Validity and application of some continuous distributions

Dr. Md. Monsur RahmanProfessor

Department of StatisticsUniversity of Rajshahi

Rajshshi – 6205E-mail: monsur_st_ru@yahoo.com

1

Normal distribution

The first discoverer of the normal probability functionwas Abraham De Moivre(1667-1754), who, in 1733,derived the distribution as the limiting form of the binomial distribution. But the same formula was derived by Karl Freidrich Gauss(1777-1855) in connection withhis work in evaluating errors of observation in astronomy.This is why the normal probability is often referred to as Gaussian distribution.

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X: Normal Variate

Density:

0,,

],)(exp[)(

2

221

21

x

xf x

2)(,)( XVarXE

Standard Normal Variate :

XZ

ZX 3

Normal distribution

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Properties of Normal distribution

Normal probability curve is symmetrical about the ordinate at x

Mean, median and mode of the distribution are equal and each of these is The curve has its points of inflection at By a point of infection, we mean a point at which the concavity changes

x

All odd order moments of the distribution about themean vanishThe values of and are 0 and 3 respectively

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includes about 68.27% of the population

includes about 95.45% of the population 2

includes about 99.73% of the population 3Application:Many biological characteristics conform to a Normal distribution - for example, heights of adult men andwomen, blood pressures in a healthy population,RBS levels in blood etc.

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Validity of Normal Distribution for a set of data

Many statistical methods can only be used if the observations follow a Normal Distribution. There are several ways of investing whether observations follow a Normal distribution. With a large sample we can inspect a histogram to see whether it looks like a Normal distribution curve. This does not work well with a small sample, and a more reliable method is the normal plot which is described below.

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8

X: Normal Variate

Density:

0,,

],)(exp[)(

2

221

21

x

xf x

2)(,)( XVarXE

Standard Normal Variate :

XZ

ZX 9

CDF OF X : F(X)

CDF OF Z : )(z )(1)( zz

P quantile of X :

P quantile of Z :

pX

pZ

,

pp

X

p

ZX

Z p

pX

pZ

is the solution of

is the solution of

pXF p )(

pZ p )(

10

Dataset nxxx ,...,, 21

• Find empirical CDF values

• Arrange the data in ascending order as

• Empirical CDF values are as follows

.,...,2,1,)( 5.0)( nixF n

ii

)()2()1( ,...,, nxxx

•Using normal table obtain the values corresponding to

)(iz)( )(ixF

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•If the given set of observations follow normal distribution, the plot (x, z) should roughly be a straight

line and the line passes through the

point and has slope .

•Graphical estimates of and may be obtained.

xz

)0,(

1

•If the data are not come from Normal distribution wewill get a curve of some sort.

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2.2 3.6 3.8 4.1 4.73.3 3.6 3.8 4.1 4.73.3 3.7 3.9 4.2 4.83.4 3.8 4.0 4.4 5.0

Table 1 : RBS levels(mmol/L) measured in the blood of 20 medical students. Data of Bland(1995), pp. 66

Bland,M.(1995): An Introductions to Medical Statistics, second edition, ELBS with Oxford University Press.

13

14

Lmmol /92.3ˆ Lmmol /642.ˆ

MLE

15

16

•Goodness of Fit Test

•We use here Kolmogorov-Smirnov (KS) test for the given data

• KS statistic=max |CDF_FIT- CDF_EMP|

• For the RBS level data we calculate KS statistic KS(cal)=0.07827

• 5% tabulated value=0.294• Conclusion: Normal distribution fit is good for the given data

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• Estimated population having RBS within the normal range (3.9 – 7.8mmol/L) is about 51%

• Estimated population having RBS below the normal range is about 49%

• Estimated population having RBS above the normal range is 0%

Results

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n

m

XXX

XXX

22221

11211

,...,,

,...,,

• Empirical CDF values of are as follows:

)1()12()11( ,...,, mXXX

miXF mi

i ,...,2,1,)( 5.0)1(

• Obtain the values corresponding to )( )1( iXF)1( iZ

• Similarly values are obtained corresponding to)2( iZ

)( )2( iXF

•Two sample case

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• If the first set of data come from normal distribution

with mean and variance , then the plot

will roughly be linear and passes

through the point with slope .

121

),( 11 ZX

1

1)0,( 1

• If the second set of data come from normal distribution

with mean and variance , then the plot

will roughly be linear and passes

through the point with slope .

2 22

),( 22 ZX

)0,( 22

1 20

• Both the lines parallel indicating different means but equal variances

• Both the lines coincide indicating equal means and equal variances

• Both the lines pass through the same point on the X-axis indicating same means but different variances

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Table 2 : Burning times (rounded to the nearest tenth of a minute) of two kinds of emergency flares. Data due to Freund and Walpole(1987), pp. 530

Brand A: 14.9,11.3,13.2,16.6,17.0,14.1,15.4,

13.0,16.9

Brand B: 15.2,19.8,14.7,18.3,16.2,21.2,18.9,

12.2,15.3,19.4

Freund, J.E. and Walpole, R.E.(1987): Mathematical Statistics, Fourth edition, Prentice-Hall Inc.

