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252015105Subgroup 0
4.3
4.2
4.1
4.0
3.9
3.8
3.7
SampleMean
1 1
0.6
0.5
0.4
0.3
0.2
0.1ampleStDev
1
Reliability DistributionsNormal & LogNormal
.
0.0
Author
Anil Kumar Ammina
Reliability Engineering
BE Analytic Solutions
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Key TerminologiesKey Terminologies
Descriptive Statistics :Numerical index that describes some characteristics of the population. Itdescribes sample or finite population, gathers and describes data in terms oftables, graphs etc. and it is subjective in nature
Inferential / Analytical Statistics :Quantitative technique that enables the experimenter to generalize about thepopulation using limited number of observations. It uses of probability and
distributions and endeavor to reduce subjectivity
Point Estimator:
The estimation of the population in terms of a single value. Specified by puttinghat on the parameter. If q the parameter, q is the point estimator
Interval Estimator:
The estimation of the population in terms of a range. Specified as Point Estimator Error If q the parameter, q D is the interval estimator^
^
Population(parameter)
Sample(Statistic)
Mean
Variance
x
s
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Types of DataTypes of Data
Variable data:Data which can be quantified, e.g., temperature, number of countries etc.
Continuous Variable: Variable that can take values in infinitesimally close
neighborhood of the previous value and yet makes some meaning, e.g., pressure,length etc.Discrete Variable: Variable that can take values at definitely distinguishableintervals of the previous value, e.g., no. of students in the class, no. of defects
etc.Binary Variable: Special type of discrete variable that can takes values as 0 and1; also Yes/No or Pass/Fail or Defective/Non defective type of data.
Attribute data :Data which cannot be quantified, e.g., fragrance of a flower, aesthetics of theproduct, pleasing appearance, conduct of the students etc. Based on humanfeelings e.g. taste of tea. Subjectivity in assessment.
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Visualizing Your DataVisualizing Your Data Box PlotBox Plot
When is it used?
You have a set or sets of continuousdata.
Maximum Observationthat falls within theUpper Limit = Q3 + 1.5 (Q3 - Q1)
Outlierany point outside the Lower orUpper Limit.
*
You wish to visually check the distributionof the data.
You wish to look for evidence of
differences between data sets.
Median/50th Percentile(Q2)
25th Percentile (Q1)
Minimum Observationthat falls within the
Lower Limit = Q1 - 1.5 (Q3 - Q1)
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Box Plot ExerciseBox Plot Exercise
A 5-weight% NaOH solution is used inlarge quantities in your chemicalprocess.
You have data for analyses on 100batches each of this solution that youhave purchased from four different
.
You are seeing problems in yourprocess and have started to wonder ifmaybe they are related to variations in
the solutions that you are purchasing.
Lets take a visual look at the data.
1 2 3 4
3.5
4.5
5.5
6.5
7.5
Vendor
NaOH-W%
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Visualizing Your DataVisualizing Your Data HistogramHistogram
When is it used?
You have a set or sets of continuousdata.
You wish to visually check the shape(frequency distribution) and spread of thedata. 3.5 4.5 5.5 6.5
0
5
10
15
Vendor 4
Frequency
How does Minitab Construct a Histogram?Counts the number of data pointsDetermines the Range, R, for the dataDetermines the # of classes, K.Determines class width = R/K
Determines how many data pointsfall into each class.Plots the number/class (frequency) inbar graph format
NoteNumber of classes,K, is user adjustable (under options)
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Visualizing Your DataVisualizing Your Data Scatter PlotScatter Plot
When is it used?
You have a set of paired data for twocontinuous variables.
You wish to visually check for evidence ofcorrelation (trends) between thevariables.
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Descriptive StatisticsDescriptive Statistics
Are we putting the correct amount of beverage into our cans?
Target Value = 500 ml
Sample Size = n = 24
485 490 495 500 505 510 515
Most Data Sets:Show a tendency to cluster about a central point
Exhibit a variability (dispersion/spread)
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Represents the nominal value of the process
Mean (x)
Central TendencyCentral Tendency
x
Median (middle data point)
Quartile values (Q1, Q3)
Q1 Q3
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Standard Deviation (s) Standard Deviation
SpreadSpread
Represents the variation in the process
Range or Span (P95 - P5)
Stability Factor (SF) = Q1/Q3
Q1 Q1Q3 Q3
P5 P95
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Descriptive StatisticsDescriptive Statistics
Click on Graphs,Select GraphicalSummary
Provides summary reportof basic statistical information
Normality Check
Basic StatisticalInformation
Box Plot
Histogram
Confidence Intervals
Minitab Output
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Using ExcelUsing Excel
Summary Statistics for Data
0
5
10
15
20
25
30
35
40
45
Series1 0 0 0 0 4 8 37 42 7 2
0 2 4 6 8 10 12 14 16 18
Mean 11.95
Standard Error 0.19
Median 12.04
Mode #N/A
Standard Deviation 1.86
Sample Variance 3.45
Kurtosis 1.00Skewness 0.03
Range 10.90
Minimum 6.89
Maximum 17.79
Sum 1195.41
Count 100
KURTOSIS
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Normal ( Gaussian ) DistributionNormal ( Gaussian ) Distribution
( )2
2
2
2
1),,(
=
x
exN
This is the distribution that is approximately followed by the random processes and
is specified by just two parameters, viz., (mu) and (sigma) as follows -
Symmetric about its mean
As increases, it becomes flatter, the peak reduces.
