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Probability density y 0 μ x Content Content Continuous Continuous random random variable variable Probability Probability density density function function Expectation Expectation and and variance variance Normal Normal distribution distribution Characteristics Characteristics of of the the normal normal distribution distribution Applications Applications of of the the normal normal distribution distribution Continuous Continuous Random Random Variable Variable AArandom randomvariable variablewhich whichcan cantake takeon onany anyvalue valueover overan an interval intervalof ofreal realnumbers numbersdefined definedby byits itssample samplespace. space. Examples Examples The Thelifetime lifetimeof ofaacertain certainbrand brandof ofbatteries batteries The Theheights heightsof ofboys boysof ofaaparticular particularage age The Theweights weightsof ofboxes boxesof ofaacertain certainbrand brandof ofeggs eggs The Thevolume volumeof ofmilk milkpacked packedin inboxes boxes Activity Activity 11 1. 1. 2. 2. 3. 3. 4. 4. Identify Identifywhich whichof ofthe thefollowing followingis/are is/arecontinuous continuousrandom random variables. variables. The Thenumber numberof ofgoals goalsscored scoredin inaafootball footballmatch. match. The Thewaiting waitingtime timefor foraabus bus by bypassengers passengersat ataabus busstop. stop. The Theposition positionwhere wherethe theroulette roulettewheel wheelstops. stops.(Fig. (Fig.1) 1) The Thesector sectornumber numberat atwhich whichthe theroulette roulettewheel wheelstops. stops.(Fig. (Fig. 2) 2) 1 Fig. 1 2 Fig. 2 4 3 Probability p.d.f.) Probability density density function function ((p.d.f.) The Theprobability probabilitydistribution distributionof ofaacontinuous continuousrandom randomvariable variable XX can canbe bedescribed describedby byits itsp.d.f.. p.d.f.. IfIfthere thereisisaareal-valued real-valuedfunction function ffsuch suchthat thatfor forevery everyinterval interval AA of ofreal realnumbers, numbers,the theprobability probabilitythat that XXtakes takesaavalue valuein in AAisisthe the integral integralof offfover overA, A,then thenffisiscalled calledthe thep.d.f. p.d.f.of ofX. X. Probability p.d.f.) Probability density density function function ((p.d.f.) For Foran aninterval intervalaa≤≤xx≤≤b, b,the theprobability probabilitythat thatXXtakes takeson onaavalue valuein inthe the interval intervalisisgiven givenby by b Probability density y P( a ≤ x ≤ b) = ∫ a f ( x) dx Note. 0 y = f (x) a b x 1. P( X = a ) = P( X = b) = 0 2. P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b) Conditions Conditions of of p.d.f p.d.f.. 1. 1. ff (x) (x) ≥≥ 00 for for all all real real numbers numbers x. x. 2. 2. ∞ ∫ − ∞ f ( x ) dx = 1 Learning Learning Difficulty Difficulty 11 Given X is a continuous random Given T is a discrete random variable and the probability variable and the probability density function of X is given by function of T is given by f ( x) = 2, 0.5 ≤ x ≤ 1 g (t ) = 0.2, for t = 1, 2, 3, 4, 5 y 0.2 f (x) = 2 2 Probability Probability density y 1.5 1 0.1 0.5 0 0.5 1 x 0 t 1 2 3 4 5 Probability p.d.f.) Probability Density Density Function Function ((p.d.f.) The Thefrequency frequencydistribution distributionof ofaacontinuous continuousvariable variablecan canbe be represented representedgraphically graphicallyby byaahistogram. histogram. Frequency density The frequency of each class is represented by the area of each rectangle. 