统计学专业英语教程王忠玉课后部分参考答案

AnswerstoSelectedProblems4.2SolutionTheoutcomesin(a)areallsequencesHHH...H,THH...H,...,TTT...Tformedby2nsubsequentlettersHorT(or,0and1).Thetotalnumberofoutcomesism=22n,eachcarriesprobability1/22n.WearelookingforoutcomeswherethenumberofHsequalsthatofTs.Thenumberkofsuchoutcomesis(2n)!/n!n!(thenumberofwaystochoosepositionsfornHsamong2nplacesavailableinthesequence).Theprobabilityinquestionis.In(b),theoutcomesarethesequencesoflength2n+1,22n+1intotal.Theprobabilityequals4.3SolutionThetotalnumberofoutcomesis365n.In(a),thenumberofoutcomesnotintheeventis365×364×…×(365n+1).So,theprobabilitythatallbirthdaysaredistinctis(365×364×…×(365n+1))/365nandthattwoormorepeoplehavethesamebirthday.Forn=22:.andforn=23:.In(b),thenumberofoutcomesnotintheeventis364nandtheprobabilityinquestion1(364/365)n.Wewantittobenear1/2,so.4.4SolutionThenumberofwaysofbeingdealt2acesfrom4is.Andthenumberofwaysofbeingdealt3jacksfrom4is.BythemultiplicationruleofTheorem2.1,therearen=(6)(4)=24handswith2acesand3jacks.Thetotalnumberof5-cardpokerhands,allofwhichareequallylikely,is.Therefore,theprobabilityofeventCofgetting2acesand3jacksina5-cardpokerhandis.5.1SolutionWedenotethenumberofgamesPaulwinsoutofthefirstfivegamesbyX1，andthenumberoutofthesecondfive(ifplayed)byX2.ThenX1～B,andX2willhavethesamedistribution.(a)WerequirePr(X1≥2),andthisismostsimplyevaluatedas(b)Wenowrequiretheprobabilitythattengamesareplayed,andthatPaulwinsatleasttwo;wedenotethiseventbyS.Fromtheconditionsstated,X1mustbe0or1,andX1+X2mustbetwoormore.WethusobtainBytheadditionlawofprobabilityformutuallyexclusiveevents.Noweachoftheseeventsisitselftheintersectionoftwoindependentevent,towhichthemultiplicationlawcanbeapplied.Hence5.2SolutionThenumberofdefectives.X,intherandomsampleofsize500hasthebinomialdistributionwithn=500andp=0.1.ToobtainanapproximatevaluefortheprobabilityPr(X>25).Weusethenormalapproximationtothisdistribution:thatis,weusethenormaldistributionwiththesamemeanandvariance.Wehavemean=np=500×0.1=50,variance=np(1p)=500×0.1×0.9=45.HenceweusetheN(50,45)distributionor,equivalently,transformtoandusethestandardizednormaldistributionN(0,1),withdistributionfunction.Sinceweareapproximatingadiscretedistribution(binomial)byacontinuousdistribution(normal)weincludeacontinuitycorrection.FromtablesofthenormaldistributionwefindthatButwerequirePr(X>25),givenbyPr(X>25)=1Pr(X≤25)=0.9999.Similarly,forPr(X<60)werequireTheprobabilitythattherearefewerthan60defectivesisthusapproximately0.922.5.3SolutionThenumberXofticketholderswhofailtoturnupfortherecitalhasabinomialdistributionwithindexn=100andparameterp=0.03,andhencemean=np=3.Weneedtofindtheprobabilitythattwoormorepeoplefailtoturnup(whentherewillbeaseatforeveryone).i.e.tofindSincenislargeandpissmallwemayapproximatethebinomialprobabilitiesusingthePoissondistributionwiththesamemean,3.Hencetherequiredprobabilityis(approximately)Notes(1)Wearetoldthat,onaverage,3%ofpeoplewhohavebookedforrecitalsfailtoturnup,andhavetomaketheassumptionthattherateofnon-attendancestaysconstantovertimewhencomputingtherequiredprobability.Inpracticethismaynotbethecaseas,forexample,differenttimesoftheyearorthevaryingfameofperformersmayaffecttheattendancerates.Wearealso,inusingthebinomialdistribution,makingtheassumptionthatpeoplewhohavebookedactindependentlywithregardtoattendanceornon-attendance.Itismostunlikelythatthisassumptionisvalid,sincemaywillbookinsmallgroupsanditwilloftenbethecasethatthefailureofonememberofagrouptoattendwillbeassociatedwiththenon-attendanceofsomeoralloftheothermembersofthatgroup.Thusasolutionbasedonthebinomialdistributioncanonlybeapproximate.Itisclear,however,thatsuchasolutionisthebestPossiblegiventheinformationatourdisposal.(2)TheuseofthePoissonapproximationtothebinomialdistributionislessnecessarynowthanitwaswhentheonlycomputationalaidsgenerallyavailableweremathematicaltablesandmechanicalcalculators.Withamodernscientificcalculatoritis(almost)asimplemattertocalculatetherequiredbinomialprobabilitiesexactly,asfollows:Thereasonwesaythatthedirectcalculationofbinomialprobabilitiesisalmostasimplematteristhat,ifnislarge,calculationofthenecessarybinomialcoefficientsbysubstitutingthevaluesofnandxinwillfailbecausesomeofthefactorialsinvolvedwilloverflowthecalculator.Thismaybeavoided,asintheabovecalculations,byappropriatecancellationbeforeenteringanyfiguresintothecalculator.Itmayalsobeavoidedbycalculatingprobabilitiessequentially,asisdoneforthePoissondistributioninProblem2A.8:beginningwiththevalueofPr(X=0),weobtaintheprobabilitiesofothervaluesofXusingtherecurrencerelationship(3)Confusionoftenarisesastowhichapproximationtothebinomialdistributiontousewhennislarge.ThetwopossibilitiesarethePoissonandnormaldistributions.ThePoissondistributionis,mathematically,thelimitingformofB(n,p)asn→∞andp→0insuchawaythattheproductnpremainsconstant.Thepresentproblemgivesanexamplewherenislargeandpissmall,andhencethePoissonapproximationisappropriate.ItisalsoworthnotingthatthenormaldistributioncanbederivedasalimitingformofthePoissonasitssingleparameter,,tendstoinfinity,sothatthenormaldistributioncanbeusedtoapproximatethePoissondistributionforlarge.WethusseehowthePoissonapproximationtothebinomialdistributionmightinturnleadtoanormalapproximation.AlthoughweseefromthisdiscussionthatitisclearlythePoissondistributionwhichisappropriatetothepresentproblem,itmakeslittledifferencenumericallywhichapproximationisused.Ifweusethenormalapproximationwefindtheanswertobe0.810ratherthan0.801.However,thisdoesnotalterthefactthat,fromthestructureoftheproblem,itisthePoisson,notthenormal,distributionwhichprovidestheappropriateapproximation.5.4Solution(a)LetthenumberofcorrectanswersgivenbyanignorantcandidatebedenotedbyX.ThenX～B,andforthisdistributionanormalapproximationisacceptable,soweconcludethat,approximately,X～

i.e.X～WearerequiredtofindthatvaluexsuchthatPr(X≥x)=0.01andsowehavetosolveforxintheequationNowimpliesthatk=2.3263,sowehaveor.Inpracticethisobviouslymeansthatthepassmarkshouldbe:setat25.(b)Asinpart(a),thedistributionofXisBSinceXisthenumberofcorrectanswersand50Xthenumberofincorrectanswers,thescoreSofthecandidatewhoattemptseveryquestionisarandomvariablesuchthatWeseethereforethatE(S)=3E(X)50andVat(S)=9Vat(X).Fromstandardpropertiesofthebinomialdistributionusedinpart(a),andVar(X)=,sowereach(c)Withonemarkpercorrectanswerandhalfthesyllabusknown,thecandidatehas25marksinthehag,withupto25furtherquestionsavailabletosecuretheextra3marks.