

average) is:

The 2nd moment of the distribution

(variance) is:

Value of random variable at F(x)=0.5 (i.e.,

Median value):

1MTTF

21Var x

2lnMedian

The most important property of the exponentialdistribution is its memoryless property. This

property indicates the risk (hazard function)

does not changes with time. If we experienced

an event that follows the Exponential

Distribution, we often refer to it as pure luck (or

bad luck)!



70/84

70Weibull Distribution

The Weibull Distribution is arguably the most popular distribution

for reliability analysis, due to its robustness in modelling thevarious failure rate behavior / phases of the Bathtub Curve. The

Weibull Distribution is given as follows.

Probability density: Cumulative probability:

Failure rate:

xxxf exp

1

xxF exp1

1

xxh

The Weibull Distribution is controlled

by 2 parameters:

The Shape Parameter

The Location / Scale Parameter



71/84


When = 1, the hazard function is constant, where

the Weibull Distribution reduces to an ExponentialDistribution with =1/.

When 1, we get a increasing hazard function

71

Example


72/84

Linear least squared methodcan be applied to determine

the best Weibull distribution fit

to the data. The first step is to

linearized the Weibull

distribution.

Example72

ii

xxF exp1

lnln1lnln ii xxF

Reliability data analysis using

Weibull distribution.

cxmy ln

=4.0775

=944750

N l (G i ) Di t ib ti


73/84

Standard normal variable z

The standard normal variable

has a mean of 0 and

standard variation of 1

Normal (Gaussian) Distribution 73

The parameter is the mean. The

parameteris its standard deviation.

The Normal distribution is a

symmetrical distribution no bias. For

this reason, the mean, median and

mode are identical.A random variable with a Gaussian

distribution is said to be normally

distributed and is called a normal

deviate. It is also often referred to as

the Bell Curve.

The Probability Density function of the

Normal Distribution is:

2

2

1exp

2

1

xxf

xzz 10

0.50.5

Normal Distribution


74/84


which is not integrate-able. In may

computer software, e.g. MS Excel, the

Normal Probability can be computed

using built-in functions, in terms of the

error function (erf) or gamma function:

Normal Distribution

Many things closely follow a NormalDistribution: heights of people size of things produced by machines errors in measurements Natural sciences

74

11

0

2

01

2

1exp

2

1x

x

x

x

dxzdxxfxF

2

21 zerfxF

E ample


75/84

Example

An average switch manufactured by Company A

lasts 300 cycles with a standard deviation of 50cycles. Assuming that switchs life is normally

distributed, what is the probability that Company

As switch will last at most 365 cycles?

75

9032.0365

where

2

502

3003651

2

21

erf

zerf

xF

x-zxF

Solution: We want to find the probability

that switch is less than or equal to 365

cycles.

Given = 300 cycles= 50 cycles

365 >

Compute this in MS Excel (or via lookup

table), the answer is: P(X365) = 0.9032.

Therefore, there is a 90.32% chance that

a switch will fail within 365 cycles.

365365 FXP

Example


76/84

Example

The average yield strength of an aerospace grade

Aluminium alloy manufactured by Company A is

560 MPa with a standard deviation of 26 MPa.

Assuming that the yield strength is normally

distributed, what is the safe yield strength for

design if a reliability of 0.999 is required?

76

96.479

001.021226560

001.0

where

1

2

2625601

2

21

erfx

x-zxF

xerf

zerf

Solution:We want to find the yield strength

corresponding to a 0.999 reliability,

or failure probability of 0.001.

Given = 560 MPa= 26 MPA

P(Xx) = 0.001

Compute this in MS Excel (or via

lookup table), z = -3.09. The answer

is: ~480 MPa.

xFxXP



77/84

Lognormal DistributionThe lognormal distribution is more versatile than the normal

distribution and is a better fit to reliability data, such as for

populations with wear-out characteristics. Also, it does not have

the Normal Distributions disadvantage of extending below zero

i.e. always positive. The lognormal distribution and the normal

distribution describe situations when the hazard function is

increasing.