0 f3 f2 f4 f1 a f5 h k f6 b x the class frequency = the frequency density of the class × class width the class frequency frequency density of a class = the class width When Wheneach eachclass classfrequency frequency(f(fi)i)isisdivided dividedby bythe thetotal totalfrequency frequency (N), (N),the thehistogram histogramshows showsthe therelative relativefrequency frequencydistribution distributionof of Relative frequency density the thecontinuous continuousrandom randomvariable. variable. 0 f3/N f2/N f4/N f1/N a f5/N h k f6/N b x As AsNNbecomes becomesvery verylarge, large, fi N = pi where theprobability probabilityof ofthe theoccurrence occurrenceof ofclass classi ioutcome(s) outcome(s)of of whereppi iisisthe the thecontinuous continuousrandom randomvariable. variable. The Thehistogram histogramthen thenshows showsthe theprobability probabilitydistribution distributionof ofthe the Probability density continuous continuousrandom randomvariable. variable. 0 p2 p3 p4 p1 a p5 h k p6 b x IfIfthe thewhole wholerange rangeof ofxxvalues valuesisisdivided dividedinto intoan aninfinitely infinitelylarge large number number(n) (n)of ofsmall smallintervals, intervals,the theprobability probabilitydistribution distributionof ofthe the continuous continuousrandom randomvariable variablecan canbe beshown shownby byaagraph graphconsisting consisting Probability density of ofaacontinuous continuouscurve curveabove abovethe the x-axis. x-axis. 0 a h k b x Learning Learning Difficulty Difficulty 22 A roulette wheel is rotated about O from A in an anticlockwise direction and stops at B. X is a random variable representing the length of arc AB measured from A to B in clockwise direction. P( X = l ) = 0 l A B O r Learning Learning Difficulty Difficulty 22 Given Given XXisisaacontinuous continuousrandom randomvariable, variable,ff(x) (x)isisthe thep.d.f. p.d.f.of ofXX and andccisisaareal realconstant. constant. 0 ≤ P( X = c) ≤ P(c ≤ X ≤ c + Δx) c + Δx 0 ≤ P( X = c) ≤ ∫ c f ( x ) dx Due Dueto toboundedness boundednessof ofthe theintegratible integratiblefunction, function, c + Δx ∫ Δx →0 c 0 ≤ P( X = c) ≤ lim f ( x ) dx = 0 Conclusion. The event impossible to occur ⇒ P(Event) = 0 P(Event) = 0 ⇒ Event impossible to occur ? Expectation Expectation and and Variance Variance IfIf XX is is aa discrete discrete random random IfIf XX is is aa continuous continuous variable, random variable, random variable, variable, ∞ E( X ) = ∫ x f ( x) dx E( X ) = ∑ x f ( x) −∞ x ∞ Var( X ) = ∑ ( x − μ ) f ( x) Var( X ) = ∫ ( x − μ ) f ( x) dx 2 x Var( X ) = ∑ x f ( x) − μ 2 x Var( X ) = E ( X 2 ) − μ 2 2 −∞ ∞ 2 = ∫ x f ( x ) dx − μ 2 −∞ Var( X ) = E ( X 2 ) − μ 2 2 Expectation Expectation and and Variance Variance The following properties hold E(a X + b) = a E( X ) + b Var (a X + b) = a 2 Var ( X ) Normal Normal Distribution Distribution AAcontinuous continuousrandom randomvariable variable XXhas hasaanormal normaldistribution distributionififthe the p.d.f. p.d.f.of ofXXisis 1 f ( x) = σ 2π ( x−μ )2 − 2 2 σ e , −∞ < x < ∞ where andσσ are areconstants constantsand andσσ>>0. 