1 special continuous probability distributions -normal distributions -lognormal distributions dr....

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Special Continuous Probability Distributions-Normal Distributions

-Lognormal Distributions

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08


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A random variable X is said to have a normal (orGaussian) distribution with parameters and ,where - < < and > 0, with probability density function

for - < x <

222

1

2

1)(

x

exf

f(x)

x

Normal Distribution


3

the effects of and

Properties of the Normal Model


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• Mean or expected value ofMean = E(X) =

• Median value of

X0.5 =

• Standard deviation

)(XVar

X

X

Normal Distribution


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Standard Normal Distribution

• If ~ N(, )

and if

then Z ~ N(0, 1).

• A normal distribution with = 0 and = 1, is calledthe standard normal distribution.

X

Z

X

Normal Distribution


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x 0 z

σ

μx'Z

f(x) f(z)

P (X<x’) = P (Z<z’)

X’ Z’

Normal Distribution


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• Standard Normal Distribution Table of Probabilities

http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html

Enter table with

and find thevalue of

• Excelz

0

z

f(z)

x

Z

Normal Distribution


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The following example illustrates every possible case of application of the normal distribution.

Let ~ N(100, 10)

Find:

(a) P(X < 105.3)

(b) P(X 91.7)

(c) P(87.1 < 115.7)

(d) the value of x for which P( x) = 0.05X

X

X

Normal Distribution - Example


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a. P( < 105.3) =

= P( < 0.53)= F(0.53)= 0.7019

10

1003.105P

X

100 x 0 z

f(x) f(z)

105.3 0.53

X

Z

Normal Distribution – Example Solution


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b. P( 91.7) =

= P( -0.83) = 1 - P( < -0.83) = 1- F(-0.83)

= 1 - 0.2033 = 0.7967

10

1007.91

X

P

100 x 0 z

f(x) f(z)

91.7 -0.83

ZZ

X



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c. P(87.1 < 115.7) = F(115.7) - F(87.1)

= P(-1.29 < Z < 1.57)= F(1.57) - F(-1.29)= 0.9418 - 0.0985 = 0.8433

7.115

10

1001.87

x

P

100

x

f(x)

87.1 115.7 0

x

f(x)

-1.29 1.57

X



12

100x

0z

f(x) f(z)

10

10064.1

x

0.05 0.05

1.64116.4



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(d) P( x) = 0.05P( z) = 0.05 implies that z = 1.64P( x) =

therefore

x - 100 = 16.4x = 116.4

10

1001

10

100P

10

100P

xxZ

xX

64.110

100

x

ZX

X



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The time it takes a driver to react to the brake lightson a decelerating vehicle is critical in helping toavoid rear-end collisions. The article ‘Fast-Rise BrakeLamp as a Collision-Prevention Device’ suggests that reaction time for an in-traffic response to abrake signal from standard brake lights can be modeled with a normal distribution having meanvalue 1.25 sec and standard deviation 0.46 sec.What is the probability that reaction time is between1.00 and 1.75 seconds? If we view 2 seconds as acritically long reaction time, what is the probabilitythat actual reaction time will exceed this value?



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75.100.1 XP

46.0

25.175.1

46.0

25.100.1XP

09.154.0 XP

54.009.1 FF

5675.02946.08621.0



16

2XP

0516.0

9484.01

63.11

63.1

46.0

25.12

F

ZP

ZP



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Statistical Quality Control is an application of probabilitistic and statistical techniques to quality control

An Application of Probability & Statistics Statistical Quality Control


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Statistical Quality Control - Elements

Analysisof processcapability

Statisticalprocesscontrol

Processimprovement

Acceptancesampling


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Statistical Tolerancing - Convention

Normal ProbabilityDistribution

LTL Nominal UTL

0.001350.00135 0.9973

+3-3


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Statistical Tolerancing - Concept

LTL UTLNominal x


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For a normal distribution, the natural tolerance limits include 99.73% of the variable, or put another way, only 0.27% of the process output will fall outside the natural tolerance limits. Two points should be remembered:

1. 0.27% outside the natural tolerances sounds small, but this corresponds to 2700 nonconforming parts per million.

2. If the distribution of process output is non normal, then the percentage of output falling outside 3 may differ considerably from 0.27%.

