Random variablesContinuous random variablesJuana [email protected] Department of StatisticsJuly 19, 2013J. Sanchez Random Variables-ContinuousAnnouncementsHomework 4 is due on Tuesday. It was posted yesterday in thehomework folder in CCLE.Quiz 3 will be taken in the TA session on Tuesday.Continuous random variables are in Chapter 5 of Ross (8th and 9thedition)J. Sanchez Random Variables-ContinuousTodayI. Introduction to Continuous Random Variables.II. Continuous random variables and their density function.Computing probabilities and percentiles.III. Computing probabilities with R, generating random numbers withR, finding percentiles with R (will be done next week in the TAsession).J. Sanchez Random Variables-ContinuousI. IntroductionThere are random variables whose set of possible values isuncountable. For example, the time that a train arrives at aspecified stop or the lifetime of a transistor.For these random variables, probabilities of events are areas under acontinuous density function and are computed using integration).J. Sanchez Random Variables-ContinuousNotesNotesNotesNotesII. Continuous random variables and their density function(pdf)Definition of continuous random variableX is a continuous random variable if there exists a nonnegative functionf, defined for all real X ∈ (−∞, ∞), having the property that for any setB of real numbers (event B),P{X ∈ B} =ZBf (x)dxThis is true for measurable sets B, which include most sets of practicalinterest.The function f is called the probability density function (pdf) of therandom variable X. Since X must assume some value, f must satisfy1 = P {X ∈ (−∞, ∞)} =Z∞−∞f (x)dxJ. Sanchez Random Variables-ContinuousComputing probabilitiesAll probability statements about X can be answered in terms of f . Forinstance, letting B = [a, b] we obtain from the definition thatP(B ) = P (a ≤ X ≤ b) =Zbaf (x)dxIf we let a = b thenP(X = a) =Zaaf (x)dx = 0. This implies that when we are computing probabilities for continuousrandom variables,P(B ) = P (a ≤ X ≤ b) =Zbaf (x)dx = P (a < X < b) =Zbaf (x)dxIn this case, X is said to be a continuous random variable.J. Sanchez Random Variables-ContinuousNotice...f (x) ≥ 0R∞−∞f (x)dx = 1 (the area under the whole curve in the range forwhich the X is defined is 1). Note that this means that if, forexample, the random variable is defined for 0 ≤ X ≤ 4, the areaunder the curve in the range 0 to 4 is 1. You don’t have to integratefrom −∞ to ∞ all pdf’s. You must pay attention to where X isdifferent from 0.J. Sanchez Random Variables-ContinuousFinding percentilesExampleLet X be a continuous random variable with the following pdff (x) =12x ; 0 ≤ x ≤ 2(a) P (1 ≤ X ≤ 1.5) =R1.5112xdx =x24|1.51= 5/16(b) Find c such thatRc012xdx = 0.5 This is what we call finding the 50thpercentile.x24|c0= 0.5 →c24= 0.5 → c2= 2 → c =√2 = 1.414(c) Could you please find the 90th percentile?x24|c0= 0.9 →c24= 0.9 → c2= 0.9(4) = 3.6 → c =√3.6 = 1.8973J. Sanchez Random Variables-ContinuousNotesNotesNotesNotesExampleLet X be a continuous random variable (r.v.) with the following pdf.f (x) =16x + k 0 ≤ x ≤ 3I want to find P (1/2 < X < 1.2) How do I do that?First find k using the property that the integral under a density curvemust be one.R30(16x + k )dx =hx212+ kxi30= 1→912+ 3k = 1 → k =112Now find the probability we wantP(1/2 < X < 1.2) =R1.21/216x + 1/12dx = 0.1575J. Sanchez Random Variables-ContinuousExampleLet X be a continuous random variable (r.v.) with the following pdf.f (x) =16x + k 0 ≤ x ≤ 3I want to find P (1/2 < X < 1.2) How do I do that?First find k using the property that the integral under a density curvemust be one.R30(16x + k )dx =hx212+ kxi30= 1→912+ 3k = 1 → k =112Now find the probability we wantP(1/2 < X < 1.2) =R1.21/216x + 1/12dx = 0.1575J. Sanchez Random Variables-ContinuousExampleSuppose that X is a continuous random variable whose probability densityfunction is given byf (x) = c (4x − 2x2) 0 ≤ X ≤ 2and 0 elsewhere. The c is a constant.(a) What is the value of c?(b) Find P (X > 1).J. Sanchez Random Variables-ContinuousExampleSuppose that X is a continuous random variable whose probability densityfunction is given byf (x) = c (4x − 2x2) 0 ≤ X ≤ 2and 0 elsewhere. The c is a constant.(a) What is the value of c?cZ20(4x − 2x2)dx = 1c83= 1c =38(b) Find P (X > 1).P(X > 1) =Z21(4x − 2x2)dx = 1/2J. Sanchez Random Variables-ContinuousNotesNotesNotesNotesExampleThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) = λe−1100xx ≥ 0This is the exponential random variable.(a) Find P (50 ≤ X ≤ 150)(b) Find Prob(X < 30).J. Sanchez Random Variables-ContinuousExample(a) Find P (50 ≤ X ≤ 150)Since by the definition of pdf,1 =Z∞0λe−x100dxthen1 = 100λandλ = 1/100(b) Find Prob(X < 30).Prob(x < 30) =Z3001100e−1100xdx = −e−1100x|300= 0.2591J. Sanchez Random Variables-ContinuousIII. Expected Value, Variance and Standard Deviation of aContinuous random variableThe expected value of a random variable is the center of gravity of thedensity function. It is representative of the average of the distributiononly if there are no extreme values skewed to one side of the distribution.Expected Value of XIf X is a random variable defined in the interval (a,b),µ = E (X ) =Rbaxf (x )dxThe variance tells us the distance of the values of x from the expected value,but squared distance.Variance of Xσ2x= Var (x) =Rba(x − µ)2f (x )dxThe standard deviation corrects the squaredness of the variance. We use it todescribe variability of the r.v.Standard Deviation of Xσx=√σ2J. Sanchez Random Variables-ContinuousExampleLet the pdf of X bef (x) = 2x 0 ≤ x ≤ 1and 0 otherwise.E (X ) =Z10x 2xdx =13E (X2) =Z10x22xdx =12σ2x=12−232=118J. Sanchez Random Variables-ContinuousNotesNotesNotesNotesExampleFor a lathe machine shop, let X denote the percentage of time out of a40-hour work week during which the lathe is actually in use. Supposethat X has a pdf given byf (x) = 3x20 ≤ X ≤ 1On average, how long is the lathe used during a week? By how much canwe expect the time of use to fluctuate around that mean?µx= E (X ) =R10x 3x2dx =R103x3dx = 3hx44i10= 0.75 or 30 hoursσ2x=R10(x − 0.75)23x2dx =380σx=q380= 0.1936 or 7.744 hoursSo,on average, the lathe is used 30 hours a week, give or takeapproximately 8 hours.J. Sanchez Random
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