0. where μμand ff(x) andσ. σ. (x)has hastwo twoparameters parameters μμand The Thenormal normaldistribution distributionwith withthese theseparameters parameterscan canbe bedenoted denoted 22). by N( μ , σ by N(μ, σ ). Importance Importance of of the the Normal Normal Distribution Distribution 1. 1. The Thenormal normaldistribution distributioncan canbe beused usedas asan anapproximation approximationto to some somediscrete discreteprobability probabilitydistributions distributionssuch suchas asbinomial binomial distribution distributionand andPoisson Poissondistribution. distribution. Example Example An Anunbiased unbiasedtetrahedral tetrahedraldie diewith withface facemarked marked1, 1,2, 2,33and and44isis thrown thrown400 400times. times. Find Findthe theprobability probabilityof ofobtaining obtainingless lessthan than90 90 ones. ones. Let LetXXbe beaarandom randomvariable variablerepresenting representingthe thenumber numberof ofones ones obtained obtainedin in400 400throws. throws. 1 Then Then X ~ Bin (400, ) 4 Importance Importance of of the the Normal Normal Distribution Distribution Method Method11 Using Usingthe thebinomial binomialdistribution, distribution, P(X P(X<<90) 90)==0.1117 0.1117 Method Method22 Using Usingthe thenormal normalapproximation approximationto tothe thebinomial binomialdistribution, distribution, 1 X ~ Bin (400, ) 4 X ~ N(100, 75) 89.5 - 100 P( X < 90) = P( X < 89.5) = P(Z < ) 75 P( Z < −1.212) = 0.1127 Importance Importance of of the the Normal Normal Distribution Distribution 2. 2. Central CentralLimit LimitTheorem Theorem The Thesample samplemean meanof ofaalarge largerandom randomsample sampleof ofrandom randomvariables variables 2 with andfinite finitevariance varianceσσ2has hasapproximately approximatelyaanormal normal withmean meanμμand 2 distribution distributionwith withmean meanμμand andvariance varianceσσ2//n. n. The Theproof proofcan canbe bedone doneby byeither eitherusing usingmoment momentgenerating generatingfunction function or orprobability probabilitydistribution distributionfunction. function. The Thedetails detailsof ofthe theproofs proofscan canbe be seen seenon on Alexander AlexanderM. M.Mood, Mood,Franklin FranklinA. A.Graybill Graybilland andDuane DuaneC. C.Boes Boes(1974). (1974). rdrdEdition) (pp.234-235). Introduction to the Theory of Statistics (3 Introduction to the Theory of Statistics (3 Edition) (pp.234-235). Mc-Graw-Hill. Mc-Graw-Hill. 第137頁。劉次華、萬建平 第137頁。劉次華、萬建平(1999)。概率論與數理統計。北京:施普 (1999)。概率論與數理統計。北京:施普 林格出版社。 林格出版社。 Activity Activity 22 –– Normal Normal Distribution Distribution Example Example11––Are Areyou youfit? fit? Source Source http://www.hked-stat.net/common/activity_as_2.htm http://www.hked-stat.net/common/activity_as_2.htm Worksheet Worksheet Word Worddocument documentversion version pdf pdfversion version BMI BMICalculator Calculator http://www.ryvita.co.uk/calculatebmi.html http://www.ryvita.co.uk/calculatebmi.html Activity Activity 22 –– Normal Normal Distribution Distribution Example Example22––Pieces Piecesof ofstring string Frequency Look Lookat at10 10cm cmon onaaruler rulerand andthen thentake takeaaball ballof ofstring. string. Every Every student studentcuts cutsaapiece pieceof ofstring stringof of10 10cm cmfrom fromthe theball ballof ofstring stringby by guessing. guessing. Measuring Measuringthe thelengths lengthsof ofall allthe thepieces piecesin inmm. mm. 0 1 2 3 4 5 6 7 8 9 10 11 cm 0 Length Notes Notes on on Activity Activity 22 The Thefollowing followingissues issuescan canbe bediscussed discussedafter afterthe theactivity. activity. 