(Theproblemwithguessingatmanyofthese25furtherquestionsisthatheistwiceaslikelytoguessincorrectly,andbepenalised,astogetananswerright.)ifthreefurtherquestionsareattempted,allmustbecorrectinordertopass,andtheprobabilityp3ofthisisp3=.Iffivequestionsareattempted,thenatleast4mustbecorrecttosecurethethreemarksneeded,andtheprobabilityp5ofthisisSince,weseethatthebetterstrategyistotryfivefurtherquestions.Note(1)Whilethenumbershavebeenkeptsimple,theset-upisamoderatelyrealisticone,andscoringsystemsofthetypedescribedareinuse.Theydo,ofcourse,offercandidatesconsiderableincentivetorevise,sinceguessworkismostunlikelytoberewarded.Inpractice,candidatesareusuallynotguessingquiteatrandom,andifthescoringschemeisknowntothemtheyhaveinterestingstrategicchoices.Forexample,tomaximizethetotalexpectedscore,acandidateshouldattemptaquestionifheassessesthechanceofansweringitcorrectlyatormoreundertheschemein(b),butunderthesystemin(c)onlyifthechanceexceeds.Otherpossibilitiesincludeallowingcandidatestocheektwo(or,ifappropriate,evenmore)answerstothesamequestion,andgainreducedcreditifthecorrectanswerisoneofthosechecked.(2)Inpart(c)thestatementoftheproblemallowsustorestrictattentiontojust3or5extraquestions.Inpracticeonewouldwishtocalculatetheoptimumnumberofextraquestionstochoose.Alittlereasoningshowsthatchoosingfourquestionsisnotsensible,sincethecandidatecanthenpassonlybyansweringallfourcorrectly,amoredifficulttaskthanguessingthreeoutofcalculationshowsthat,ifsevenextraquestionsareattempted,theprobabilityp7ofgettingfiveormorecorrectis11/35,andthisstrategyisthusjustasgoodaschoosingfiveextraquestions.Tryingmorethansevenquestionsisnotadvisable,though.Ifeightorninequestionsaretried,thenatleastsixmustbeansweredcorrectly.Tryingninequestionsisthereforeabetterstrategythantryingeight,andtheprobabilityofsuccessfromninequestionsis835/39,whichtssmallerthanp5andp7.5.5Solution(a)WeassumethatthenumberofincomingcallsinoneminutehasthePoissondistributionwithmean5.Hence,ifXisthenumberofincomingcallsinoneminute,Then(b)Thenumberofcallsina2-minuteperiod,Y,hasthePoissondistributionwithmean10.Hence(c)Thenumberofcalls,W,in20minuteshasthePoissondistributionwithmean100.WerequirePr(W≤102)and,sincethemeanofthedistributionislarge,areabletouseanormalapproximation.SincethemeanandvarianceofthePoissondistributionarethesame,theappropriatenormaldistributionisN(100,102).Usingacontinuitycorrection(连续校正),wefind(d)Then,usingthebinomialdistributionB(5,0.9596),theprobabilitythatexactlyfouroutoffivel-minuteperiodscontain2ormorecallsis5(0.9596)4(10.9596)=0.171.Notes(1)Althoughitisnotexplicitlystatedintheproblem,theintentionisthatweshouldassumethatthenumberoftelephonecallsinoneminutehasaPoissondistribution.Thisarisesifweassumethatcallsaredistributedatrandomovertime,withnotendencyeithertoarriveingroupsortobeevenlyspaced.Wealsoassumethatthereisnochangeovertimeintherateatwhichcallsarereceived.AmathematicallypreciseformulationoftheseassumptionsisprovidedbythePoissonprocess,amodelforcompletelyrandomoccurrencesintime,withthefollowingproperties:(i)eventsrelatingtonon-overlappingintervalsoftimearestatisticallyindependent,(ii)inanysmalltimeinterval,theprobabilityofanoccurrenceintheintervalisproportionaltothelengthoftheinterval,(iii)inanysmalltimeinterval,theprobabilityoftwoormoreoccurrencesisproportionalto,i.e.isnegligible.Whilewehavegiventheexactdefinition,forcompleteness,theimportantfactissimplythat,whenoccurrencesarecompletelyhaphazardintime,thenumberoftheseoccurrencesinafixedtimeisarandomvariablewithaPoissondistribution.ItisthiswhichjustifiesuseofthePoissondistributioninsomanycasesinvolvingaccidents,infrequentoccurrences,etc.Inmanyapplications,butinparticularinthefieldofecology,thePoissondistributionisoftenusedtodescribethedistributionoforganismsintwo-orthree-dimensionalspace.Alternativedistributions(iftheorganismsarenotdistributedrandomly)canbedescribedasregularorclustered,andinterestoftencentresonfittingoneoftheso-calledcontagiousdistributions(污染分布)toclustered,clumpedoraggregateddata.Anexampleofacontagiousdistributionisthenegativebinomialdistribution;allhavethepropertythatthevarianceisgreaterthanthemean.(2)Inpart(d)weneedtocomputeprobabilitiesfromtwodistributions(Poissonandbinomial),andtocombinethese.Itisnotuncommontofindproblemsofthistype,wheretheprobabilitypof'success'inaBernoullitrialiscalculatedfromsomestandarddistribution,andthenusedinthecalculationofabinomialprobability.5.6Solution(a)Asanillustration,considerthecasex=5,whenweobserve:TTTTHSincethetossesareindependent,weobtainPr(X=5)=(1p)4p,and,ingeneral,Pr(X=x)=(1p)x1p,x=1,2,3,…WesaythattherandomvariableXhasageometricdistribution,withparameterp.(b)Inthiscase,toobtaintheresultY=y,wemusthaveasequenceoftossesofthefollowingform:Tossnumber 12345 …y1 yResult TTHHT …T HThefirsty1tossesresultinm1headsandy-mtails,withtheseheadsandtailsarrangedinanyorder.ThusthesequenceTTHHT...Tillustratedcorrespondstojustoneofthepossibleorderingsforthefirsty1tossesinwhichthereareexactlym1heads;theprobabilityofobtainingm1headsfromthesetossesisfoundfromthebinomialB(y1,p)distribution.Theprobabilitythattheythtossisaheadissimplyp.WeseethereforethatTherandomvariableYissaidtohaveanegativebinomialdistribution,withparametersmandp.Wenotethatwhenm=1wehavethespecialcaseofthegeometricdistributionfoundinpart(a).Notes(1)Wemayreadilycheckthat,forthegeometricdistribution,theprobabilityfunctionsumsto1,sinceThereadermaycaretoperformthesamecheckforthenegativebinomialdistribution.(2)Byconsideringthenumberofheadsinn+mindependenttossesofacoinwecanreadilyverifythefollowingresult,whichconnectsnegativebinomialandbinomialrandomvariables: (*)forn=0,1,2,…,whereYhasthenegativebinomialdistributiongivenintheproblemandZisabinomialrandomvariablewiththeB(n+m,p)distribution.Theevent{Z≥m}occursifthereareatleastmheadsinthefirstn+mtosses,whichimpliesthatthenumberoftossesYuntilthem=thheadmustbenolargerthann+m.Thetwoevents(Y≤n+m}and{Z≥m}areidentical,andthushavethesameprobability,sothatequation(*)holds.Itisinstructive(andfarmoredifficult)toproveequation(*)usingalgebra.Werequirei.e.Aftercancellingpm,weseethat,fori=0,1,...,n,thecoefficientof(1-p)'onthelefthandsideissimplywhileontherighthandsidethecoefficientisandtheequalityofthesetwoexpressionsfollowsfromconsideringthecoefficientofw'onbothsidesoftheidentitywherewisadummyvariable.6.1SolutionInordertosolveforthetwounknownvaluesaandb,weneedtwoequationsinvolvingaandb.Theseequationsareobtainedfromthetwoitemsofinformationthatwearegiven,asfollows.