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Exponential Distribution. = mean interval between consequent events = rate = mean number of counts in the unit interval > 0 X = distance between events >0 X ~ e( ) f ( x) e x F ( x) 1 e f(x) E( X ) 1/ x F(x) V ( X ) 1 / 2 Memoryless property P( x t1 t2 | x t1 ) P( x t2 ) Exponential and Poisson relationship Unit Matching between x and IME 312 ! 1/ Exponential Dist. 1/ X ~ Expo( ) f ( x) e Y ~ Pois ( ) x F ( x) 1 e E( X ) x e f ( y) y! 1 V (X ) IME 312 Poisson Dist. 1 2 E (Y ) V (Y ) y Relation between Exponential distribution ↔ Poisson distribution Xi : Continuous random variable, time between arrivals, has Exponential distribution with mean = 1/4 X X1=1/4 X2=1/2 0 X3=1/4 X4=1/8 X5=1/8 X6=1/2 1:00 Y1=3 X7=1/4 X8=1/4 i 1 12 X9=1/8 X10=1/8 2:00 Y2=4 Yi : Discrete random variable, number of arrivals per unit of time, has Poisson distribution with mean = 4. (rate=4) Y ~ Poisson (4) IME 301 and 312 12 Xi X11=3/8 1 4 X12=1/8 3:00 Y3=5 Y Y1 Y2 Y3 4 3 Continuous Uniform Distribution a xb x ~ u ( a, b) 1 f ( x) ba 0 xa x a F ( x) a xb b a 1 bx ab E( X ) 2 IME 312 (b a) 2 V (X ) 12 f(x) a b F(x) a b Gamma Distribution K = shape parameter >0 = scale parameter >0 f ( x) x k x e ( k ) (k ) x 0 k 1 k 1 x e dx For Gamma Function, you can use: and if K is integer (k’) then: IME 312 E( X ) k / Var( X ) k / 2 (1) 1 (k ' ) (k '1)! Application of Gamma Distribution K = shape parameter >0 = number of Yi added = scale parameter >0 = rate if X ~ Expo ( ) and Y = X1 + X2 + ……… + Xk then Y ~ Gamma (k, ) i.e.: Y is the time taken for K events to occur and X is the time between two consecutive events to occur IME 312 Relation between Exponential distribution ↔ Gamma distribution Xi : Continuous random variable, time between arrivals, has Exponential distribution with mean = 1/4 X X1=1/4 0 X2=1/2 X3=1/4 X4=1/8 X5=1/8 X6=1/2 X7=1/4 1:00 X8=1/4 12 i 1 Xi 12 X9=1/8 X10=1/8 X11=3/8 2:00 Y1=1 Y2=7/8 Y3=1/2 Y4=3/4 Yi : Continuous random variable, time taken for 3 customers to arrive, has Gamma distribution with shape parameter k = 3 and scale=4 IME 312 1 4 X12=1/8 3:00 Weibull Distribution a = shape parameter >0 = scale parameter >0 a 1 ( x ) a f ( x) a x e a f ( x) 0 F ( x) 1 e IME 312 ( x ) a for x0 for x0 Normal Distribution x x ~ N ( , ) 1 f ( x) e 2 ( x )2 z ~ N (0,1) F ( x) ( z ) IME 312 0 V (X ) E( X ) (x ) 2 2 2 Standard Normal Use the table in the Appendix Normal Approximation to the Binomial Use Normal for Binominal if n is large X~Binomial (n, p) np 2 np (1 p ) where : X X or 0.5 ~ Norm X np Z , where, Z ~ Norm(0,1) np(1 p) Refer to page 262 IME 312 Central Limit Theorem x1 , x2 , x3 ,....., xn : random sample from a population with Then Z X X n as and : sample mean has standard normal distribution N(0, 1) n commonly n 30 IME 312 2 What does Central Limit Theorem mean? Consider any distribution (uniform, exponential, normal, or …). Assume that the distribution has a mean of and a standard deviation of . Pick up a sample of size “n” from this distribution. Assume the values of variables are: x1 , x2 , x3 ,....., xn Calculate the mean of this sample . Repeat this X process and find many sample means. Then our sample means will have a normal distribution with a mean of and a standard deviation of n . IME 312 u , v = degrees of freedom = probability Distribution Definition v Chi-Square X Z i 1 t dist. F dist. Z T u2 F 2 v Where: Z ~ N (0,1) IME 312 2 i 2 v u v Notation , ,v 2 v 2 t v , t ,v v Fu ,v , F ,u ,v