Frequency Distribution Tables and Graphs - SelfStudys

Chapter7

FrequencyDistributionTablesandGraphsFrequencyDistributionTablesandGraphs7.0Introduction

Jagadeesh iswatchingsportsnews.Avisualappearedon theT.V. screengivingdetailsof themedalswonbydifferentcountriesinOlympics2012.

Olympics2012-MedalsTallyRankCountryGoldSilverBronzeTotal1UnitedStates4629291042China382723883GreatBritain291719654Russia242632825Korea138728

Theabovetableprovidesdataaboutthetopfivecountriesthatgotthehighestnumberofmedalsintheolympics2012aswellasthenumberofmedalstheywon.Information,availableinthenumericalformorverbalformorgraphicalformthathelpsintakingdecisionsordrawingconclusionsiscalledData.

lWhichcountryhasgotthehighestnumberofmedals?lWhichcountryhasgotthehighestnumberofbronzemedals?lWritethreemorequestionsbasedondataprovidedinthetable.

TryThis

Giveanythreeexamplesofdatawhichareinsituationsorinnumbers.7.1Basicmeasuresofcentraltendency

Usuallywecollectdataanddrawcertainconclusionsbasedonthenatureofadata.Understandingitsnature,wedocertaincomputationslikemean,medianandmodewhicharereferredasmeasuresofcentraltendency.Letusrecall.7.1.1ArithmeticMeanItisthemostcommonlyusedmeasureofcentraltendency.Forasetofnumbers,themeanissimplytheaverage,i.e.,sumofallobservationsdividedbythenumberofobservations.Arithmeticmeanofx1,x2,x3,x4,.......xnis

Arithmeticmean=

= (shortrepresentation)∑xirepresentsthesumofallxiswhereitakesthevaluesfrom1ton

Example 1:Ashok got the following marks in different subjects in a unit test. 20, 11, 21, 25, 23 and 14. What isarithmeticmeanofhismarks?

Solution:Observations=20,11,21,25,23and14

Arithmeticmean =

= ==19

Example2:Arithmeticmeanof7observationswasfoundtobe32.Ifonemoreobservation48wastobeaddedtothedatawhatwouldbethenewmeanofthedata?

Solution:Meanof7observations =32Sumof7observationsis xi=32×7=224Addedobservation=48Sumof8observations xi=224+48=272

∴Meanof8observations = =34Example3:Meanageof25membersofaclubwas38years.If5memberswithmeanageof42yearshavelefttheclub,

whatisthepresentmeanageoftheclubmembers?Solution:Meanageof25membersoftheclub=38years

Totalageofallthe25members=38×25=950Meanageof5members=42yearsTotalageof5members=42×5=210Totalageofremaining20members=950–210=740

∴Presentmeanageofclubmembers = = =37yearsExample4:Arithmeticmeanof9observationswascalculatedas45.Indoingsoanobservationwaswronglytakenas42

for24.Whatwouldthenbethecorrectmean?Solution:Meanof9observations=45

Sumof9observations=45×9=405Whencomputingmean42wastakeninsteadof24∴Correctsumof9observations=405–42+24=387

Actualmeanof9observations= = =43Weobserve,

lFromtheaboveexampleswecanseethatArithmeticMeanisarepresentativevalueoftheentiredata.lArithmeticmeandependsonbothnumberofobservationsandvalueofeachobservationinadata.lItisuniquevalueofthedata.lWhenalltheobservationsofthedataareincreasedordecreasedbyacertainnumber,themeanalsoincreasesor

decreasesbythesamenumber.lWhenall theobservationsofthedataaremultipliedordividedbyacertainnumber, themeanalsomultipliedor

dividedbythesamenumber.7.1.2ArithmeticMeanbyDeviationMethodTherearefiveobservationsinadata,7,10,15,21,27.WhentheteacheraskedtoestimatetheArithmeticMeanofthedatawithoutactualcalculation,threestudentsKamal,NeelimaandLekhyaestimatedasfollows:Kamalestimatedthatitliesexactlybetweenminimumandmaximumvalues,i.e.17,Neelimaestimatedthatitisthemiddlevalueoftheordered(ascendingordescending)data;15,Lekhyaaddedalltheobservationsanddividedbytheirnumber,i.e.16.Wecalleachoftheseestimationsas‘estimatedmean’or‘assumedmean’isrepresentedwith‘A’.Letusverifywhichoftheestimationscoincideswiththeactualmean.Case1:ConsiderKamal’sestimatedarithmeticmeanA=17

Theiractualarithmeticmeanis = = =ScoreAinTermsofDeviations7177=17−10101710=17−7151715=17−2211721=17+4271727=17+10

IfeachobservationiswrittenintermsofdeviationfromassumedmeanA,wehave

=

=

=17+ =17–1=16∴Arithmeticmean=Estimatedmean+Averageofdeviations

Case2:ConsiderNeelima’sestimatedarithmeticmeanA=15

Theirarithmeticmeanis = =

intermsofdeviations=

=

=15+ =15+1=16Case3:ConsiderLekhya’sestimatedarithmeticmeanA=16

Theirarithmeticmeanis = =

intermsofdeviations=

=

=16+ =16

TryThesePrepareatableofestimatedmean,deviationsof theabovecases.Observe theaverageofdeviationswiththedifferenceofestimatedmeanandactualmean.Whatdoyouinfer?[Hint:Comparewithaveragedeviations]

It is clear that the estimated mean becomes the actual arithmetic mean if the sum (or average) of deviations of allobservationsfromtheestimatedmeanis‘zero’.WemayusethisverificationprocessasameanstofindtheArithmeticMeanofthedata.Fromtheabovecases,itisevidentthatthearithmeticmeanmaybefoundthroughtheestimatedmeananddeviationofallobservationsfromit.