(i)Thefunctionfx(x)isaprobabilitydensityfunction.Henceandsoresultingintheequationab3=3.(ii)ThemeanofXisunity.SoandthusSplittingtermsinthesecondbracketgivesandtheseintegralsareevaluatedasabovetoyield,whichsimplifiestogiveab4=12.Thisistobesolvedtogetherwiththeequationab3=3obtainedearlier.Wenowseethatb=4anda=.NoteInsomerespectsthisproblemmightbethoughtratherunrealistic,inthatonerarelyfindsarandomvariablewithaquadraticdensityfunctioninnature.(Suchrandomvariablesare,however,ofvalueinsimulation,wheretheyareusedasabasisforgeneratingotherdistributions,andinparticularthenormaldistribution.)Ifahistogramofobservationsonsomerandomvariableshowedsymmetry,andtherangewasrestricted,onemightperhapsconsiderusingaquadraticdensityfunctionasaroughapproximation.Themostdirectvalueinthisproblemresidesmainlyinthereinforcementitcartgive,usingonlyelementarycalculus,totheresultthatallprobabilitydensityfunctionsintegrateto1overtherelevantrange,andtotheformulafortheexpectationofarandomvariableX.6.2wheretheintegralistakenovertherangeoftherandomvariableX.Wecanclearlyusethisresulttoevaluatek,asfollows.i.e.Thenandthisgivesor.ThemeanofXisdefinedasThevarianceofXmaybewrittenasVar(X)=E(X2){E(X)}2.ThuswerequireleadingtoThedensityfunctioninthisproblemisparabolic,andsowehaveasinglemode,m,givenbythevalueofxwhichmaximizesfx(x)overtherange2≤x≤5.Nowandsowhen,Thusthemodeofthedistributionisatx=.(Differentiatingagainshowsthat.Thisisalwaysnegative,andinparticularwhenx=,providingconfirmationthatthestationaryvalueisamaximum.)Inordertodeterminethemedian,M,weneedtosolvetheequationfromthedefinitionofthemedian.Thussothatleadingto.ThisequationreducestothecubicTheequationhassolution(seeNote2)andsowemayfactorisethecubicasTheequation2M214M+11=0hasroot0.902and6.098,bothofwhichareunacceptablesincewemusthave2≤M≤5.Themedianisthereforegivenby.Notes(1)Thisproblemisusefulinindicatingthetypeofmanipulationneededtocalculatethepopulationmean,variance,modeandmedianforasimpleprobabilitydensityfunction.(2)Inthisexamplethemean,modeandmedianallcoincideatThereasonforthisresultissimplythattheprobabilitydensityfunctionfx(x),beingparabolic,issymmetrical,asshowninfollowfigure.Hadthisgraphbeendrawninitiallythenthecommonidentityofthemean,modeandmedianatwouldhavebeenspottedinstantly.Indeed,hadthisfactnotbeenknownthentheroottothecubicequation(*)wouldnothavebeenobvious.Weseethereforethevalueofaroughgraph,insheddinglightonotherwisepurelyalgebraicmanipulationswhichmightwellhaveproducedthewronganswerhornanarithmeticerror.(3)AsFx(x)isacontinuous,increasing,functionofxintherange2≤x≤5,theequationFx(x)=canonlyhaveonerootforxinthisrange.Oncetherootx=isknownthen,strictly,nofurtheranalysisisnecessarytofindthemedian.FigureProbabilitydensityfunctionforProblem6.26.3SolutionForeachcomponent,theprobabilitythatthecomponentwearsoutwithinthefirst1000hoursofuseisgivenbyTofindtherequiredprobabilitywenowusethebinomialdistribution.Sincethereare5components,actingindependently,andallhaveprobability0.6321ofwearingout,thenumberXwhichdowearoutisbinomiallydistributedwithindex(i.e.n)5andparameter(i.e.p)0.6321.WerequiretheprobabilitythatXexceeds2,i.e.Notes(1)Thefirstpartofthesolutionisadirectapplicationoftheexponentialdistribution,althoughtosolvetheproblemonedoesnothavetoknowthis.Theexponentialdistributioniscommonlyusedasawaitingdinedistribution,oratime-to-failuredistribution.