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Understanding Continuous Probability Distributions in IE

School
Pennsylvania State University**We aren't endorsed by this school
Course
IE 322
Subject
Industrial Engineering
Date
Dec 11, 2024
Pages
30
Uploaded by ConstableWaterYak5
IE 322: Probabilistic Models in IELecture 19-Chapter 6: Some Continuous Probability DistributionsPrakash ChakrabortyThe Pennsylvania State UniversityFall 2024Keywords:Uniform, Normal, Normal curve, Lognormal, Chi-squared, Exponential, GammaPrakash Chakraborty(PSU)Lecture 19-Fall 20241 / 30
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6.1 Uniform distribution with parametersA,BOne of the simplest continuous distributions. This distribution is characterized by being‘flat,’ and thus the density is uniform in a closed interval, say[A,B].Figure 1:pdf of Uniform[1,3]DefinitionThe density function of the continuous uniform random variableXon the interval[A,B]isf(x;A,B) =1B−A,A≤x≤B,0,elsewhere.Prakash Chakraborty(PSU)Lecture 19-Fall 20242 / 30
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ExampleSuppose that a large conference room at a certain company can be reserved for no morethan 4 hours. Both long and short conferences occur quite often. In fact, it can beassumed that the lengthXof a conference has a uniform distribution on the interval[0,4].(a) What is the probability density function?(b) What is the probability that any given conference lasts at least 3 hours?Prakash Chakraborty(PSU)Lecture 19-Fall 20243 / 30
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6.1 Basic properties of Uniform distributionUsually, a uniform distribution is used to represent a random variable that we know thelower(A)and upper bounds(B), but not much about which values are more likely.TheoremThe mean and variance of the uniform distribution areµ=A+B2,andσ2=(B−A)212.Prakash Chakraborty(PSU)Lecture 19-Fall 20244 / 30
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6.2 Normal distribution with parametersµ,σOne of the most important probability distributions.The graph of its pdf is called a normal curve.The shape of a normal curve is determined by its meanµand varianceσ2.DefinitionThe density of the normal random variableX, with meanµand varianceσ2, isf(x;µ, σ) =1√2πσe−(x−µ)22σ2,−∞<x<∞.Prakash Chakraborty(PSU)Lecture 19-Fall 20245 / 30
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6.2 Normal distribution: propertiesWe denote that random variableXfollows a normal distribution with meanµandvarianceσ2byX∼N(µ, σ2).The following are characteristics of a normal distribution.1The mode, the value ofxwheref(x)is at its maximum, occurs atx=µ.2The curve is symmetric around the mean,µ.3The normal curve approaches the horizontal axis asxapproaches∞or−∞.4The total area under the curve is equal to 1.5[Important]IfXis a normal random variable with meanµand varianceσ2,aX+bis a normal random variable with meanaµ+band variancea2σ2.Prakash Chakraborty(PSU)Lecture 19-Fall 20246 / 30
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6.2 Normal curves comparisonThevalue ofµdecides where thecenterof the distribution is located→because a normal distribution is symmetric around the mean.Thevalue ofσdecides howspreadout the distribution is.(a)Sameσ, butµ1̸=µ2(b)Sameµ, butσ1< σ2(c)µ1̸=µ2andσ1< σ2Figure 2:Normal CurvesPrakash Chakraborty(PSU)Lecture 19-Fall 20247 / 30
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6.3 Area under the normal curveSuppose we would like to computeP(x1<X<x2)whereX∼N(µ, σ2)P(x1<X<x2) =Px1−µσ<Z<x2−µσ,whereZ∼N(0,1).(Table A.3 in the textbook containsP(Z<z) =F(z)is given for different values ofz. )Prakash Chakraborty(PSU)Lecture 19-Fall 20248 / 30
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ExampleGiven the standard normal table, find the area under the curve that lies(a) to the right ofz=1.84 and(b) betweenz=−1.97 andz=0.86and findksuch that(c)P(Z>k) =0.3015 and(d)P(k<Z<−0.18) =0.4197.Prakash Chakraborty(PSU)Lecture 19-Fall 20249 / 30
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ExampleGiven a random variableXhaving a normal distribution withµ=50 andσ=10, findthe probability thatXassumes a value between 45 and 62.Prakash Chakraborty(PSU)Lecture 19-Fall 202410 / 30
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6.3 Using the normal curve in reverseSometimes we are given probabilitypand asked to findxthat satisfiesP(X<x) =p. Inthis case we can also use the table of standard normal distribution. Using the relationshipbetweenXandZ, if we findzsuch thatp=P(Z<z) =PX−µσ<z=P(X< σz+µ).Therefore,x=σz+µ.Prakash Chakraborty(PSU)Lecture 19-Fall 202411 / 30