1. 1. Describe Describethe thedistribution distributionof ofdata. data. 2. 2. Is Isititsymmetrical symmetricalabout aboutthe themean? mean? IfIfnot, not,are arethere thereany any factors factorscontributing contributingto tothe theshape shapeof ofthe thedistribution? distribution? 3. 3. Can Canwe wemanipulate manipulatethe thedata dataso sothat thatthese thesefactors factorscan canbe be removed removedafter aftermanipulation? manipulation?IfIfyes, yes,how? how?IfIfnot, not,why? why? 4. 4. Describe Describethe thedistribution distributionof ofthe themanipulated manipulateddata. data. History History of of the the Normal Normal Distribution Distribution Abraham De Moivre (1667 – 1754) The normal curve was developed mathematically in 1733 by De Moivre as an approximation to the binomial distribution. His paper was not discovered until 1924 by Karl Pearson. Pierre Simon Laplace (1749 – 1827) Laplace used the normal curve in 1783 to describe the distribution of errors. In 1810, he proved the Central Limit Theorem. History History of of the the Normal Normal Distribution Distribution Johann Carl Friedrich Gauss (1777 – 1855) Gauss used the normal curve to analyze astronomical data in 1809. The normal curve is often called the Gaussian distribution. In Germany, the portrait of Gauss, the normal curve and its p.d.f. have been put on their 10 Deutschmark Note. Properties Properties of of the the Normal Normal Distribution Distribution 1. 1. ff (x) (x) >> 00 for for all all values values of of x. x. 2. 2. ∞ ∫ − ∞ f ( x ) dx = 1 3. 3. The The parameters parameters μμ and and σσ of of the the normal normal distribution distribution are are the the mean mean and and the the standard standard deviation deviation of of the the distribution distribution respectively. respectively. Expected Value of the Normal Distribution A random variable X follows a normal distribution with mean and variance 2, find E(X). f ( x) 1 Let z E ( x) x 2 e 2 E( X ) E ( x) 2 2 , ∞ x ∞ x f ( x ) dx x 1 2 1 2 , we get ∞ ∫ (z )e ∞ ∞ ∫ ze ∞ z2 2 z2 2 dz dz 1 2 z2 ∞ 2 ∫e ∞ dz 2 2 Variance Variance of of the the Normal Normal Distribution Distribution AArandom randomvariable variableXXfollows followsaanormal normaldistribution distributionwith withmean meanμμand andvariance variance σσ22,,find findE(X). E(X). Var( X ) = ∫ Let z = ∞ 2 ( x − μ) −∞ x−μ , we get σ σ2 Var ( X ) = 2π σ2 Var ( X ) = 2π Var ( X ) = σ 2 1 e σ 2π ∫ z2 − ∞ 2 2 z e −∞ 2π = σ 2 dz ( x − μ )2 − 2σ 2 dx Properties Properties of of the the Normal Normal Distribution Distribution Normal curve is symmetrical - two halves identical - Tail Tail Theoretically, the curve extends to -∞ Mean, median, and mode are equal Theoretically, the curve extends to +∞ Properties Properties of of the the Normal Normal Distribution Distribution The The location location of of the the normal normal distribution distribution is is determined determined by by μμ .. probability density Normal Normaldistributions distributionswith withdifferent different means meansbut butequal equalstandard standarddeviation. deviation. X1 ~ N(μ1, σ2) X2 ~ N(μ2, σ2) X3 ~ N(μ3, σ2) μμ22>>μμ11>>μμ33 μ3 μ1 μ2 x Properties Properties of of the the Normal Normal Distribution Distribution The The