Thedifferencebetweenanyscoreofdataandassumedmeaniscalleddeviations.Arithmeticmean=Estimatedmean+Averageofdeviations

=Estimatedmean+

=A+Example5:Findthearithmeticmeanof10observations14,36,25,28,35,32,56,42,50,62byassumingmeanas40.

Alsofindmeanbyregularformula.Doyoufindanydifference.Solution:Observationsofthedata=14,25,28,32,35,36,42,50,56,62LettheassumedmeanisA=40

∴Arithmeticmean=A+

=40+

=40+

=40+

=40–=40−2=38

Byusualformula = =

= =38Inboththemethodswegotthesamemean.

This way of computing arithmetic mean by deviation method is conveniently used for data with large numbers anddecimalnumbers.

Considerthefollowingexample.Example6:Marketvalue(inrupees)ofasharethroughaweekischangingas3672,3657,3673,3665,3668.Findthe

arithmeticmeanofthemarketvalueoftheshare.Solution:Observationsofthedata=3657,3665,3668,3672,3673Estimatedmean=3668

Arithmeticmean =A+

=3668+

=3668+ =3668+ =3668-1=`3667.

TryThese

1.Estimatethearithmeticmeanofthefollowingdata(i)17,25,28,35,40(ii)5,6,7,8,8,10,10,10,12,12,13,19,19,19,20Verifyyouranswersbyactualcalculations.

Projectwork

1.Collectmarksof10ofyourclassmatesindifferentsubjectsintherecentexaminations.Estimatethearithmeticmeanofmarksineachsubjectandverifythembyactualcalculations.Howmanyofyourestimationsrepresentexactmean?

2.Measuretheheightsofstudentsofyourclassandestimatethemeanheight.Verifytheirmeanfromrecordsofyourphysicaleducationteacher.Doyounoticeanydifference?

7.1.3Median

Medianisanotherfrequentlyusedmeasureofcentraltendency.Themedianissimplythemiddletermofthedistributionwhenitisarrangedineitherascendingordescendingorder,i.e.thereareasmanyobservationsaboveitasbelowit.Ifnnumberofobservationsinthedataarrangedinascendingordescendingorder

•Whennisodd, observationisthemedian.

•Whenniseven,arithmeticmeanoftwomiddleobservations and isthemedianofthedata.Example7:Findthemedianof9observations14,36,25,28,35,32,56,42,50.Solution:Ascendingorderofthedata=14,25,28,32,35,36,42,50,56

Noofobservationsn=9(oddnumber)

Medianofthedata= observation=5thobservation=35∴Median=35

Example8:Ifanotherobservation61isalsoincludedtotheabovedatawhatwouldbethemedian?Solution:Ascendingorderofthedata=14,25,28,32,35,36,42,50,56,61

Noofobservationsn=10(evennumber)Thentherewouldbetwonumbersatthemiddleofthedata.

Medianofthedata=arithmeticmeanof and observations=arithmeticmeanof5thand6thobservations

= =35.5DoThisHerearetheheightsofsomeofIndiancricketers.Findthemedianheightoftheteam.

S.No.PlayersNameHeights1.VVSLaxman5’11’’2.ParthivPatel5’3’’3.HarbhajanSingh6’0’’4.SachinTendulkar5’5’’5.GautamGambhir5’7’’6.YuvrajSingh6’1’’7.RobinUthappa5’9’’8.VirenderSehwag5’8’’9.ZaheerKhan6’0’’10.MSDhoni5’11’’

5’10’’means5feet10inchesNote:

lMedianisthemiddlemostvalueinordereddata.lItdependsonnumberofobservationsandmiddleobservationsoftheordereddata.Itisnoteffectedbyanychange

inextremevalues.

TryThese1.Findthemedianofthedata24,65,85,12,45,35,15.2.Ifthemedianofx,2x,4xis12,thenfindmeanofthedata.3.Ifthemedianofthedata24,29,34,38,xis29thenthevalueof‘x’is

(i)x>38(ii)x<29(iii)xliesinbetween29and34(iv)none7.1.4ModeWhenweneedtoknowwhatisthefavouriteuniformcolour inaclassormostsellingsizeof theshirt inshop,weusemode.Themodeissimplythemostfrequentlyoccurringvalue.Considerthefollowingexamples.Example9:Inashoemartdifferentsizes(ininches)ofshoessoldinaweekare;7,9,10,8,7,9,7,9,6,3,5,5,7,10,7,

8,7,9,6,7,7,7,10,5,4,3,5,7,8,7,9,7.Whichsizeoftheshoesmustbekeptmoreinnumberfornextweektosell?Givethereasons.

Solution:Ifwewritetheobservationsinthedatainorderwehave3,3,4,5,5,5,5,6,6,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,9,9,9,9,9,10,10,10.Fromthedataitisclearthat7inchsizeshoesaresoldmoreinnumber.Thusthemodeofthegivendatais7.

So7inchsizeshoesmustbekeptmoreinnumberforsale.Example10:Thebloodgroupof50donors,participatedinblooddonationcampareA,AB,B,A,O,AB,O,O,A,AB,

B,A,O,AB,O,O,A,B,A,O,AB,O,O,A,AB,B,O,AB,O,B,A,O,AB,O,O,A,AB,B,A,O,AB,O,A,AB,B,A,O,AB,O,O.Findthemodeoftheaboveverbaldata.

Solution:Byobserving the datawe can find thatA group is repeated for 12,B group is repeated for 7,ABgroup isrepeatedfor12,Ogroupisrepeatedfor19times.∴Modeofthedatais=‘O’group.

Think,DiscussandWrite

Istheiranychangeinmode,ifoneortwomoreobservations,equaltomodeareincludedinthedata?Note:

lModeisthemostfrequentobservationofthegivendata.lItdependsneitheronnumberofobservationsnorvaluesofallobservations.lItisusedtoanalysebothnumericalandverbaldata.lTheremaybe2or3ormanymodesforthesamedata.

Exercise-7.1

1.Findthearithmeticmeanofthesalesperdayinafairpriceshopinaweek.`10000,`.10250,`.10790,`.9865,`.15350,`.10110

2.Findthemeanofthedata;10.25,9,4.75,8,2.65,12,2.353.Meanofeightobservationsis25.Ifoneobservation11isexcluded,findthemeanoftheremaining.4.Arithmeticmeanofnineobservationsiscalculatedas38.Butindoingso,anobservation27ismistakenfor72.Find

theactualmeanofthedata.5.Fiveyearsagomeanageofafamilywas25years.Whatisthepresentmeanageofthefamily?6.Twoyearsagothemeanageof40peoplewas11years.Nowapersonleftthegroupandthemeanageischangedto12

years.Findtheageofthepersonwholeftthegroup.7Findthesumofdeviationsofallobservationsofthedata5,8,10,15,22fromtheirmean.8.Ifsumofthe20deviationsfromthemeanis100,thenfindthemeandeviation.9.Marksof12studentsinaunittestaregivenas4,21,13,17,5,9,10,20,19,12,20,14.Assumeameanandcalculate

thearithmeticmeanofthedata.Assumeanothernumberasmeanandcalculatethearithmeticmeanagain.Doyougetthesameresult?Comment.

10.Arithmeticmeanofmarks(outof25)scoredby10studentswas15.Oneofthestudent,namedKarishmaenquiredtheother9studentsandfindthedeviationsfromhermarksarenotedas 8, 6, 3, 1,0,2,3,4,6.FindKarishma’smarks.

11.Thesumofdeviationsof‘n’observationsfrom25is25andsumofdeviationsofthesame‘n’observationsfrom35is25.Findthemeanoftheobservations.

12.Findthemedianofthedata;3.3,3.5,3.1,3.7,3.2,3.813.Themedianofthefollowingobservations,arrangedinascendingorderis15.10,12,14,x 3,x,x+2,25.Thenfindx.14.Findthemodeof10,12,11,10,15,20,19,21,11,9,10.15.Modeofcertainscoresisx.Ifeachscoreisdecreasedby3,thenfindthemodeofthenewseries.16.Findthemodeofalldigitsusedinwritingthenaturalnumbersfrom1to100.17.Observationsofarawdataare5,28,15,10,15,8,24.Addfourmorenumberssothatmeanandmedianofthedata

remainthesame,butmodeincreasesby1.18.Ifthemeanofasetofobservationsx1,x2,....,....,x10is20.Findthemeanofx1+4,x2+8,x3+12,........,x10+40.19.Sixnumbersfromalistofnineintegersare7,8,3,5,9and5.Findthelargestpossiblevalueofthemedianofallnine

numbersinthislist.20.Themedianofasetof9distinctobservationsis20.Ifeachofthelargest4observationsofthesetisincreasedby2,

findthemedianoftheresultingset.7.2OrganisationofGroupedDataWehavelearnttoorganizesmallerdatabyusingtallymarksinpreviousclass.Butwhathappensifthedataislarge?Weorganizethedatabydividingitintoconvenientgroups.Itiscalledgroupeddata.Letusobservethefollowingexample.Aconstructioncompanyplannedtoconstructvarioustypesofhousesfortheemployeesbasedontheirincomelevels.Sotheycollectedthedataaboutmonthlynetincomeofthe100employees,whowishtohaveahouse.Theyare(inrupees)15000,15750,16000,16000,16050,16400,16600,16800,17000,17250,17250………………75000.This is a largedataof100observations, ranging from`15000 to` 75000.Even ifwemake frequency table for eachobservationthetablebecomeslarge.Insteadthedatacanbeclassifiedintosmallincomegroupslike10001 20000,2000130000,...,70001 80000.

Thesesmallgroupsarecalled‘classintervals’.Theintervals10001 20000hasalltheobservationsbetween10001and20000includingboth10001and20000.Thisformofclassintervaliscalled‘inclusiveform’,where10001isthe‘lowerlimit’,20000isthe‘upperlimit’.