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Above plot indicates that both the samples come from normal population with unequal means and variances 23

Log-normal distribution

In probability theory, a log-normal distribution is aprobability distribution of a random variable whoselogarithm is normally distributed. If X is a random variable with a normal distribution, then Y = exp(X) hasa log-normal distribution; likewise, if Y is log-normallydistributed, then X = log(Y) is normally distributed. It is occasionally referred to as the Galton distribution.

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Density:

0,,0

],)(exp[)(

2

2log21

21

x

xf x

x

Mean =

Variance=

Median=

Mode=

25

Log-normal density function

x

f(x)

26

ApplicationCertain physiological measurements, such as blood pressure of adult humans (after separation on male/female subpopulations), vitamin D level in blood etc. follow lognormal distribution.Subsequently, reference ranges for measurements in healthy individuals are more accurately estimated byassuming a log-normal distribution than by assuming a symmetric distribution about the mean.

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Table 3 : Vitamin D levels(ng/ml) measured in the blood of 26 healthy men. Data due to Bland(1995), pp. 113

14 25 30 42 54 17 26 31 43 54

20 26 31 46 63 21 26 32 48 67

22 27 35 52 83 24

Bland,M.(1995): An Introductions to Medical Statistics, Second edition, ELBS with Oxford University Press.

28

29

449.ˆ

509.3ˆ ng/ml

ng/ml

• MLE

30

31

•Goodness-of-fit test

• KS statistic=max |CDF_FIT- CDF_EMP|

• For the vitamin D level data we calculate KS statistic KS(cal)=0.0967• 5% tabulated value=0.274

• Conclusion: Lognormal distribution fit is good for the given vitamin D data

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• Estimated population having vitamin D level within the normal range (30 – 74 ng/ml) is about 56%

• Estimated population having vitamin D level below the normal range is about 40%

• Estimated population having vitamin D level above the normal range is about 4%

Results

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Weibull Distribution

Weibull distribution is used to analyze the lifetime dataT: Lifetime variable• Density function

0,,0],)(exp[)()( 1

ttf tt

• : Scale parameter(.632 quantile)

• : Shape parameter(<1 or >1 or =1)

• CDF : ])(exp[1)( ttF

])(exp[)( ttR

• Reliability (or Survival) function:

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1)()(

tth

•Increasing hazard rate : for

•Decreasing hazard rate: for

•Constant hazard rate : for

tth )( 1

1tth )(

1)( th 1

])}1({)1([)(

)1()(2122

1

TV

TE

:pt p quantile, which is the solution of ptF p )(

•Accordingly,1

)]1log([ pt p

•Hazard Function :

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Density function:

0,0),exp()( 1 ttf t

• : Scale parameter(.632 quantile)

• CDF : )exp(1)(

ttF

)exp()( ttR

• Reliability (or Survival) function:

•Weibull distribution reduces to exponential distribution when 1

Exponential distribution

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1)( th

2)(

)(

TVar

TE

:pt p quantile, which is the solution of ptF p )(

•Accordingly, )]1log([ pt p

•Hazard Function :

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The red curve is the exponential density

The red line is theexp. hazard function

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From the Weibull CDF we get

)log()log())](1log(log[ ttF,XAY

where

)log(

)log(

))](1log(log[

A

tX

tFY•

)()2()1( ,...,, nttt• Ordered lifetimes are:

• values are obtained through the empirical CDF values as given below

)(iY

,)( 5.0)( n

iitF

ni ,...,2,1

Validity of Weibull distribution for a set of data

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•

• If the data follow Weibull distribution with scale parameter and shape parameter , the plot of (X,Y) will roughly be linear with slope and passes through the point .

)0),(log(

• Accordingly, the graphical estimates of and may be obtained.

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Table 4: Specimens lives (in hours) of a electrical insulation at temperature appear below. Data due to Nelson(1990), pp. 154

Nelson,W.(1990): Accelerated Testing: Statistical Models, Test Plans, and Data Analyses, John Wiley and Sons.

Co200

2520, 2856, 3192, 3192, 3528

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42

• MLE of and

• Log-likelihood function of and based on observed data

nttt ,...,, 21

)()log()1()log( ii ttnLogL • MLE of and by maximizing the log-Likelihood with respect to and using numerical method.

• Graphical estimates may be used as starting values required for the numerical method

• The MLEs of and are denoted by and respectively.

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For the insulation fluid data given in table 4 the following results (based on MLEs) are obtained:

78.3)ˆ(.

56.142)ˆ(.

61.10ˆ

49.3208ˆ

ES

ES

hours

hours

Estimated median life= 3099.548 hours

])(exp[)(ˆˆ

ˆ

ttR •ML estimate of R(t)

Time (hour): 3000 3500 3700 4000Reliability : .6124 .0807 .0107 .0000311

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Weibull versus Exponential Model

•Suppose we want to test whether we accept exponential or Weibull model for a given set of data

•The above test is equivalent to test whether the shape parameter of Weibull distribution is unity or not i.e. vs 1:0 H 1:1 H

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•Test Procedure(LR test)

•Under the log-likelihood function is

which yields , MLE of .

