Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 1 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 1 2 1 f ( x) e 2 2 f(x) 2 x for - < x < x 2 Properties of the Normal Model the effects of and 3 Normal Distribution • Mean or expected value of Mean = E(X) = • Median value of X X X0.5 = • Standard deviation Var(X ) 4 Normal Distribution Standard Normal Distribution • If X~ N(, ) Z and if X then Z ~ N(0, 1). • A normal distribution with = 0 and = 1, is called the standard normal distribution. 5 Normal Distribution P (X<x’) = P (Z<z’) f(z) f(x) X’ x x'μ Z σ Z’ 0 z 6 Normal Distribution • Standard Normal Distribution Table of Probabilities http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html Enter table with Z x f(z) and find the value of • Excel 0 z z 7 Normal Distribution - Example The following example illustrates every possible case of application of the normal distribution. Let X ~ N(100, 10) Find: (a) P(X < 105.3) (b) P(X 91.7) (c) P(87.1 < X 115.7) (d) the value of x for which P(X x) = 0.05 8 Normal Distribution – Example Solution f(x) f(z) 100 105.3 a. P( X< 105.3) = x 0 0.53 z X 105.3 100 P 10 = P(Z< 0.53) = F(0.53) = 0.7019 9 Normal Distribution – Example Solution f(x) f(z) 91.7 100 b. P( X 91.7) x -0.83 0 z X 91.7 100 = P 10 = P( Z -0.83) = 1 - P( Z < -0.83) = 1- F(-0.83) = 1 - 0.2033 = 0.7967 10 Normal Distribution – Example Solution f(x) f(x) x x 87.1 100 115.7 -1.29 0 1.57 c. P(87.1 < X 115.7) = F(115.7) - F(87.1) 87.1 100 x P 115.7 10 = P(-1.29 < Z < 1.57) = F(1.57) - F(-1.29) = 0.9418 - 0.0985 = 0.8433 11 Normal Distribution – Example Solution f(x) f(z) 0.05 100 116.4 0.05 x 1.64 z 0 1.64 x 100 10 12 Normal Distribution – Example Solution (d) P( X x) = 0.05 P( Z z) = 0.05 P( X x) = implies that z = 1.64 x 100 X x 100 x 100 P P Z 1 10 10 10 therefore x 100 1.64 10 x - 100 = 16.4 x = 116.4 13 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? 14 Normal Distribution – Example Solution P1.00 X 1.75 1.75 1.25 1.00 1.25 P X 0.46 0.46 P 0.54 X 1.09 F 1.09 F 0.54 0.8621 0.2946 0.5675 15 Normal Distribution – Example Solution 2 1.25 P X 2 P Z 0.46 PZ 1.63 1 F 1.63 1 0.9484 0.0516 16 Lognormal Distribution 17 Lognormal Distribution Definition - A random variable X is said to have the Lognormal Distribution with parameters and , where > 0 and > 0, if the probability density function of X is: f (x) f(x) 0 1 x 2 0 e 1 2 ln x 22 x , for x > 0 , for x 0 18 Lognormal Distribution - Properties • Rule: then If X ~ LN(,), Y = ln ( X ) ~ N(,) • Probability Distribution Function ln x F ( x) F where F(z) is the cumulative probability distribution function of N(0,1) 19 Lognormal Distribution - Properties Mean or Expected Value 1 2 2 E( X ) e • Median x0.5 e • Standard Deviation 2μ σ 2 σ 2 e 1 SD(X) e 1 2 20 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(X) and Var(X) (b) Compute P(X > 120) (c) Compute P(110 X 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? 21 Lognormal Distribution –Example Solution a) E( X ) e u 2 Var ( X ) e b) 2 e 2u 2 5.005 2 (e e 5.005 149.16 1) 223 P( X 120) 1 P( X 120) ln 120 5.0 1 P( Z ) 0.1 1 F (2.13) 1 0.0166 0.9834 22 Lognormal Distribution –Example Solution ln 110 5.0 ln 130 5.0 c) P (110 X 130) P ( Z ) 0.1 0.1 P(2.99 Z 1.32) F (1.32) F (2.99) 0.0934 0.0014 0.092 d) X 0.5 median e e 148.41 u 5 23 Lognormal Distribution –Example Solution e) P P( X 120) 1 P( X 120) ln 120 5.0 1 P( Z ) 0.1 1 F (2.12) 1 0.0170 0.983 Let Y=number of items tested that have strength of at least 120 y=0,1,2,…,10 24 Lognormal Distribution –Example Solution Y ~ B(10,0.983) E (Y ) np 10 * 0.983 9.83 f) The value of x, say xms, for which P( X xms ) 0.05 is determined as follows: ln xms 5.0 P( Z ) 0.05 0.1 P ( Z 1.64) 0.05 ln xms 5.0 1.64 0 .1 xms 125.964 25