7.2.1InterpretationofGroupedfrequencydistribution:Example11:Marksof30studentsinmathematicstestaregivenintheadjacentgroupedfrequencydistribution.

Sl.NoMarksNoofStudents10–5525–107310–1510415–206520–252

(i)Intohowmanygroupsthedataisclassified?(ii)Howmanystudentsarethereinthethirdgroup?(iii)Ifastudentgets10marks,shouldhebeincludedin2ndor3rdclass?(iv)Whatarethemarksof6studentswhoarein4thclassinterval?(v)Whataretheindividualmarksof2studentsinthefifthgroup?

Answers(i)Thedataisclassifiedinto5groupsor5classes.(ii)Thereare10studentsinthethirdgroup.(iii)Here10istheupperlimitof2ndclassandlowerlimitof3rdclass.Insuchcaseupperlimitisnotincludedin

theclass.So10isincludedinthe3rdclassinterval.(iv)Marksof6studentsin4thclassintervalvariesfrom15andbelow20.(v)Individualmarksofstudentscan’tbeidentifiedfromthisfrequencydistribution,theymaybefrom20andbelow

25.

DoThis

Agesof90peopleinanapartmentaregivenintheadjacentgroupedfrequencydistributionAgesNoofPeople1–101511–201421–301731–402041–501851–60461–702

(i)HowmanyClassIntervalsarethereinthetable?(ii)HowmanypeoplearethereintheClassInterval21-30?(iii)Whichagegrouppeoplearemoreinthatapartment?(iv)Canwesaythatbothpeoplethelastagegroup(61-70)areof61,70oranyotherage?

7.2.2LimitsandBoundariesSupposewehavetoorganizeadataofmarksinatest.Wemakeclassintervalslike1-10,11-20,......Ifastudentgets10.5marks,wheredoesitfall?Inclass1-10or11-20?Inthissituationwemakeuseofreallimitsorboundaries.Considertheclassintervalsshownintheadjacenttable.

LimitsBoundaries1–100.5-10.511–2010.5–20.521–3020.5–30.531–4030.5–40.5

lAverageofUpperLimit (UL)offirstclassandLowerLimit (LL)ofsecondclassbecomes theUpperBoundary

(UB)ofthefirstclassandLowerBoundary(LB)ofthesecondclass.i.e.,Averageof10,11; =10.5istheboundary.

lNowalltheobservationsbelow10.5fallintogroup1-10andtheobservationsfrom10.5tobelow20.5willfallintonextclassi.e11-20havingboundaries10.5to20.5.Thus10.5fallsintoclassintervalof11-20.

lImaginetheULofthepreviousclassinterval(usuallyzero)andcalculatetheLBofthefirstclassinterval.Average

of0,1is =0.5istheLB.lSimilarly imagine theLLof theclassafter the lastclass intervalandcalculate theUBof the lastclass interval.

Averageof40,41is =40.5istheUB.Theseboundariesarealsocalled“trueclasslimits”.Observelimitsandboundariesforthefollowingclassintervals.ClassintervalLimitsBoundaries

InclusiveclassesLowerlimitUpperlimitLowerboundariesUpperboundaries1-101100.510.511-20112010.520.521-30213020.530.5

ClassintervalLimitsBoundaries

exclusiveclassesLowerlimitUpperlimitLowerboundariesUpperboundaries

1-1011001011-201120102021-3021302030

Thereintheaboveillustrationwecanobservethatincaseofdiscreteseries(Inclusiveclassintervals)limitandboundariesaredifferent.But incaseofcontinuousseries (exclusiveclass intervals) limitsandboundariesare thesame.Differencebetweenupperandlowerboundariesofaclassiscalled‘classlength’,representedby‘C’.

DoThese

1.Longjumpmadeby30studentsofaclassaretabulatedasDistance(cm)101–200201–300301–400401–500501–600Noofstudents471531

I.Arethegivenclassintervalsinclusiveorexclusive?II.Howmanystudentsareinsecondclassinterval?III.Howmanystudentsjumpedadistanceof3.01mormore?IV.Towhichclassintervaldoesthestudentwhojumpedadistanceof4.005mbelongs?

2.Calculatetheboundariesoftheclassintervalsintheabovetable.3.Whatisthelengthofeachclassintervalintheabovetable?7.2.3ConstructionofgroupedfrequencyDistribution

Consider themarksof50students inMathematicssecured inSummativeassessment Ias31,14,0,12,20,23,26,36,33,41,37,25,22,14,3,25,27,34,38,43,32,22,28,18,7,21,20,35,36,45,9,19,29,25,33,47,35,38,25,34,38,24,39,1,10,24,27,25,18,8.After seeing the data, you might be thinking, into how many intervals the data could be classified? How frequencydistributiontablecouldbeconstructed?Thefollowingstepshelpinconstructionofgroupedfrequencydistribution.

ClassintervalsTallymarksFrequency(Noofstudents)

0–7||||408–15 |616–23 ||||924–31 |||1332–39 ||||1440–47||||4

Step1:Findtherangeofthedata.Range=Maximumvalue–Minimumvalue

=47–0=47Step2:Decidethenumberofclassintervals.(Generallynumberofclassintervalsare5to8)

Ifnoofclassintervals=6

Lengthoftheclassinterval= 8(approximately)Step3:Writeinclusiveclassintervalsstartingfromminimumvalueofobservations.

i.e0-7,8-15andsoon...Step4:Usingthetallymarksdistributetheobservationsofthedataintorespectiveclassintervals.Step5:Countthetallymarksandwritethefrequenciesinthetable.Nowconstructgroupedfrequencydistributiontableforexclusiveclasses.

Think,DiscussandWrite

1.Makeafrequencydistributionofthefollowingseries.1,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,6,6,6,6,7,7.

2.Constructafrequencydistributionforthefollowingseriesofnumbers.2,3,4,6,7,8,9,9,11,12,12,13,13,13,14,14,14,15,16,17,18,18,19,20,20,21,22,24,24,25.(Hint:Use

inclusiveclasses)3.Whatarethedifferencesbetweentheabovetwofrequencydistributiontables?4.Fromwhichofthefrequencydistributionswecanwritetherawdataagain?

7.2.4CharacteristicsofGroupedFrequencyDistribution1.Itdividesthedataintoconvenientandsmallgroupscalled‘classintervals’.2.Inaclassinterval5-10,5iscalledlowerlimitand10iscalledupperlimit.3.Classintervalslike1-10,11-20,21-30....arecalledinclusiveclassintervals,becausebothlowerandupper

limitsofaparticularclassbelongtothatparticularclassinterval.4.Classintervalslike0-10,10-20,20-30...arecalledexclusiveclassintervals,becauseonlylowerlimitofa

particularclassbelongstothatclass,butnotitsupperlimit.5.Averageofupperlimitofaclassandlowerlimitofthenextclassiscalledupperboundofthefirstclass

andlowerboundofthenextclass.6.In exclusive class intervals, both limits andboundaries are equal but in caseof inclusive class intervals

limitsandboundariesarenotequal.7.Differencebetweenupperandlowerboundariesofaclassiscalled‘lengthoftheclass’.8.Individualvaluesofallobservationscan’tbeidentifiedfromthistable,butvalueofeachobservationofa

particularclassisassumedtobetheaverageofupperandlowerboundariesof thatclass.Thisvalue iscalled‘classmark’or‘midvalue’(x).

Example12:Thefollowingmarksachievedby30candidatesinmathematicsofSSCexaminationheldintheyear2010.45,56,75,68,35,69,98,78,89,90,70,56,59,35,46,47,13,29,32,39,93,84,76,79,40,54,68,69,60,59.

Constructthefrequencydistributiontablewiththeclassintervals;failed(0–34),thirdclass(35–49),secondclass(50–60),firstclass(60–74)anddistinction(75–100).

Solution:ClassintervalsTallymarksFrequency(Noofstudents)

0–34|||335–49||||||750–59||||560–74|||||675–100||||||||9

Classintervalsarealreadygiven.Soproceedfromstep3Step3:Writeclassintervalsasgiven.

Step 4: These are inclusive class intervals. Recall that upper limits also belong to the class. Using the tally marks,distributetheobservationsofthedataintodifferentclassintervals.

Step5:Countthetallymarksandwritethefrequenciesinthetable.(Note:Thelengthsofclassintervalsarenotsameinthiscase)Example13:Agroupedfrequencydistributiontableisgivenbelowwithclassmark(midvaluesofclassintervals)and

frequencies.Findtheclassintervals.Classmarks71523313947Frequency5111921126

Solution:Weknowthatclassmarksarethemidvaluesofclassintervals.Thatimpliesclassboundariesliebetweeneverytwosuccessiveclassmarks.Step1:Findthedifferencebetweentwosuccessiveclassmarks;h=15-7=8.(Findwhetherdifferencebetweeneverytwosuccessiveclassesissame)Step2:Calculatelowerandupperboundariesofeveryclasswithclassmark‘x’,asx–h/2andx+h/2

Forexampleboundariesoffirstclassare7– =3or7+ =11ClassMarksClassintervalsFrequency

7(7–4)–(7+4)=03–11515(15–4)–(15+4)=11–191123(23–4)–(23+4)=19–271931(31–4)–(31+4)=27–352139(39–4)–(39+4)=35–431247(47–4)–(47+4)=43–516

7.3CumulativeFrequencyIn a competitive examination 1000 candidates appeared for a written test. Theirmarks are announced in the form ofgroupedfrequencydistributionasshownintheadjacenttable.

ClassInterval(Marks)NoofCandidates0–102510–204520–306030–4012040–5030050–6036060–705070–802580–901090–1005

TwocandidatesSarath,Sankararelookingatthetableanddiscussinglike…Sarath:Howmanycandidateshaveappearedforthetest?Sankar:Itseems1000candidatesappearedforthetest.Sarath:See,360candidatesachieved50-60marks.Sankar:If60isthecutoffmark,howmanycandidatesareeligibletogetcallletter?Sarath:Doyoumeanhowmanygot60andabovemarksinaltogether?Sankar:Itis50+25+10+5,thatis90candidateswillbeeligible.Sarath:Butthereareonly105jobs.Howmanycandidatesareeligible,ifcutoffmarkas50.Sankar : In that case, 360 + 50 + 25 + 10 + 5, that is totally 450 candidates are eligible to get call letter for

interview.Similarlywecanmakesomemoreconclusions.Numberofcandidates,whogotequalormorethan90(Lowerboundary)=5Numberofcandidates,whogotequalormorethanLBofninthCI=10+5=15Numberofcandidates,whogotequalormorethanLBofeighthCI=25+15=40Numberofcandidates,whogotequalormorethanLBofseventhCI=50+40=90

Wearegettingthesevaluesbytakingprogressive totaloffrequenciesfromeither thefirstor lastclass to theparticularclass.Thesearecalledcumulativefrequencies.Theprogressivesumoffrequenciesfromthelastclassofthetothelowerboundaryofparticularclassiscalled‘GreaterthanCumulativeFrequency’(G.C.F.).ClassIntervaLBFrequencyGreaterthan(Marks)(NoofCandidates)cumulativefrequency

0–1002525+975=100010–20104545+930=97520–30206060+870=93030–4030120120+750=87040–5040300300+450=75050–6050360360+90=45060–70605050+40=9070–80702525+15=40

80–90801010+5=1590–1009055

Watchouthowwecanwritethesegreaterthancumulativefrequenciesinthetable.1.Frequencyinlastclassintervalitselfisgreaterthancumulativefrequencyofthatclass.2.Add the frequencyof theninthclass interval to thegreater thancumulative frequencyof the tenthclass

intervaltogivethegreaterthancumulativefrequencyoftheninthclassinterval3.Successivelyfollowthesameproceduretogettheremaininggreaterthancumulativefrequencies.

The distribution that represent lower boundaries of the classes and their respective Greater than cumulativefrequenciesiscalledGreaterthanCumulativeFrequencyDistribution.

Similarlyinsomecasesweneedtocalculatelessthancumulativefrequencies.Forexampleifateacherwantstogivesomeextrasupportforthosestudents,whogotlessmarksthanaparticularlevel,weneedtocalculatethelessthancumulativefrequencies.Thus theprogressive totalof frequenciesfromfirstclass to theupperboundaryofaparticularclass iscalledLess thanCumulativeFrequency(L.C.F.).

ClassIntervalUBNoofLessthan(Marks)Candidatescumulativefrequency0–55775–10101010+7=1710–15151515+17=3215–202088+32=4020–252533+40=43

Considerthegroupedfrequencydistributionexpressingthemarksof43studentsinaunittest.1.Frequencyinfirstclassintervalisdirectlywrittenintolessthancumulativefrequency.2.Addthefrequencyofthesecondclassintervaltothelessthancumulativefrequencyofthefirstclassintervalto

givethelessthancumulativefrequencyofthesecondclassinterval3.Successivelyfollowthesameproceduretogetremaininglessthancumulativefrequencies.The distribution that represents upper boundaries of the classes and their respective less than cumulativefrequenciesiscalledLessthanCumulativeFrequencyDistribution.

TryThese

1.Lessthancumulativefrequencyisrelatedto_______________2.Greaterthancumulativefrequencyisrelatedto_______________3.WritetheLessthanandGreaterthancumulativefrequenciesforthefollowingdata

ClassInterval1-1011-2021-3031-4041-50Frequency471252

4.Whatistotalfrequencyandlessthancumulativefrequencyofthelastclassaboveproblem?Whatdoyouinfer?Example14:Givenbeloware themarksofstudents ina less thancumulativae frequencydistribution table..Write the

frequenciesoftherespectiveclasses.AlsowritetheGreaterthancumulativefrequencies.Howmanystudents’marksaregiveninthetable?

ClassInterval(Marks)1-1011-2021-3031-4041-50L.C.F.(Noofstudents)1227546775Solution:

ClassIntervalL.C.F.FrequencyG.C.F.(Marks)(Noofstudents)1–10121212+63=7511–202727-12=1515+48=6321–305454-27=2727+21=4831–406767-54=1313+8=2141–507575-67=88

Totalnumberofstudentsmentionedinthetableisnothingbuttotaloffrequenciesorlessthancumulativefrequencyofthelastclassorgreaterthancumulativefrequencyofthefirstclassinterval,i.e.75.

Exercise-7.2

1.Givenbelowaretheagesof45peopleinacolony.

3387253126550254856332822156259161419243526912461542633252211422352486210244351374836Constructgroupedfrequencydistributionforthegivendatawith6classintervals.

2.Numberofstudentsin30classroomsinaschoolaregivenbelow.Constructafrequencydistributiontableforthedatawithaexclusiveclassintervalof4(students).253024182124323422202232402830222631341538282016152024302518

3.Classintervalsinagroupedfrequencydistributionaregivenas4–11,12–19,20–27,28–35,36–43.Writethenexttwoclassintervals.(i)Whatisthelengthofeachclassinterval?(ii)Writetheclassboundariesofallclasses,(iii)Whataretheclassmarksofeachclass?

4.Inthefollowinggroupedfrequencydistributiontableclassmarksaregiven.ClassMarks102234465870Frequency614202195

(i)Constructclassintervalsofthedata.(Exclusiveclassintervals)(ii)Constructlessthancumulativefrequenciesand(iii)Constructgreaterthancumulativefrequencies.

