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1 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 Program Department of Engineering Management, Information and Systems
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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 Presentation

### Transcript of Special Continuous Probability Distributions Normal Distributions Lognormal Distributions

• *Special Continuous Probability DistributionsNormal DistributionsLognormal Distributions

Dr. Jerrell T. Stracener, SAE FellowLeadership in EngineeringEMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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

for - < x < Normal Distribution

• *

the effects of and

Properties of the Normal Model

• *

Mean or expected value ofMean = E(X) =

Median value of

X0.5 =

Standard deviationNormal Distribution

• *

Standard Normal Distribution

If ~ N(, )

and if

then Z ~ N(0, 1).

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

• *x0zf(x)f(z)P (X
• *Normal Distribution

• *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) = 0.05Normal Distribution - Example

• *

a. P( < 105.3) =

= P( < 0.53)= F(0.53)= 0.7019100x0zf(x)f(z)105.30.53Normal Distribution Example Solution

• *

b. P( 91.7) =

= P( -0.83) = 1 - P( < -0.83) = 1- F(-0.83) = 1 - 0.2033 = 0.7967100x0zf(x)f(z)91.7-0.83Normal Distribution Example Solution

• *

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 = 0.8433100xf(x)87.1115.70xf(x)-1.291.57Normal Distribution Example Solution

• *100x0zf(x)f(z)0.050.051.64116.4Normal Distribution Example Solution

• *

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

therefore

x - 100 = 16.4x = 116.4Normal Distribution Example Solution

• *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

• *Normal Distribution Example Solution

• *Normal Distribution Example Solution

• *Lognormal Distribution

• *

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

Lognormal Distribution

• *

Rule:If ~ LN(,),

then = ln ( ) ~ N(,)

Probability Distribution Function

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

• *Mean or Expected Value Standard Deviation MedianLognormal Distribution - Properties

• *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?Lognormal Distribution - Example

• *Lognormal Distribution Example Solutiona) b)

• *Lognormal Distribution Example Solutionc)d)

• *Lognormal Distribution Example Solutione)Let Y=number of items tested that have strength of at least 120y=0,1,2,,10

• *Lognormal Distribution Example Solutionf) The value of x, say xms, for which is determined as follows: