Random variablesContinuous random variablesJuana [email protected] Department of StatisticsJ. Sanchez Random Variables-ContinuousThis LectureI. Introduction to Continuous Random Variables.II. Computing probabilities and percentiles.III. Expected Value, Variance and Standard DeviationIV. Cumulative Distribution FunctionJ. 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-ContinuousContinuous random variables and their density function(p df)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-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-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-ContinuousExampleLet 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-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) =Z2138(4x − 2x2)dx = 1/2J. Sanchez Random Variables-ContinuousPercentilesDefinitionThe 100 xqth percentile of a random variable X is the value c of therandom variable such thatP(X ≤ c) =Rc−∞f (x)dxWhere it says −∞ you put the lowest value of X for which f (X ) isdefined.There are some percentiles that are particularly important in Statistics(a) 50th percentile or medianRc−∞f (x)dx = 0.5. Solve for c .(b) 25th percentile or lower quartileRc−∞f (x)dx = 0.25. Solve for c .(c) 75th percentile or third quartileRc−∞f (x)dx = 0.75. Solve for c .The difference between the third and first quartiles is called theInterquartile range. IT tells us how spread out is the middle 50% area ofthe distribution.J. Sanchez Random Variables-ContinuousExampleThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) =1100e−1100xdx , x ≥ 0This is the exponential random variable.Find the 90th percentile and the Interquartile Range.90th percentileRc01100e−1100xdx = 0.9. Solve for c.1−e−1100c= 0.9 → −1100c = log 0.1 → c = −100 log(0.1) = 230.25825th percentile or lower quartileRc01100e−1100xdx = 0.25. Solve for c .1 −e−1100c= 0.25 → −1100c = log 0.75 → c = −100 log(0.75) =28.768J. Sanchez Random Variables-Continuous75th percentile or third quartileRc01100e−1100xdx = 0.75. Solve for c .1 −e−1100c= 0.75 → −1100c = log 0.25 → c = −100 log(0.25) =138.628So IQR= 138.628-28.768=109.86J. Sanchez Random Variables-ContinuousExampleThe amount of time, in hours, that a computer functions before breakingdown is a continuous r.v. with pdff (x) =1100e−1100xx ≥ 0This is the exponential random variable.(a) Find P (50 ≤ X ≤ 150)(b) Find Prob(X < 30).J. Sanchez Random Variables-ContinuousI II. 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 =23E (X2) =Z10x22xdx =12σ2x=12−232=118J. Sanchez Random Variables-ContinuousExampleFor 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
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