5.Themarksobtainedby35studentsinatestinstatistics(outof50)areasbelow.35115354523314021131520474842344345333711132718123739381613185414743

Constructafrequencydistributiontablewithequalclassintervals,oneofthembeing10-20(20isnotincluded).6.Construct theclassboundariesof thefollowingfrequencydistribution table.Alsoconstruct less thancumulativeand

greaterthancumulativefrequencytables.Ages1-34-67-910-1213-15Noofchildren101215139

7.Cumulativefrequencytableisgivenbelow.Whichtypeofcumulativefrequencyisgiven.Trytobuildthefrequenciesofrespectiveclassintervals.Runs0-1010-2020-3030-4040-50Noofcricketers38192530

8.Number of readers in a library are given below.Write the frequency of respective classes.Alsowrite the less thancumulativefequencytable.

Numberofbooks1-1011-2021-3031-4041-50GreaterthanCumulativefrequency4236231467.4GraphicalRepresentationofData:

Frequencydistributionisanorganiseddatawithobservationsorclassintervalswithfrequencies.Wehavealreadystudiedhowtorepresentofdiscreteseriesintheformofpictographs,bargraphs,doublebargraphandpiecharts.Letusrecallbargraphfirst.7.4.1BarGraph

Adisplayofinformationusingverticalorhorizontalbarsofuniformwidthanddifferentlengthsbeingproportionaltotherespectivevaluesiscalledabargraph.Letusseewhatabargraphcanrepresent.Studythefollowingverticalbargraph.

Achievementinexam(i)Whatdoesthisbargraphrepresent?(ii)HowmanystudentssecuredA,BorCgrades?(iii)Whichgradeissecuredbymorenumberofthestudents?(iv)Howmanystudentsarethereintheclass?

Itiseasytoanswerthequestionsfromthegraph.Similarly insomegraphsbarsmaybedrawnhorizontally.Forexampleobserve thesecondbargraph. Itgives thedataaboutnumberofvehiclesinavillageSangaminNelloredistrict.

Think,DiscussandWrite1.Allthebars(orrectangles)inabargraphhave

(a)samelength(b)samewidth(c)samearea(d)equalvalue2.Doesthelengthofeachbardependonthelengthsofotherbarsinthegraphs?3.Doesthevariationinthevalueofabaraffectthevaluesofotherbarsinthesamegraph?4.Wheredoweuseverticalbargraphsandhorizontalbargraphs.

7.5GraphicalRepresentationofGroupedFrequencyDistribution

Letuslearnthegraphicalrepresentationofgroupedfrequencydistributionsofcontinuousseriesi.e.withexclusiveclassintervals.Firstoneofitskindishistogram.7.5.1Histogram

7.5.1.1InterpretationofHistogram:Observethefollowinghistogramforthegivengroupedfrequencydistribution.

ClassIntervalFrequency(Marks)(NoofStudents)0–10310–20520–30930–401040–501550–601960–701370–801180–90990-1006

(i)Howmanybarsarethereinthegraph?(ii)Inwhatproportiontheheightofthebarsaredrawn?(iii)Widthofallbarsissame.Whatmaybethereason?(iv)Shallweinterchangeanytwobarsofthegraph?

Fromthegraphyoumighthaveunderstoodthat(i)Thereare10barsrepresentingfrequenciesof10classintervals.(ii)Heightsofthebarsareproportionaltothefrequencies,(iii)Widthofbarsissamebecausewidthrepresentstheclassinterval.Particularlyinthisexamplelengthof

allclassintervalsissame.(iv)Asit isrepresentingacontinuousseries, (withexclusiveclass intervals),wecan’t interchangeanytwo

bars.

TryTheseObservetheadjacenthistogramandanswerthefollowingquestions-

(i)Whatinformationisbeingrepresentedinthehistogram?(ii)Whichgroupcontainsmaximumnumberofstudents?(iii)HowmanystudentswatchTVfor5hoursormore?(iv)Howmanystudentsaresurveyedintotal?

7.5.1.2ConstructionofaHistogramATVchannelwantstofindwhichagegroupofpeoplearewatchingtheirchannel.Theymadeasurveyinanapartment.Representthedataintheformofahistogram.Step 1 : If the class intervals given are inclusive (limits) convert them into the exclusive form (boundaries) since the

histogramhastobedrawnforacontinuousseries.ClassFrequencyClassInterval(NoofIntervals(Agegroup)viewers)11–201010.5–20.521–301520.5–30.531–402530.5–40.541–503040.5–50.551–602050.5–60.561–70560.5–70.5LimitsBoundaries

Step2:ChooseasuitablescaleontheX-axisandmarktheclassintervalsonit.Step3:ChooseasuitablescaleontheY-axisandmarkthefrequenciesonit.(Thescalesonboththeaxesmaynotbe

same)Scale:X-axis1cm=oneclassintervalY-axis1cm=5peopleStep4:Drawrectangleswithclassintervalsasbasesandthecorrespondingfrequenciesasthecorrespondingheights.