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ExampleGiven a normal distribution with meanµ=40 andσ=6, find the value ofxthat has(a) 40.9%of the area to the left and(b) 14%of the area to the rightPrakash Chakraborty(PSU)Lecture 19-Fall 202412 / 30
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Using normal curve in reverse (2)Sometimes, you may not be able to findpyou are looking for in the standard normalCDF table. If that is the case, findp1andp2closest topsuch thatp1<p<p2and thecorrespondingz1andz2. Then, find theinterpolatedzvalue byz=z1×p−p1p2−p1+z2×p2−pp2−p1For instance, suppose you need to findzsuch thatP(Z<z) =0.51. Then,0.5080<0.51<0.5120 and the correspondingZ’s are 0.02 and 0.03. Therefore,z=0.02×0.51−0.50800.5120−0.5080+0.03×0.5120−0.510.5120−0.5080=0.025.Prakash Chakraborty(PSU)Lecture 19-Fall 202413 / 30
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6.4 Application of the normal distributionA typical iPhone battery lasts, on average 10 hours with a standard deviation of 2 hours.Assuming the battery life is normally distributed, find the probability that a given batterywill last less than 7.5 hours.Prakash Chakraborty(PSU)Lecture 19-Fall 202414 / 30
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6.4 ExampleA company pays its employees an average wage of 15.90 an hour with a standarddeviation of 1.5. If the wages are approximately normally distributed, find the minimumwage of the top 5%highest-paid employees.Prakash Chakraborty(PSU)Lecture 19-Fall 202415 / 30
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6.5 Normal approximation to the BinomialEarlier we discussed how a Poisson distribution can approximate a binomial distributionwith parameternandp. In fact, a binomial distribution can also be approximated by anormal distribution whennis large.TheoremIf X is a binomial random variable with meanµ=np and varianceσ2=np(1−p), thenthe distribution ofX−npqqqqqqqnp(1−p),becomes closer to that of the standard normal random variable Z∼N(0,1), as nincreases.This approximation works well whennis large andpis not too close to 0 or 1(works best whenp=0.5)Whennis not too large, we performcontinuity correctionby adding 0.5 tox:P(X≤x)≈PZ≤x+0.5−µσ.whereX∼Binomial(n,p), andµ=np,σ=qqqqqqqnp(1−p).Prakash Chakraborty(PSU)Lecture 19-Fall 202416 / 30
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ExampleThe probability that a patient recovers from a rare blood disease is 0.4. If 100 people areknown to have contracted this disease, what is the probability that fewer than 30 survive?Prakash Chakraborty(PSU)Lecture 19-Fall 202417 / 30
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6.6 Exponential distribution with parameterβPlays a key role in reliability engineering and designing service systems.DefinitionThe continuous random variableXhas an exponential distribution, with parameterβ, ifits density function is given byf(x;β) =1βe−x/β,x>0,0,elsewhere,whereβ >0.Basic Properties:Prakash Chakraborty(PSU)Lecture 19-Fall 202418 / 30
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6.6 Connection with PoissonThe exponential distribution is closely related to the Poisson distribution.Recall that the Poisson distribution represents the number of events happening in thettime window given the average number of events in unit time isλ. The pdf isp(x;λt) =e−λt(λt)xx!.The time between two events follow an exponential distribution withβ=1/λ.Prakash Chakraborty(PSU)Lecture 19-Fall 202419 / 30
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ExampleSuppose that the time until the engine failure of a certain model of a car is representedby an exponential random variableTwith mean time to failureβ=5 years.(a) What is the probability that an engine of the car model is functioning at the end ofyear 8?(b) Suppose a company bought 5 of these models 8 years ago. What is the probabilitythat at least 2 of them are functioning at the end of year 8?Prakash Chakraborty(PSU)Lecture 19-Fall 202420 / 30
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6.6 The memoryless propertyThe exponential distribution has its distinct statistical property known as thememoryless property.Suppose an electronic component where lifetime has an exponential distribution, theprobability that the component lasts forthours can be found byP(X≥t) =[[[[[[[∞tλe−λxdx=−e−λx∞t=e−λt.Suppose that the component has been working fort0time period, then the probability oflasting an additionalthours can be computed asP(X≥t+t0|X≥t0) =P(X≥t+t0)P(X≥t0)=e−λ(t+t0)e−λt0=e−λt.Prakash Chakraborty(PSU)Lecture 19-Fall 202421 / 30
