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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

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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

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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

Normal Distribution – Example Solution

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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

Normal Distribution – Example Solution

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100x

0z

f(x) f(z)

10

10064.1

x

0.05 0.05

1.64116.4

Normal Distribution – Example Solution

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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

Normal Distribution – Example Solution

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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?

Normal Distribution – Example Solution

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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

Normal Distribution – Example Solution

16

2XP

0516.0

9484.01

63.11

63.1

46.0

25.12

F

ZP

ZP

Normal Distribution – Example Solution

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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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Lognormal Distribution –Example Solution

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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Lognormal Distribution –Example Solution

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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Lognormal Distribution –Example Solution

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