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Copyright unreserved and unlimited Probability Models 1 UNIVERSITY OF PRETORIA DEPARTMENT OF INDUSTRIAL AND SYSTEMS ENGINEERING ENGINEERING STATISTICS BES221 PROBABILITY MODELS The Chancellor of the Exchequer is a man whose duties make him more or less of a taxing machine. He is entrusted with a certain amount of misery which it is his duty to distribute as fairly as he can. Robert Lowe, Viscount Sherbrooke, British Liberal politician Histograms Probabilities Distributions Distributions Probability Models Sampling DISCRETE DISTRIBUTIONS : CONCEPTS Random variables – Variates Output from a stochastic process Discrete and continuous variates and distributions Histograms and distributions Stochastic models – mathematical The raison d’être of probability models BINOMIAL DISTRIBUTION (Discrete) Variate : Number of “successes” in n Bernoulli trails Number of “successes” in a “small” sample of size n taken from a “large” population of size N Jean (or Johann) Bernoulli (1667 – 1748) Jacques (or Jakop) Bernoulli (1654 – 1705)(brother of Jean) P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 2 Daniel Bernoulli (1700 – 1782) Johann Bernoulli (1667 – 1748) (Son of Johann) Assumptions : 1. Only two possible outcomes – “success” and “failure” 2. Probability of a “success” stays the same and constant = p 3. There are n trails with n a constant 4. Trails are mutually independent P(X=x) = b(x;n,p) = nCx px (1-p)n-x x = 0,1,2,…, n 0p1 n > 0 and integer Mean : =np Variance : 2 = n p (1-p) Symmetrical Binomial Distributions (n large and p 0,5) The Normal approximation to the Binomial distribution (In general, if np > 5 and n(1-p) > 5 this approximation may be used) Cumulative probabilities Binomial probability tables Applications of the Binomial distribution P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 3 POISSON DISTRIBUTION (Discrete) Siméon-Denis Poisson (1781 – 1840) (Fr : Fish) The Poisson Process (One of the most important stochastic processes) Assumptions : 1. The probability of a “success” during a small interval of time is proportional to the length of the time interval 2. The probability of more than one “success” during a small interval of time is negligible 3. The probability of a “success” during a small interval of time does not depend on what happened during any of the previous time intervals (no memory property) Variate : A discrete number of occurrences per time/space/entity (Goals per soccer game, Drop-goals per rugby game, Typing errors per page, Defectives per assembly, Arrivals per hour, Inclusions per m3, Particles per liter, Breakdowns per day, Dropouts per class, Accidents per kilometer, Sixes per cricket innings, Freckles per nose, etc.) P(X=x) = f(x;) = x e- / x! for x = 0,1,2, … Mean : = Variance : 2 = Use Poisson as a good approximation to the Binomial if n 20 and p 0,05 or if n 100 then np 10 Relationship to the Exponential distribution – Queuing Theory P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 4 MEAN AND VARIANCE OF DISTRIBUTIONS Every distribution has a mean and variance that are functions of the parameters of the distribution In general : = Σ x f(x) 2 = Σ (x - )2 f(x) Binomial : =np 2 = n p (1-p) = 2 = There are many other somewhat specialized discrete distributions Poisson : The most widely used, and therefore most important, discrete distributions are the Binomial and Poisson MONTE CARLO SIMULATION Generating and transforming random numbers (using a pseudorandom generator and appropriate transformations) may be used to simulate many statistical experiments (with the help of a computer and appropriate software !) (For further information you may enroll for BAN313 !) P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 5 CONTINUOUS DISTRIBUTIONS : CONCEPTS Continuous random variables Probability is an area : P(a X b) = ∫ f(x) dx for all values of x between a and b = ∫ x f(x) dx for all values of x 2 = ∫(x - )2 f(x) dx for all values of x Please note : P(X = x) = 0 if X is continuous !!! THE NORMAL DISTRIBUTION (Continuous) Developed and applied mainly by Abraham de Moivre (16671754), Carl Friedrich Gauss (1777-1855), Pierre Simon Laplace (1747-1827) and Sir Francis Galton (1822 -1911) Carl Friedrich Gauss P S Kruger/2011 Abraham de Moivre Pierre Simon Laplace Sir Francis Galton Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 6 f(x; ,2) = [ 1 / {(2) }] [exp { –(x - )2 / 22 }] - < x < + This function is NOT analytical integratable ! Mean is and Standard Deviation is The Standard Normal distribution : = 0 and 2 = 1 One may transform ANY Normal distribution to the Standard Normal distribution by using the following transformation (transforming the variate X to the variate Z) : Z = (X - ) / P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 7 B A X X1 X2 B A z1 = (x1 - ) / 0 Z z2 = (x2 - ) / The use of the Standard Normal Tables (examples) The Normal approximation to the Binomial. Approximation is acceptable if both np and n(1-p) are greater than 15 The Normal distribution is probably the most important distribution in mathematical statistics ! P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 8 OTHER CONTINUOUS DISTRIBUTIONS The Uniform or Rectangular distribution The Log-Normal distribution The Gamma distribution The Beta distribution The Weibull distribution The Exponential Distribution (special form of the Gamma) f(x) = (1 / ) e –(x / ) x0 Mean is and Standard Deviation is Cumulative distribution function : F(t) = 1 - e –(t / ) Therefore P(0 x t) = F(t) F(t) f(x) 0 P S Kruger/2011 t x Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 9 LINEAR COMBINATIONS OF RANDOM VARIABLES If X1 and X2 are two independent random variables with means 1 and 2 and standard deviations 1 and 2 respectively and Y = a1 X1 a2 X2 Then Y = a1 1 a2 2 (Y)2 = (a1)2 (1)2 + (a2)2 (2)2 Furthermore If X1 and X2 are both Normally distributed then Y is also Normally distributed Y will approach a Normal distribution, independent of the distributions of Xi, if i is fairly large (For i > 30 the Normality assumption can be made with confidence, mainly because of the Central Limit Theorem) There are a large number of other continuous distributions of special nature and application, but the Normal and Exponential distributions are probably the most important {Special relationship between the Poisson and Exponential distributions} ******************* P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 10 WORKED EXAMPLES The Poisson Distribution The rate at which people arrives at the ground floor of a building, with the purpose of using the lift, is Poisson distributed with a mean of 8 persons per minute. The maximum capacity of the lift is 14 people. Determine the probability that more people will arrive during a 2-minute interval than what the lift can accommodate. 1 = 8 persons per minute therefore, during a two minute interval 2 = 16 persons per two minutes (The Poisson Distribution may be scaled !) P(X = x) = x e- / x! for P(X 15) = 1 - P(X 14) = 1 - (x e- / x!) = 1 - 0,367527 = 0,6325 x = 0,1,2, … for x = 0 to 14 and = 16 The Exponential Distribution A manufacturer of colour television sets offers a guarantee consisting of replacing the TV-tube if it fails within one year. The lifetime of television tubes is Exponentially distributed with a mean of 4 years. Determine the expected percentage of all television sets sold that will have to be repaired under the guarantee. P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited 11 Probability Models F(t) f(X) x=t=1 X = 4 (mean) and t = 1 F(t) = 1 - e –(t / ) = 1 – 0,7788 = 0,2212 22% of the tubes will have to be replaced under guarantee The Normal Distribution The time it takes for a driver of a motor vehicle to react to the brake lights of a preceding vehicle is Normally distributed (Ergonomics, 1993, pp. 390-395) with a mean of 1,25 seconds and a standard deviation of 0,46 seconds. Determine the probability that the reaction time will be more than 1,75 seconds. = 1,25 and = 0,46 x = 1,75 z = (x - ) / = (1,75 – 1,25) / 0,46 = 1,0869 P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 12 0 z=1,0869 x=1,75 Z P(X 1,75) = P(Z 1,0869) = 1 - P(Z 1,09) = 1 - 0,8621 = 0,1379 Linear Combinations of Random Variables Two resistors are installed in series. The nominal resistance of each component is 10 Ohm. It is known that the resistance is Normally distributed. Determine the value of the standard deviation of the resistance to ensure that the probability for the total resistance of the assembly to exceed 21 Ohm is 0,01 or less. Y = a1 X1 a2 X2 with both a1 and a2 equal to 1 Y = a1 1 a2 2 = 10 + 10 = 20 z = (x - ) / = (21 – 20) / Y P S Kruger/2011 Engineering Statistics BES221 Copyright unreserved and unlimited Probability Models 13 0,01 0 z = 1 / Y Z 1 / Y= 2,33 (from tables) Y = 0,4292 (Y)2 = (a1)2 (1)2 + (a2)2 (2)2 = 0,4292 2()2 = 0,4292 and = 0,4632 ************************ P S Kruger/2011 Engineering Statistics BES221