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Statistical DistributionsUniform DistributionUniform Distribution Probability FunctionProbability that x is between 2 and 7.5? Probability that x = 8?Slide 5Uniform Distribution Cumulative FunctionUniform Distribution Discrete vs. ContinuousUniform DistributionUniform - ExampleUniform SolutionArithmetic MeanVarianceVariance ExampleNormal DistributionNormal Dist.Slide 16Normal - ExampleExponential DistributionSlide 19Exponential RatesExponential Distribution Probability FunctionExponential Distribution Cumulative FunctionExponential Distribution Cumulative Function (<=)Forgetfulness PropertyForgetfulness ExampleForgetfulness Examples The following all have the same probabilitySolution to ExampleHow do we apply these?Statistical Statistical DistributionsDistributionsStatistical Statistical DistributionsDistributionsUniform DistributionA R.V. is uniformly distributed on the interval (a,b) if it probability functionFully defined by (a,b) P(x) = 1/(b-a) for a <= x <= b = 0 otherwiseUniform Distribution Probability Function 1 10 1 1/9Probability that x is between 2 and 7.5? Probability that x = 8? 1 10 1 1/9Uniform DistributionThe cumulative distribution of a uniform RV is F(x) = 0 for x < a = (x-a)/(b-a) for a <= x <= b = 1 otherwiseUniform Distribution Cumulative Function 1 10 1Uniform DistributionDiscrete vs. Continuous•Discrete RV –Number showing on a die•Continuous RV–Time of arrival –When programming, make it discrete to some number of decimal placesUniform 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-aUniform - ExampleA 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 5 10 15 20 25 30 40 11/30X = RV, Uniform (0,30) -- i.e. 7:00 – 7:30Bus: 7:00, 7:20, 7:40Yellow Box <= 5 minute wait A B CP (x > 5) = A + C = 1 – B = 5/6Arithmetic MeanGiven a set of measurements y1, y2, y3,… ynMean = (y1+y2+…yn) / nVarianceVariance 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)2i=1..nVariance ExampleYi= 12, 10, 9, 8, 14, 7, 15, 6, 14, 10m = 10.5V= σ2 = (1/10) ((12-10.5)2 + (10-10.5)2 +…. = (1/10) (1.52 + .52 + 1.52….) = (1/10) (88.5) = 8.85Standard Deviation = σ = 2.975Normal Distribution•Has 2 parameters–Mean - μ–Variance – σ2–Also, Standard deviation - σNormal Dist.0-3 -2 -1 1 2 3Mean +- n σ.3413 .1359 .0215 .0013Normal Distribution•Standard Normal Distribution has–Mean = 0 StdDev = 1•Convert non-standard to standard to use the tablesZ value = # of StdDev from the meanZ is value used for reading table Z = (x – m) σNormal - ExampleThe 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 = - 0.5Z(90) = (90 – 75)/10 = 1.5.6915 - .5 = .1915 + .9332 - .5 = .4332 = .6247 or 62.47%Exponential DistributionA RV X is exponentially distributed with parameter > 0 if probability function Mean = 1/Variance = 1 / 2 e = 2.71828182 exP(x) =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 outExponential 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 DistributionProbability Functionxf(x)See handout for various graphs.Exponential DistributionCumulative FunctionGiven Mean = 1/ Variance = 1/ 2F(x) = P (X <=x) = 1 – e - xExponential DistributionCumulative Function (<=)x1F(x)Forgetfulness PropertyGiven: 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 ExamplesThe 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


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MSU CMPS 4223 - Lecture Notes

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