dispersion dispersion of of the the normal normal distribution distribution is is determined determined by by σσ.. Normal Normaldistributions distributionswith withequal equalmean mean but butdifferent differentstandard standarddeviations. deviations. probability density X1 ~ N(μ, σ12) X2 ~ N(μ, σ22) X3 ~ N(μ, σ32) μ x σσ33>>σσ22>>σσ11>>00 Properties Properties of of the the Normal Normal Distribution Distribution X ~ N(μ, σ2) 68% 95% 99.7% μ −3σ μ − 2σ μ−σ μ μ+σ μ + 2σ μ + 3σ Standardization Standardization of of aa Normal Normal Variable Variable The Thestandard standardnormal normaldistribution distributionisisaanormal normaldistribution distributionwith with mean mean00and andvariance variance1. 1. Standardization Standardization of of aa Normal Normal Variable Variable The Thestandard standardnormal normaldistribution distributionisisaanormal normaldistribution distributionwith with mean mean00and andvariance variance1. 1. X −μ 2 2 ,,then IfIfXX~~NN((μμ,,σσ ))and and Z = then σ E( Z ) = E( E( Z ) = E( Z ) = 1 σ 1 σ E( Z ) = 0 X −μ σ ) [E( X ) − E(μ )] (μ − μ ) Var ( Z ) = Var ( Var(Z ) = Var ( Z ) = 1 σ 2 1 X −μ [Var( X − μ )] Var( X ) σ σ2 Var(Z ) = 2 = 1 σ 2 σ ) Example Example The life time of the batteries for a notebook computer under normal usage is normally distributed with mean 210 minutes and standard deviation 15 minutes. a) What percentage of these batteries will have a life time between 195 minutes and 240 minutes? b) What percentage of batteries will have a life time less than 180 minutes? (a) (a) Let LetXXbe bethe thelifetime lifetimeof ofthe thebatteries batteriesfor foraanotebook notebook computer. computer. 2 XX~~NN(210, (210,15 152)) P(195 P(195≤≤XX≤≤240) 240)==?? x 150 165 180 195 210 225 240 255 270 Table Table of of the the Standard Standard Normal Normal Distribution Distribution Area Area under under the the Standard Standard Normal Normal Curve Curve zz 0.00 0.00 0.01 0.01 0.02 0.02 0.03 0.03 0.04 0.04 0.05 0.05 0.06 0.06 0.07 0.07 0.08 0.08 0.09 0.09 0.0 0.0 0.1 0.1 0.2 0.2 0.3 0.3 0.4 0.4 0.0000 0.0000 0.0398 0.0398 0.0793 0.0793 0.1179 0.1179 0.1554 0.1554 0.0040 0.0040 0.0438 0.0438 0.0832 0.0832 0.1217 0.1217 0.1591 0.1591 0.0080 0.0080 0.0478 0.0478 0.0871 0.0871 0.1255 0.1255 0.1628 0.1628 0.0120 0.0120 0.0517 0.0517 0.0910 0.0910 0.1293 0.1293 0.1664 0.1664 0.0160 0.0160 0.0557 0.0557 0.0948 0.0948 0.1331 0.1331 0.1700 0.1700 0.0199 0.0199 0.0596 0.0596 0.0987 0.0987 0.1368 0.1368 0.1736 0.1736 0.0239 0.0239 0.0636 0.0636 0.1026 0.1026 0.1406 0.1406 0.1772 0.1772 0.0279 0.0279 0.0675 0.0675 0.1064 0.1064 0.1443 0.1443 0.1808 0.1808 0.0319 0.0319 0.0714 0.0714 0.1103 0.1103 0.1480 0.1480 0.1844 0.1844 0.0359 0.0359 0.0753 0.0753 0.1141 0.1141 0.1517 0.1517 0.1879 0.1879 0.5 0.5 0.6 0.6 0.7 0.7 0.8 0.8 0.9 0.9 0.1915 0.1915 0.2257 0.2257 0.2580 0.2580 0.2881 0.2881 0.3159 0.3159 0.1950 0.1950 0.2291 0.2291 0.2611 0.2611 0.2910 0.2910 0.3186 0.3186 0.1985 0.1985 0.2324 0.2324 0.2642 0.2642 0.2939 0.2939 0.3212 0.3212 0.2019 0.2019 0.2357 0.2357 0.2673 0.2673 0.2967 0.2967 0.3238 0.3238 0.2054 0.2054 0.2389 0.2389 0.2704 0.2704 0.2995 0.2995 0.3264 0.3264 0.2088 0.2088 0.2422 0.2422 0.2734 0.2734 0.3023 0.3023 0.3289 0.3289 P(0 ≤ z ≤ 1.00) 0.2123 0.2123 0.2454 0.2454 0.2764 0.2764 0.3051 0.3051 0.3315 0.3315 0.2157 0.2157 0.2486 0.2486 0.2794 0.2794 0.3078 0.3078 