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6.6 Gamma distribution with parametersαandβThe exponential distribution is a special case of a distribution known as the gammadistribution. The gamma distribution represents the time untilαnumber of Poissonevents happen.DefinitionThe continuous random variableXhas a gamma distribution, with parametersαandβ,if its density function is given byf(x;α, β) =1βαΓ(α)xα−1e−x/β,x>0,0,elsewhere,whereα >0, β >0, and the gamma functionΓ(α)is given byΓ(α) =[[[[[[[∞0xα−1e−xdx,forα >0.Prakash Chakraborty(PSU)Lecture 19-Fall 202422 / 30
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6.6 Basic properties of gamma distributionThe gamma distribution can be defined for anyα >0 (integer condition forαis notnecessary).Some properties of the gamma function:1Γ(α) = (α−1)Γ(α−1)for positive integerα.2Γ(α) = (α−1)!for a positive integerα.3Γ(1) =14Γ(1/2) =√π.Basic properties of the gamma distribution:Prakash Chakraborty(PSU)Lecture 19-Fall 202423 / 30
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ExampleSuppose that the number of calls a call center receives is a Poisson process with anaverage 5 calls a minute. What is the probability that up to a minute will elapse by thetime 2 calls have come in to the call center?Prakash Chakraborty(PSU)Lecture 19-Fall 202424 / 30
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6.6 Gamma distribution: Memoryless?The gamma distribution does not have the memoryless property. The only continuousdistribution that has the memoryless property is the exponential distribution.Although the gamma distribution is originated from the Poisson distribution, it has otherwide applications in biomedical science and reliability engineering to model thesurvivaltimedistribution.Prakash Chakraborty(PSU)Lecture 19-Fall 202425 / 30
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ExampleIt is known, from the market data, that the length of time in months between customercomplaints about a certain product is a gamma distribution withα=2 andβ=4.Changes were made to tighten quality control requirements. Following these changes, 20months passed before the first complaint. Does it appear as if the quality controltightening was effective?Prakash Chakraborty(PSU)Lecture 19-Fall 202426 / 30
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6.9 Lognormal distribution with parametersµ,σThe lognormal distribution has many applications in finance (e.g. return on financialinvestment).DefinitionThe continuous random variableXhas alognormal distributionif the random variableY= log(X)has a normal distribution with meanµand standard deviationσ. Theresulting density function ofXisf(x;µ, σ) =1√2πσxexp−(log(x)−µ)22σ2,x≥0,0,x<0.Basic Properties:Prakash Chakraborty(PSU)Lecture 19-Fall 202427 / 30
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ExampleConcentrations of pollutants produced by chemical plants is under governmentregulations. Suppose the concentration of a pollutant in parts per million has a lognormaldistribution with parametersµ=3.2 andσ=1. What is the probability that theconcentration exceeds 8 parts per million?Prakash Chakraborty(PSU)Lecture 19-Fall 202428 / 30
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6.7 Chi-squared distribution with parameterνThe chi-square distribution is another special case of the gamma distribution by selectingα=ν/2 andβ=2, whereνis a positive integer. Sinceβis fixed, the chi-squareddistribution has only one parameter,ν, called thedegrees of freedom.DefinitionThe continuous random variableXhas achi-squared distribution, withνdegrees offreedom, if its density function is given byf(x;ν) =12ν/2Γ(ν/2)xν/2−1e−x/2,x>0,0,elsewhere,whereνis a positive integer.Basic Properties:Prakash Chakraborty(PSU)Lecture 19-Fall 202429 / 30
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6.7 Relation to normalThe chi-squared distribution is closely related to the normal distribution. It is known thatifX1,X2, . . . ,XνhaveN(0,1)distribution, thenX21+X22+. . .+X2ν∼χ2(ν),whereχ2(ν)represents the chi-squared distribution withνdegrees of freedom. This iswhyνis called the degrees of freedom parameter.Prakash Chakraborty(PSU)Lecture 19-Fall 202430 / 30
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