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Statistical Distributions Uniform Distribution A R.V. is uniformly distributed on the interval (a,b) if it probability function Fully defined by (a,b) P(x) = 1/(b-a) for a <= x <= b = 0 otherwise Uniform Distribution Probability Function 1 1/9 1 10 Probability that x is between 2 and 7.5? Probability that x = 8? 1 1/9 1 10 Uniform Distribution The cumulative distribution of a uniform RV is F(x) = 0 for x < a = (x-a)/(b-a) for a <= x <= b = 1 otherwise Uniform Distribution Cumulative Function 1 1 10 Uniform Distribution Discrete vs. Continuous • Discrete RV – Number showing on a die • Continuous RV – Time of arrival – When programming, make it discrete to some number of decimal places Uniform Distribution • Mean = (a+b)/2 • Variance = (b-a)2 /12 • P (x < X < y) = F (y) – F (x) = (y-a) - (x-a) = y – x – a + a = y - x b-a b-a b–a b-a Uniform - Example A bus arrives at a bus stop every 20 minutes starting at 6:40 until 8:40. A passenger does not know the schedule but randomly arrives between 7:00 and 7:30 every morning. What is the probability the passenger waits more than 5 minutes. Uniform Solution X = RV, Uniform (0,30) -- i.e. 7:00 – 7:30 Bus: 7:00, 7:20, 7:40 Yellow Box <= 5 minute wait 1 1/30 A B C 5 10 15 20 25 30 40 P (x > 5) = A + C = 1 – B = 5/6 Arithmetic Mean Given a set of measurements y1, y2, y3,… yn Mean = (y1+y2+…yn) / n Variance Variance of a set of measurements y1, y2, y3,… yn is the average of the deviations of the measurements about their mean (m). V = σ2 = (1/n) Σ (yi – m)2 i=1..n Variance Example Yi= 12, 10, 9, 8, 14, 7, 15, 6, 14, 10 m = 10.5 V= σ2 = (1/10) ((12-10.5)2 + (10-10.5)2 +…. = (1/10) (1.52 + .52 + 1.52….) = (1/10) (88.5) = 8.85 Standard Deviation = σ = 2.975 Normal Distribution • Has 2 parameters – Mean - μ – Variance – σ2 – Also, Standard deviation - σ Normal Dist. .3413 .1359 .0215 .0013 -3 -2 -1 1 2 3 0 Mean +- n σ Normal Distribution • Standard Normal Distribution has – Mean = 0 StdDev = 1 • Convert non-standard to standard to use the tables Z value = # of StdDev from the mean Z is value used for reading table Z = (x – m) σ Normal - Example The scores on a college entrance exam are normally distributed with a mean of 75 and a standard deviation of 10. What % of scores fall between 70 & 90? Z(70) = (70 – 75)/10 = - 0.5 Z(90) = (90 – 75)/10 = 1.5 .6915 - .5 = .1915 + .9332 - .5 = .4332 = .6247 or 62.47% Exponential Distribution A RV X is exponentially distributed with parameter > 0 if probability function Mean = 1/ Variance = 1 / 2 e = 2.71828182 P(x) = x e For x >= 0 = 0 Otherwise Exponential Distribution • Often used to model interarrival times when arrivals are random and those which are highly variable. • In these instances lambda is a rate – e.g. Arrivals or services per hour • Also models catastrophic component failure, e.g. light bulbs burning out Exponential Rates • Engine fails every 3000 hours – Mean: Average lifetime is 3000 hours – = 1/3000 = 0.00033333 • Arrivals are 5 every hour – Mean: Interarrival time is 12 minutes – = 1 / 5 = 0.2 • Mean = 1 / Exponential Distribution Probability Function f(x) x See handout for various graphs. Exponential Distribution Cumulative Function Given Mean = 1/ Variance = 1/ 2 F(x) = P (X <=x) = 1 – e - x Exponential Distribution Cumulative Function (<=) 1 F(x) x Forgetfulness Property Given: the occurrence of events conforms to an exponential distribution: The probability of an event in the next x- unit time frame is independent on the time since the last event. That is, the behavior during the next x-units of time is independent upon the behavior during the past y-units of time. Forgetfulness Example • The lifetime of an electrical component is exponentially distributed with a mean of . • What does this mean?? Forgetfulness Examples The following all have the same probability • Probability that a new component lasts the first 1000 hours. • Probability that a component lasts the next 1000 hours given that it has been working for 2500 hours. • Probability that a component lasts the next 1000 hours given that I have no idea how long it has been working. Solution to Example • Suppose the mean lifetime of the component is 3000 hours. • = 1/3000 • P(X >= 1000) = 1 – P(X <= 1000) 1 – (1-e -1/3* 1) = e -1/3 = .717 How do we apply these? 1. We may be given the information that events occur according to a known distribution. 2. We may collect data and must determine if it conforms to a known distribution.

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