0.3340 0.3340 0.2190 0.2190 0.2517 0.2517 0.2823 0.2823 0.3106 0.3106 0.3365 0.3365 0.2224 0.2224 0.2549 0.2549 0.2852 0.2852 0.3133 0.3133 0.3389 0.3389 1.0 1.0 1.1 1.1 1.2 1.2 1.3 1.3 1.4 1.4 0.3413 0.3413 0.3643 0.3643 0.3849 0.3849 0.4032 0.4032 0.4192 0.4192 0.3438 0.3438 0.3665 0.3665 0.3869 0.3869 0.4049 0.4049 0.4207 0.4207 0.3461 0.3461 0.3686 0.3686 0.3888 0.3888 0.4066 0.4066 0.4222 0.4222 0.3485 0.3485 0.3708 0.3708 0.3907 0.3907 0.4082 0.4082 0.4236 0.4236 0.3508 0.3508 0.3729 0.3729 0.3925 0.3925 0.4099 0.4099 0.4251 0.4251 0.3531 0.3531 0.3749 0.3749 0.3944 0.3944 0.4115 0.4115 0.4265 0.4265 0.3554 0.3554 0.3770 0.3770 0.3962 0.3962 0.4131 0.4131 0.4279 0.4279 0.3577 0.3577 0.3790 0.3790 0.3980 0.3980 0.4147 0.4147 0.4292 0.4292 0.3599 0.3599 0.3810 0.3810 0.3997 0.3997 0.4162 0.4162 0.4306 0.4306 0.3621 0.3621 0.3830 0.3830 0.4015 0.4015 0.4177 0.4177 0.4319 0.4319 1.5 1.5 1.6 1.6 1.7 1.7 1.8 1.8 0.4332 0.4332 0.4452 0.4452 0.4554 0.4554 0.4641 0.4641 0.4345 0.4345 0.4463 0.4463 0.4564 0.4564 0.4649 0.4649 0.4357 0.4357 0.4474 0.4474 0.4573 0.4573 0.4656 0.4656 0.4370 0.4370 0.4484 0.4484 0.4582 0.4582 0.4664 0.4664 0.4382 0.4382 0.4495 0.4495 0.4591 0.4591 0.4671 0.4671 0.4394 0.4394 0.4505 0.4505 0.4599 0.4599 0.4678 0.4678 0.4406 0.4406 0.4515 0.4515 0.4608 0.4608 0.4686 0.4686 0.4418 0.4418 0.4525 0.4525 0.4616 0.4616 0.4693 0.4693 0.4429 0.4429 0.4535 0.4535 0.4625 0.4625 0.4699 0.4699 0.4441 0.4441 0.4545 0.4545 0.4633 0.4633 0.4706 0.4706 X − 210 Let Let Z = 15 P(195 P(195≤≤XX≤≤240) 240) == P(-1 P(-1≤≤ZZ≤≤2) 2) == 0.3413 0.3413++0.4772 0.4772 == 0.8185 0.8185 x 150 165 180 195 210 225 240 255 270 z -4 -3 -2 -1 0 1 2 3 4 (a) (a) Let LetXXbe bethe thelifetime lifetimeof ofthe thebatteries batteriesof ofaanotebook notebook computer. computer. 22) XX~~NN(210, 15 (210, 15 ) X − 210 Let Let Z = 15 P(195 P(195≤≤XX≤≤240) 240) == P(-1 P(-1≤≤ZZ≤≤2) 2) == 0.3413 0.3413++0.4772 0.4772 == 0.8185 0.8185 (b) (b) P(X P(X≤≤180) 180)==?? x 150 165 180 195 210 225 240 255 270 References References rdrd DeGroot, M.H., & Schervish, M.J. (2002). Probability and statistics (3 DeGroot, M.H., & Schervish, M.J. (2002). Probability and statistics (3 ed.). ed.). New NewYork: York:Addison-Wesley. Addison-Wesley. Hald, Hald,A. A.(1990). (1990). AAhistory historyof ofprobability probabilityand andstatistics statisticsand andtheir their applications applicationsbefore before1750. 1750. New NewYork: York:John JohnWiley Wiley&&Sons. Sons. James JamesW. W.Tankard, Tankard,Jr. Jr. The TheStatistical StatisticalPioneers. Pioneers. Massachusetts: Massachusetts: Schenkman SchenkmanPublishing PublishingCompany, Company,Inc. Inc. 劉次華、萬建平 劉次華、萬建平(1999)。概率論與數理統計。北京:施普林格出版社。 (1999)。概率論與數理統計。北京:施普林格出版社。 Normal NormalDistribution Distribution http://www.stat.wvu.edu/SRS/Modules/Normal/normal.html http://www.stat.wvu.edu/SRS/Modules/Normal/normal.html The TheMacTutor MacTutorHistory Historyof ofMathematics Mathematicsarchive archive http://turnbull.mcs.st-and.ac.uk/~history/ http://turnbull.mcs.st-and.ac.uk/~history/ References References LiLiKam-yuk Kam-yukand andChow ChowWai-keung Wai-keung(2002). (2002). New NewWay WayMathematics Mathematics&& Statistics Statisticsfor forHong HongKong KongAS-Level AS-Level(2nd (2ndEdition). Edition). Hong HongKong: Kong:SNP SNP MANHATTAN MANHATTANPRESS PRESS(H.K.) (H.K.)LTD. LTD.