Caution


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The diameter of a metal shaft used in a disk-drive unitis normally distributed with mean 0.2508 inches andstandard deviation 0.0005 inches. The specificationson the shaft have been established as 0.2500 0.0015 inches. We wish to determine what fraction ofthe shafts produced conform to specifications.



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spec meetingP

91924.0

0000.091924.0

60.41.40

0005.0

0.2508-0.2485

0.0005

0.2508-0.2515

x0.2485P0.2515xP

0.2515x0.2485P

0.2508 0.2515 USL

0.2485 LSL

0.2500

f(x)

xnominal

Normal Distribution - Example Solution


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Thus, we would expect the process yield to be approximately 91.92%; that is, about 91.92% of the shafts produced conform to specifications. Note that almost all of the nonconforming shafts are too large, because the process mean is located very near to the upper specification limit. Suppose we can recenter the manufacturing process, perhaps by adjusting the machine, so that the process mean is exactly equal to the nominal value of 0.2500. Then we have



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0.2515x0.2485P

9973.0

00135.099865.0

00.33.00

0005.0

0.2500-0.2485

0.0005

0.2500-0.2515

x0.2485P0.2515xP

0.2500 0.2515 USL

0.2485 LSL

f(x)

xnominal



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27


28


29


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Using a normal probability distribution as a model for a quality characteristic with the specification limits at three standard deviations on either side of the mean. Now it turns out that in this situation the probability of producing a product within these specifications is 0.9973, which corresponds to 2700 parts per million (ppm) defective. This is referred to as three-sigma quality performance, and it actually sounds pretty good. However, suppose we have a product that consists of an assembly of 100 components or parts and all 100 parts must be non-defective for the product to function satisfactorily.



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The probability that any specific unit of product is non-defective is

0.9973 x 0.9973 x . . . x 0.9973 = (0.9973)100 = 0.7631

That is, about 23.7% of the products produced underthree sigma quality will be defective. This is not anacceptable situation, because many high technology products are made up of thousands of components.An automobile has about 200,000 components andan airplane has several million!



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Lognormal Distribution


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Definition - A random variable is said to have the Lognormal Distribution with parameters and , where > 0 and > 0, if the probability density function of X is:

, for x > 0

, for x 0

22

xln2

1

e2x

1 )x(f

0

x

f(x)

0

X

Lognormal Distribution


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• Rule: If ~ LN(,),

then = ln ( ) ~ N(,)

• Probability Distribution Function

where F(z) is the cumulative probability distribution function of N(0,1)

xFxF

ln )(

Y

X

X

Lognormal Distribution - Properties


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Mean or Expected Value

22

1

)(

eXE

2

1

12σe

2σ2μeSD(X)

• Standard Deviation

• Median

ex 5.0

Lognormal Distribution - Properties


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A theoretical justification based on a certain materialfailure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. Suppose the parameters are = 5 and = 0.1

(a) Compute E( ) and Var( )(b) Compute P( > 120)(c) Compute P(110 130)(d) What is the value of median ductile strength?(e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120?(f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

XX

XX

Lognormal Distribution - Example


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Lognormal Distribution –Example Solution

a)

223)1()(

16.149)(22

2

2

005.5005.52

eeXVar

eeeXEu

u

b) )120(1)120( XPXP

9834.0

0166.01

)13.2(1

)1.0

0.5120ln(1

F

ZP


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c) )1.0

0.5130ln

1.0

0.5110ln()130110(

ZPXP

092.0

0014.00934.0

)99.2()32.1(

)32.199.2(

FF

ZP

d) 41.14855.0 eemedianX u


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e) )120( XPP

983.0

0170.01

)12.2(1

)1.0

0.5120ln(1

)120(1

F

ZP

XP

Let Y=number of items tested that have strength of at least 120Y=0,1,2,…,10


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83.9983.0*10)(

)983.0,10(~

npYE

BY

f) The value of x, say xms, for which is determined as follows:

05.0)( msxXP

964.125

64.11.0

0.5ln

05.0)64.1(

05.0)1.0

0.5ln(

ms

ms

ms

x

x

ZP

xZP

1 special continuous probability distributions -normal distributions -lognormal distributions dr....

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