Gaussian Distribution
0
0.02
0.04
0.06
0.08
0.1
-50 -30 -10 10 30 50 70 90 110 130 150 170
x
Probability
Density
Function
sigma =5
sigma = 10
sigma = 20
sigma = 30
Majority of the processesnaturally follow
Normal Distribution.
Non-assignable reasons
of variation are attributed
to Normal distribution
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( )( )
XE X x f x dx
= =
The expectedor meanvalue of a continuous random variable Xwith pdf f (x) is
MeanMean
n
For continuous data
== iX
n1For discrete data
Mean is simply the numerical average of the data
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VarianceVariance
The varianceof continuous random variable Xwith pdf f(x) and mean is
2 2( ) ( ) ( )X V x x f x d x
= =
( )2
[ ]E X =
=
ni xx
22 )(
For continuous data
The standard deviationis ( ).X V x =
Short-cut Formula for Variance ( ) [ ]22
( ) ( )V X E X E X =
= i n1 1
The Variance is a measure of the average squared deviation from the mean
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Z (Standardised Normal) DistributionZ (Standardised Normal) Distribution
==
zezNN
z
for
2
1)1,0,()1,0( 2
2
Put the following substitution in the normal distribution
Symmetric about its mean z=0
( )= xz
The probability density function is given as follows -
Basis for 6 and z-score
zx +=
Binds the family of normal distributions into a single
z Distribution
0
0.1
0.2
0.3
0.4
0.5
-4 -3 -2 -1 0 1 2 3 4
z
Probability
Density
Function
Z = 6 means
Z-score is 6
and
The process is 6
.
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NormalDistribution
StandardizedNormal Distribution
Z=X
f(X)
f(Z)
Standardize the Normal Distribution
Z=0
z=
Z
One table!
X
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6.2 50.12
10
XZ
= = =
Normal Distribution Standardized
Standardized Normal Distribution
Shaded Area Exaggerated
10= 1Z =
5=
6.2 X Z
0Z=
0.12
f(X)
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P(X > 8) = .3821P(X > 8) = .3821
Normal Distribution Standardized
8 5.30
10
XZ
= = =
Shaded Area Exaggerated
10=1Z =
5=X
Z0Z =
?
.308
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P(X > 8) = .3821P(X > 8) = .3821 (continued)
Z .00 .01 .02
Cumulative Standardized Normal
Distribution Table (Portion) 0 1Z Z= =
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.
0.1 .5478
Shaded Area Exaggerated
0 Z
P(X>8) = 1 .6179 = .3821
.30
.3821
Standardized Normal DistributionStandardized Normal Distribution
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Standardized Normal DistributionStandardized Normal Distribution
(Different Type of Tables)(Different Type of Tables)
12.010
52.6=
=
=
XZ
Normal Standardized
ZZ= 0
Z
= 1
.12X
= 5
= 10
6.2Shaded Area Exaggerated
.0478
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Normal Distribution Thinking Challenge
You work in Quality Control for, Light bulb
life has a normal distribution with
= 2000 hours & = 200 hours. Whats
the probability that a bulb will last?
a. between 2000 & 2400
hours?
b. less than 1470 hours
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Solution: P(2000Solution: P(2000 XX 2400)2400)
Normal Standardized
Z
X
=
=
=
2400 2000
200 2.0
Z Z= 0
Z = 1
2.0
.4772
X = 2000
= 200
2400
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Solution: P(XSolution: P(X 1470)1470)
Normal Standardized
Z
X
=
=
=
1470 2000
200 2.65
ZZ= 0
Z = 1
-2.65
.0040
X = 2000
= 200
1470
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Z .00 0.2.01
Cumulative Standardized
Normal Distribution Table
What is ZGiven
Probability = 0.6217 ?
0 1= =
Finding Z Values for Known Probabilities
.6217 0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
.6179 .62550.3Shaded AreaExaggerated
.6217
.31Z =
0 Z
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Recovering X Values for Known Probabilities
Normal Distribution StandardizedNormal Distribution
10=1Z =.6217
( ) ( )5 .31 10 8.1X Z = + = + =
5=X Z
0Z =0.31?