77

Probability density:

Standard normal variable z

basewhere

log

2

1exp

ln2

12

b

x

xbxf b

xbz log



78/84

Lognormal Distribution 78

2

21 zerfxF

2

2

exxE

bb base_ln


For exponential base random variables, the 1st

moment of the distribution (mean or average) is:

Note that for base b random variables, the

relationship of : is:

basewhere

21exp

ln2111

0

2

01

b

dxzxb

dxxfxF xx

x

x

b

b

base_

base_

Example


79/84

Example

Solution:

1: = ln(560)-0.152/2 = 6.317

Note that F(x) = 0.001 < 0.5.

2: F(x) = 1-R(x) = 0.001 = 0.5 x (1 - erf( z / sqrt(2)))

3: -z = sqrt(2) x erf-1(1 - 0.001 x 2) = -3.090

4: ln(x) = + z = 6.317 - 0.15 x 3.090 = 6.069

5: x = e(6.069) = 432.45 MPa

Based on the lognormal distribution, the yield strength corresponding

to a probability of 0.001 is ~ 432 MPa

79

The average yield strength of an aerospace grade

Aluminium alloy manufactured by Company A is

560 MPa. The distribution of the material isknown to follow a lognormal distribution (base e)

with a standard deviation of 0.08. What is the safe

yield strength for design if a reliability of 0.999 is

required?


80/84

Probability Distribution Summary 8


81/84

Probability Distribution Summary 81

The lognormal distribution Very powerful and can be applied to

describe various wear-out failure

processes

Applications include Electronics,

material, structure etc.

The Weibull Distribution Very powerful and can be applied to

describe all 3 phases of the bathtubcurve failure processes

Applications include Electronics,

mechanical components, material,

structure, random failure etc.

The Normal Distribution: Very straightforward and widely

used for approximations Applications include Natural

sciences, Electronics, mechanical

components etc.

The Exponential Distribution Special case of Weibull Distribution. Very commonly used for large

systems and field failures

Applications include accidents,

electronics, mechanical

components, natural disasters

(CFR), luck.

The Uniform DistributionThis distribution is a key in risk analysis for generating random variables from

other distributions in computer software.Little real-life applications, include pressure in a room, density across a block of

material, waiting time.

Issues in Distribution Modeling


82/84

Issues in Distribution Modeling

Without knowing the underlying distribution of the

variable, no model is absolutely correct. Only,some models are more useful or more physically-

sounded than others.

The model should be sufficiently simple to be

handled by available mathematical and statisticalmethods, and be sufficiently realistic such that the

deducted results are of practical use.

The low confidence in the tails of the distribution

were no data exists (extrapolation) is a major barrier to the application of reliability methods

especially where high level of reliability is

demanded, e.g. nuclear, aerospace and civil

industries.

MS Excel 83


83/84

MS Excel

Compute the CDF or Failure Probability of the normal

distribution given mean and the standard deviation.

NORMDIST(x, mean, standard deviation, [ true, false])

Note: True = CDF, False = pdf

Compute the normal random variable x of given mean, the

standard deviation and CDF.

NORMDISTINV(CDF, mean, standard deviation)

83

Failure Prob Mean Sig x

0.999 40 1.5 ---> 44.63534846

0.12 40 1.5 ---> 38.23751981

0.5 40 1.5 ---> 40

x Mean Sig [true, false] CDF or pdf

36 40 1.5 TRUE ---> 0.00383038136 40 1.5 FALSE ---> 0.007597324

NORM DIST Analysis 84


84/84

NORM DIST Analysis 84

15.0

5.00

1,5.0,12GAMMA.INVsqrt2

1,5.0,21GAMMA.INVsqrt2

Pfor

Pfor

P

Pxz

15.0

5.00

12erf2

21erf21

1

Pfor

Pfor

P

Px

z

In terms of the error function:

In terms of the Gamma function:

Exercise:

With the aid of MS Excel, determine the best

normal distribution fit to the dataset shown in

the table.

me4504 02 basic reliability

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