special continuous probability distributions normal distributions lognormal distributions
DESCRIPTION
Systems Engineering Program. Department of Engineering Management, Information and Systems. EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS. Special Continuous Probability Distributions Normal Distributions Lognormal Distributions. - PowerPoint PPT PresentationTRANSCRIPT
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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
25
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