1 FORMULA-SHEET Arithmetic Mean, Geometric Mean, Harmonic Mean & Median AM = (1 / N) x i i GM = [ x i ] 1/ N i HM = [(1 / N) (1 / x i )] 1 i If the data set contains negative values, the following good approximation is indicated: GM = (AM x HM)1/2 Probability distribution ...Properties (1) (2) Normalization property i = 1, 2, ...N p(xi) = 1 Expected value property Expected Mean x1 to x2 [p(x)dx] = 1 Discrete variable xi i = 1, 2, ...N {(xi) p(xi)} = E[xi] Expected Mean Continuous variable, x x1 to x2 [x p(x)dx] = E[x] MOMENTS OF A VARIABLE ... Ist moment: Expected mean = m1 2nd moment m2 = i = 1, 2, ...N {(xi)2 p(xi)} : Discrete variable case m2 = x1 to x2 [(x )2 p(x)dx] : Continuous variable case Variance Discrete variable xi i = 1, 2, ...N {(xi - )2 p(xi)} = V[xi] & 2 = V 2 V = (m2) - (m1)2 Continuous variable, x x1 to x2 [(x - )2 p(x)dx] = V[x] & 2 = V = V = (m2) - (m1)2 Statistical arrangement of epochs... PERMUTATION - CASE 1 ------------------------------------------------------PERMUTATION - CASE 2 PERMUTATION - CASE 3 COMBINATIONS 3 BINOMIAL THEOREM Combinatorial Problems Number of different groups of size r that can be selected from a set of n objects without any regard to the order of selection is: n = n C = n! /(n r)! r! r r Binomial Theorem n n ( x + y ) n = x k y nk k =0 k B(X; N, p) = [ N C X ] p X q ( N X ) where q = (1 - p); and, N N N! =[ C X ] = : Binomial coefficient. [X!] [( N X)!] p Mean of binomial distribution: N p Variance of binomial distribution: N p q Bernoulli process This discrete event process corresponds to Bernoulli trials of a RV, X depicting the number of successes. Considering N (fixed integer) trials and each trial is independent of the other, the possible outcomes are, success (HEAD) or failure (TAIL). Suppose: There are n trials and all outcomes are independent Each trial has two possible out comes: ALL or NONE; HEAD or TAIL; 0 or 1; SUCCESS or FAILURE p(SUCCESS) = p p(FAILURE) = (1 - p) = q 4 Given the binomial distribution: b(x; n, p), suppose n tends to and n p = , a constant > 0. Then, f (x; ) represents the pdf of Poisson distribution. f(x; ) = x exp() / x! with, mean = variance = Sampling distribution: FINITE POPULATION N: Size of finite population with E[x] = and : Standard deviation n: Size of the sample (< N) xs: sampled data variable Mean of the sampled data: Standard deviation of the sampled population set: Normalized sample set variable: E[xs] = s = s = /(n)1/2 Zs = (xs - s)/ s Between x1 and x2 Probability = P(x1 < xs < x2) = P{[Z1: (x1- s)/s] < Zs < [Z2: (x2 - s)/s]} 5 = Area = (x 2 - s )/s (x1 - s )/s 2 1 Z exp dZ 2 2