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Special Continuous Probability Distributions Normal Distributions Lognormal Distributions

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.

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Special Continuous Probability Distributions Normal Distributions Lognormal Distributions

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  1. 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 Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering

  2. f(x) x Normal Distribution A random variable X is said to have a normal (or Gaussian) distribution with parameters  and , where - < <  and  > 0, with probability density function for - < x < 

  3. Properties of the Normal Model • the effects of  and 

  4. Normal Distribution • Mean or expected value of • Mean = E(X) =  • Median value of • X0.5 =  • Standard deviation

  5. Normal Distribution • Standard Normal Distribution • If ~ N(, ) • and if • then Z ~ N(0, 1). • A normal distribution with  = 0 and  = 1, is called • the standard normal distribution.

  6. Normal Distribution P (X<x’) = P (Z<z’) f(z) f(x) x z 0  Z’ X’

  7. Standard Normal Distribution Table of Probabilities • http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html • Enter table with • and find the • value of  • Excel f(z)  z 0 z Normal Distribution

  8. Normal Distribution - Example 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) = 0.05

  9. a. P( < 105.3) = = P( < 0.53) = F(0.53) = 0.7019 Normal Distribution – Example Solution f(x) f(z) x z 105.3 100 0.53 0

  10. b. P(  91.7) = = P( -0.83) = 1 - P( < -0.83) = 1- F(-0.83) = 1 - 0.2033 = 0.7967 Normal Distribution – Example Solution f(x) f(z) x z 0 100 91.7 -0.83

  11. 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 Normal Distribution – Example Solution f(x) f(x) x x 87.1 100 115.7 -1.29 0 1.57

  12. Normal Distribution – Example Solution f(x) f(z) 0.05 0.05 x z 1.64 116.4 100 0

  13. Normal Distribution – Example Solution (d) P(  x) = 0.05 P(  z) = 0.05 implies that z = 1.64 P(  x) = therefore x - 100 = 16.4 x = 116.4

  14. Normal Distribution – Example Solution The time it takes a driver to react to the brake lights on a decelerating vehicle is critical in helping to avoid rear-end collisions. The article ‘Fast-Rise Brake Lamp as a Collision-Prevention Device’ suggests that reaction time for an in-traffic response to a brake signal from standard brake lights can be modeled with a normal distribution having mean value 1.25 sec and standard deviation 0.46 sec. What is the probability that reaction time is between 1.00 and 1.75 seconds? If we view 2 seconds as a critically long reaction time, what is the probability that actual reaction time will exceed this value?

  15. Normal Distribution – Example Solution

  16. Normal Distribution – Example Solution

  17. Lognormal Distribution

  18. f(x) x 0 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

  19. Lognormal Distribution - Properties • Rule: If ~ LN(,), • then = ln ( ) ~ N(,) • Probability Distribution Function • where F(z) is the cumulative probability distribution • function of N(0,1)

  20. Lognormal Distribution - Properties Mean or Expected Value • Median • Standard Deviation

  21. Lognormal Distribution - Example A theoretical justification based on a certain material failure 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?

  22. Lognormal Distribution –Example Solution a) b)

  23. Lognormal Distribution –Example Solution c) d)

  24. Lognormal Distribution –Example Solution e) Let Y=number of items tested that have strength of at least 120 y=0,1,2,…,10

  25. Lognormal Distribution –Example Solution f) The value of x, say xms, for which is determined as follows:

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