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Normal Distribution ExampleNormal Distribution Example
Solution :
The bolt has mean diameter of 10.5mm and its standard deviation is 0.8mm.What is the probability that in a random sample, the diameter is between 8.8mmand 12.5mm?
5.12;8.8;8.0;5.10 ==== xx UL
)5.2125.2Pr()5.128.8Pr(
)Pr()Pr(
5.2
8.0
5.105.12
125.28.0
5.108.8
=
=
=
=
=
=
=
=
zx
zzzxxx
xz
xz
ULUL
UU
LL
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Normal Distribution ExampleNormal Distribution Example
=
)125.2Pr()5.2Pr()5.2125.2Pr( = zzz
z
0.97700.01680.0062)1()5.2125.2Pr(
)125.2Pr())5.2Pr(1()5.2125.2Pr(
==
=
z
zzz
Single-Tail Z Table (values from 0.00 to 7.99)
Total area under z
distribution is 1.0000 Z distribution is symmetric
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Assessing Normality
Construct charts
For small- or moderate-sized data sets, do stem-and-leaf display and
box-and-whisker plot look symmetric? For large data sets, does the histogram or polygon appear bell-shaped?
Do the mean, median and mode have similar values?
Check the P-Value
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Central Limit TheoremCentral Limit Theorem
For majority of the distributions, the mean of the samples follow almost a normaldistribution as the sample size becomes at least 30.
When the inference is to be drawn about the sample mean E.g. Sachin
Tendulkars average batting performance in test series. It may be possible that hisbatting statistics for all the tests is a non-normal distribution. In such situation, onemay like to consider a series average and use this data of means to analyze hisperformance.
It is used as the basis to calculate confidence interval when the sample size isgreater than 30
For all practical purposes, Central Limit Theorem is true
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It is the most widely-used general purpose distribution.
The pdfof the normal distribution
NormalNormal ProbabilityProbability DensityDensity FunctionFunction
N l S i i l P iN l S i i l P i
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Normal Statistical PropertiesNormal Statistical Properties
The Normal Mean= Median = Mode
The reliability for a mission of time
There is no closed-form solution for the normal reliability function
The instantaneous normal failure rate
Th L l Di t ib tiTh L l Di t ib ti
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The Lognormal DistributionThe Lognormal Distribution
The lognormal distribution is commonly used to model, if the logarithm of therandom variable is normally distributed
The pdffor this distribution
Where, = ln(T), where the Tvalues are the times-to-failure.
equal probabilities under the normal and lognormal pdfs, incremental areas should also be equal
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The mean of the lognormal distribution, , is
The mean of the natural logarithms of the times-to-failure,
The standard deviation of the lognormal distribution
The standard deviation of the natural logarithms of the times-to-failure
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The Lognormal Reliability Function
Eff t f
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The lognormal distribution is a distribution skewed to the right.The pdfstarts at zero, increases to its mode, and decreases thereafter.
The degree of skewness increases as increases for a givenFor the same ,the pdfs skewness increases as increases.
Effect of
Effect of
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Effect of
For values significantly greater than 1, the pdf rises very sharply in thebeginning, i.e. for very small values of Tnear zero, and essentially follows theordinate axis, peaks out early, and then decreases sharply like an exponentialpdf or a Weibull pdf with 0 < < 1.
The parameter, , in terms of thelogarithm of the T s is also the scaleparameter, and not the location parameteras in the case of the normal pdf.
The parameter , or the standard deviationof the T's in terms of their logarithm or oftheir , is also the shape parameter and
not the scale parameter, as in the normalpdf, and assumes only positive values.
Example 1
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Example 1
Concentration of pollutants produced by chemical factory historically known toexhibit lognormal distribution. Suppose it is assumed that the concentration of acertain pollutant, in parts per million, has a lognormal distribution with parameters
=3.2 and = 1. What is the probability that the concentration exceeds8 parts per million.
Let the random variable Xbe pollutant concentration
[ ] 1314.0)12.1(1
2.3)8ln(8 ==
=XP
Since ln(X)has a normal distribution with mean=3.2=3.2=3.2=3.2and standard deviation ====1111,,,,
Here, we use the notation to denote the cumulative distribution function of thestandard normal distribution. As a result, the probability that the pollutant
concentration exceeds 8 parts per million is 0.1314.
Example 2
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Example 2
The Life, in thousands of miles, of a certain type of electronic control for locomotivesHas an lognormal distribution with= 5.149 and = 0.737. Find the 5th percentile ofthe life of such locomotive.
We know, from Z-table that 05.0)645.1( =
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Questions ?
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Thank You
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