(2)Thesecondpartofthesolutionmakesuseofthebinomialdistribution.Asinseveralotherproblemsdiscussedinthisunit,calculationsinvolvinganotherdistributionhavetobeusedfirsttoobtainthevalueofthebinomialparameterp.(3)Despitethemeanlifetimebeing1000hours,theprobabilitythatmorethantwocomponentsfailduringthisperiodisquitehigh,at0.74.Itwouldplainlybeverysillyofthemakerstooffertheproposedguarantee!Thereasonfortheprobabilitybeingashighasthisissimplythattheexponentialdistributionisveryskew.Althoughthemeanlifetimeis1000hours,theprobabilitythatanindividuallifetimeislessthanthisisashighas0.6321.Hadthedistributionbeensymmetrical,theprobabilitythatanindividuallifetimewouldbelessthan1000hourswouldhavebeenandthedistributionofXwouldthushavebeenItiseasytoshowthatforthisdistribution，distinctlylessthan0.74,althoughstilltoohighavaluetocontemplateofferingaguarantee.(4)Althoughinthestatementoftheproblemitwasmadeclear(twice!)thatfailuresofreplacementcomponentscouldbeignored,inpracticethisisanimportantconsideration.Insuchaprocess,'occurrences'(herefailures)takeplacerandomlyintime,andthetotalnumberoftheseoccurrencesinafixedtimehasaPoissondistribution(andtheintervalfromsomestarting-pointtoanoccurrencehasanexponentialdistribution,asgivenabove).Itcanbeshownthat,foranysingleoriginalcomponent(andanyreplacementsneededforit)thenumberoffailuresin1000hourshasthePoissondistributionwithmean1;sincetherearefivecomponents,actingindependently,thetotalnumberoffailuresYwillhavethePoissondistributionwithmean5,so7.11SolutionHereXistherandomvariablethatgivestheaverageofoneroll.OfcoursetheaverageforjustonerollisthesameasthewinningsforthatonerollandsoLetX2000betherandomvariablecorrespondingtotheaverageearningsfor2000rolls.TheCLTsaysthatthecorrespondingapproximatingnormaldistributionhasInfriendlyterms2000-DiceRollAveragenaxning8DistributionNormalDistributionwithandOurdesiredvalueisP(<X2000≤0.75)butbytheCLTwecanapproximatewiththenormaldistribution.Wefindthecorrespondingz-scoresandsoNotethatthismakessense.Iftheaveragewinningsperrollis$2/3≈$0.66thenafteralotofrollsweoughttobeprettyclosetothat.Havingunder$0.75onaverageisprettylikely!7.12SolutionLetXbetherandomvariablecorrespondingtothelifespanofanindividualbacteriumandletX160correspondtotheaverageofthelifespansofthe160bacteria.SinceXisalreadygiventobedistributedwiththeCLTsaysthatthecorrespondingapproximatingnormaldistributionhasInfriendlyterms160-BacteriaAverageLifespanDistributionNormalDistributionwithandOurdesiredvalueisP(4.8≤X2000<)butbytheCLTwecanapproximatewiththenormaldistribution.Wefindthecorrespondingz-scoresandsoP(4.8≤X160<)P(1.26≤Z<)10.1038=0.8962=89.62%AppendixAMathematicalSymbolsA.1MathematicalSymbols+ plus,positive加，正 ≥ greaterthan,equalto大于等于 minus,negative减，负 ≤ lessthan,equalto小于等于 plusorminus,positiveornegative muchgreaterthan绝对大于加或减，正或负 muchlessthan绝对小于× multipliedby乘以 squareroot平方根 