0H

itnl 10 )log(

in t1

•Maximum of is given by

0l

itnl ˆ1

0 )ˆlog(ˆ

46

•Similarly, under the maximum of the log-likelihood is given by

1H

ˆ

ˆˆˆ

ˆ

1 )()log()1ˆ()log(

ˆii ttnl

where and are the MLE s of and

under .

ˆ

1H

•LR test implies follows chi-square distribution with 1 df.

)ˆˆ(2 01 ll

•If , accept (use) exponential Model

)1,1()ˆˆ(2 2

01 ll47

• If , accept (use) Weibull model

)1,1()ˆˆ(2 2

01 ll

• For the insulation fluid data given in table 4

87.17)1293.451877.36(2)ˆˆ(2 01 ll

34.3)1,95(.2

Conclusion: Weibull model may be accepted at 5% level of significance

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Accelerated Life Testing (ALT) for Weibull Distribution

• Stress: Temperature, Voltage, Load, etc.• Under operating (used) stress level, it takes a lot of time to get sufficient number of failures

• Lifetimes obtained under high stress levels

• Aim: (i) To estimate the lifetime distribution under used stress level, say, (ii) To estimate reliability for a specified time under (iii) To estimate quantiles under

0S0S

0S

Sampling scheme(under constant stress testing)• Divide n components into k groups with number

of components

respectively, where

• components exposed under stress levels • , j-th lifetime corresponding to • Obtain the equation for the lifetime corresponding

to i-th group

knnn ,...,, 21

k

iinn

1

in iS

ijT iS

50

• If the data corresponding to the i- th group follow , the plot will roughly be linear with slope and passes through the point

• If the plots are linear and parallel, then lifetimes under different stress levels are Weibull with common slope and different scale which implies that depends on the stress levels

)log()log())](1log(log[ iiijiij ttF

ijiiij XAY

),( iiW ),( ii YX

i )0),(log( i

ii

iS

•The equation for the lifetime corresponding to i-th group

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• If the k plots are linear and parallel, the lifetimes under different stress levels are Weibull with different slopes and different scales which implies that both and both depend on the stress levels . In this case modeling is difficult.

• For the first case the relationship between the life and stress will be identified

• Plot log(.632 quantile) against the stress levels• If the plot yields a straight line then the life-stress relationship will be

i ii

iSi

S10)log( 52

&, 10• Estimation of

• Likelihood function under stress level iS

i

i

ij

i

ij

i

n

j

tt

iL1

1 }])(exp{))([(

where

,

)exp( 10 ii S

• Total log-likelihood

k

iiLLogLogL

110 )(),,(

using ML method

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• Using numerical method MLEs of may be obtained

&, 10

• MLE of at , say, , is obtained through the relationship

)ˆˆexp(ˆ 0100 S

0S 0

• Hence ML estimate of Weibull density under used stress level is obtained. Accordingly, estimate of reliability for a specified time, median life and other desired percentiles may also be obtained

0S

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Table 5: Specimens lives (in hours) of a electrical insulation at three temperatures appear below, data of Nelson(1990), pp. 154

2520

2856

3192

3192

3528

816

912

1296

1392

1488

300

324

372

372

444

Co200 Co225 Co250

Nelson,W.(1990): Accelerated Testing: Statistical Models, Test Plans, and Data Analyses, John Wiley and Sons. 55

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Above three plots of the data given in table5 are roughly linear and parallel, so the lifetimes under three stress levels are Weibull with common slope and different scale parameters which implies that the scale parameters depend on the stress levels

• Arrhenious life-stress relationship (temperature stress)

),/1()log( 10 W where W is the temperature in degree kelvin

• Temperature in degree kelvin= temperature in degree centigrade plus 273.16

57

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• Results based on MLEs for the data given in table 5 with respect to the Arrhenious-Weibull model

39707.13ˆ0

9961.105961 68566.ˆ

98.21754ˆ0

9008.269ˆlog2 L

At used stress(180 deg. Centigrade) the followingresults are obtained

Estimated median lifetime=12747.08 hours

and 68566.ˆ

59

17807.13ˆ0

98923.105961

87.27080ˆ0

9037.273ˆlog2 L

At used stress(180 deg. Centigrade) the followingresults are obtained

Estimated median lifetime=18771.03 hours

• Results based on MLEs for the data given in table 5 with respect to the Arrhenious-Exponential model

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• Weibull versus Exponential Model for ALT

1:0 H 1:1 Hvs

9037.273ˆlog2 0 L

9008.269ˆlog2 1 L

(For Exponential model)

(For Weibull model)

)1,95(.34.30029.4)ˆlogˆ(log2 201 LL

Conclusion: Accept Weibull model at 5% level of significance

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