dividedby除以 ∞ infinity无限大= equalto等于 proportionalto与……成比例≡ identicallyequalto恒等于 sumof和 notequalto不等于 productof积 notidenticallyequalto不恒等于 difference差≈ approximatelyequalto约等于 ∴ therefore所以～ oftheorderof,similarto约，近似 ∠ angle角> greatethan大于 || parallelto平行< lessthan小于 ⊥ perpendicularto垂直≯ notgreaterthan不大于 : isto比≮ notlessthan不小于 A.2MathematicalRelationsCommonPhrasesisgraterthan Islessthanisabove IsbelowIshigherthan IslowerthanIslongerthan IsshorterthanIsbiggerthan IssmallerthanIsincreased Isdecreasedorreducedfrom ≥ ≤Isgreaterthanorequalto IslessthanorequaltoIsatleast IsatmostIsnotlessthan Isnotmorethan = Isequalto IsnotequaltoIsexactlythesameas IsdifferentfromHasnotchangedfrom HaschangedfromIsthesameas IsnotthesameasAppendixBRomanNumerals1____I 24____XXIV 99____XCIX2____II 25____XXV 100____C3____III 26____XXVI 101____CI4____IV 27____XXVII 144____CXLIV5____V 28____XXVIII 200____CC6____VI 29____XXIX 400____CD7____VII 30____XXX (orCCCC)8____VIII 31____XXXI 500____D9____IX 32____XXXII 900____CM10____X 33____XXXIII (orDCCCC)11____XI 34____XXXIV 1000____M12____XII 35____XXXV 1900____MCM(or13____XIII 36____XXXVI 1995____MDCCCC)14____XIV 37____XXXVII 1995____MCMXCV15____XV 38____XXXVIII 1999____MCMXCIX16____XVI 39____XXXIX 2000____MM17____XVII 40____XL 2005____MMV18____XVIII 49____XLIX 2010____MMX19____XIX 50____L20____XX 60____LX21____XXI 70____LXX22____XXII 80____LXXX23____XXIII 90____XCAppendixCEmergenceofaFourthResearchParadigm♦Before1600,EmpiricalScience—Descriptionofnaturalphenomena♦1600-1950,TheoreticalScience■Eachdisciplinehasgrownatheoretiponet.Theoreticalmodelsoftenmotivateexperimentsandgeneralizeourunderstanding.—Newton’sLaws,Maxwell’sEquations...♦1950s-1990s,ComputationalScience■Overthelast50years,mostdisciplineshavegrownathird,computationalbranch(a.g.empirical,theoretical,andcomputationaleconogoly,orphysics,orlinguistics.)■ComputationalSciencetraditionallymeantsimulation.Itgrewoutofourinabllitytofindclosed-fromsolutionforcomplexmathematicalmodels.—Simulationofcomplexphenomena♦1900-now,DataScience■Thefloodofdatafromnewscientificinstrumentsandsimulations■Theabilitytoeconomicallystoreandmanagepetanytesofdataonline■TheInternetandcomputingGridthatmakesallthesearchivesuniversallyaccessible■Scienticinfo.Management,acquisition,organization,query,andvisualizationtasksscalealmostlinearlywithdatavolumes.Dataminingisamajornewchanllege!—Scientistsoverwhelmedwithdatasetsfrommanydifferentsources●Datacapturedbyinstruments●Datageneratedbysimulations●DatageneratedbysensornetworksInthe1970’sPhysicsNobelPrizewinnerKenWilsoncalledcomputationalsciencethe‘thridparadigmofscience,supplementingtheoryandexperimentation’.NeedforcomputationalscientiststrainedinalgorithmsandnumericalmethodsandabletoprogramparallelHPCclustersandsupercomputers.JimGrayidentifiedthecomingofageof‘fourthparadigm’ofdata-intensivescience.Bothscienceandindustrynowneedsscientiststrainedinmanagingandmining‘BigData’.JimGrayandAlexSzalay,TheWorldWideTelescope:AnArchetypeforOnlineScience,Comm.ACM,45(11):50-54,Nov.2002eScienceisthesetoftoolsandtechnologiestosupportdatafederationandcollaboration♦Foranalysisanddatamining♦Fordatavisualizationandexploration