7.5.1.3HistogramwithVaryingBaseWidthsConsiderthefollowingfrequencydistributiontable.CategoryClassIntervals(Marks)PercentageofStudentsFailed0-3528ThirdClass35-5012SecondClass50-6016FirstClass60-10044

Youhavenoticedthatfordifferentcategoriesofchildrenperformancetherangeofmarksforeachcategoryisnotuniform.Ifweobservethetable,thestudentswhosecuredfirstclassis44%whichspreadsovertheclasslength40(60to100).Whereasthestudentwhohavesecuredsecondclassis16%ofthestudentsspreadovertheclasslength10(50to 60) only. Therefore to represent the above distribution table into histogramwe have take thewidths of classintervalsalsointoaccount.In such cases frequency per unit class length (frequency density) has to be calculated and histogram has to beconstructedwithrespectiveheights.Anyclassintervalmaybetakenasunitclassintervalforcalculatingfrequencydensity.Forconvenienceleastclasslengthistakenasunitclasslength.∴Modifiedlengthofanyrectangleisproportionaltothecorrespondingfrequency

Density=ClassintervalsPercentageClasslengthLengthoftherectangle(Marks)ofstudents

0–352835

35–501215

50–601610

60–1004440Withthemodifiedlengthshistogramhastobeconstructedasinthepreviousexample.

Step1:ChooseasuitablescaleontheX-axisandmarktheclassintervalsonit.Step2:ChooseasuitablescaleontheY-axisandmarkthefrequenciesonit.(Thescalesonboththeaxesmaynot

besame)Scale:X-axis1cm=1Min.classinterval

Y-axis1cm=2%Step3:Drawrectangleswithclassintervalsasbasesandthecorrespondingfrequenciesastheheights.

7.5.1.4Histogramforgroupedfrequencydistributionwithclassmarks

Example15:Constructahistogramfromthefollowingdistributionoftotalmarksobtainedby65studentsofclassVIII.Marks(Midpoints)150160170180190200Noofstudents810251273Solution:Asclassmarks(midpoints)aregiven,classintervalsaretobecalculatedfromtheclassmarks.Step1:Findthedifferencebetweentwosuccessiveclasses.h=160-150=10.

(Findwhetherdifferencebetweeneverytwosuccessiveclassesissame)

Step2:Calculatelowerandupperboundariesofeveryclasswithclassmark‘x’,asx– andx+ .Step3:Chooseasuitablescale.X-axis1cm=oneclassinterval

Y-axis1cm=4studentsStep4:Drawrectangleswithclassintervalsasbasesandthecorrespondingfrequenciesastheheights.

ClassClassFrequencyMarks(x)Intervals(Noofstudents)150145–1558160155–16510170165–17525180175–18512190185–1957200195–2053

Think,DiscussandWrite

1.Classboundariesaretakenonthe‘X’-axis.Whynotclasslimits?2.Whichvaluedecidesthewidthofeachrectangleinthehistogram?3.Whatdoesthesumofheightsofallrectanglesrepresent?

7.5.2FrequencyPolygon

7.5.2.1InterpretationofFrequencyPolygonFrequencypolygonisanotherwayofrepresentingaquantitativedataanditsfrequencies.Letusseetheadvantagesofthisgraph.

Considertheadjacenthistogramrepresentingweightsof33peopleinacompany.Letusjointhemid-pointsoftheuppersidesoftheadjacentrectanglesofthishistogrambymeansoflinesegments.Letuscallthesemid-pointsB,C,D,E,FandG.When joinedby line segments,weobtain the figureBCDEFG.Tocomplete thepolygon,weassume that there isa

class interval with frequency zero before 30.5-35.5 and one after 55.5 - 60.5, and their mid-points are A and H,respectively.ABCDEFGHisthefrequencypolygon.Although,thereexistsnoclassprecedingthelowestclassandnoclasssucceedingthehighestclass,additionofthetwoclass intervalswith zero frequency enables us tomake the area of the frequency polygon the same as the area of thehistogram.Whyisthisso?

Think,DiscussandWrite

1.Howdowecompletethepolygonwhenthereisnoclassprecedingthefirstclass?2.Theareaofhistogramofadataanditsfrequencypolygonaresame.Reasonhow.3.Isitnecessarytodrawhistogramfordrawingafrequencypolygon?4.Shallwedrawafrequencypolygonforfrequencydistributionofdiscreteseries?

7.5.2.2ConstructionofaFrequencyPolygon

Considerthemarks,(outof25),obtainedby45studentsofaclassinatest.Drawafrequencypolygoncorrespondingtothisfrequencydistributiontable.ClassFrequencyMidInterval(No.ofValues(Marks)students)0-572.55-10107.510-151412.515-20817.520-25622.5Total45

StepsofconstructionStep1:Calculatethemidpointsofeveryclassintervalgiveninthedata.Step2:Drawahistogramforthisdataandmarkthemid-pointsofthetopsoftherectangles(hereinthisexampleB,

C,D,E,Frespectively).Step3:Jointhemidpointssuccessively.Step4:Assumeaclassintervalbeforethefirstclassandanotherafterthelastclass.Alsocalculatetheirmidvalues

(AandH)andmarkontheaxis.(Here,thefirstclassis0–5.So,tofindtheclasspreceding0-5,weextendthehorizontalaxisinthenegativedirectionandfindthemid-pointoftheimaginaryclass-interval–5–0)

Step5:JointhefirstendpointBtoAandlastendpointFtoGwhichcompletesthefrequencypolygon.

Frequencypolygoncanalsobedrawnindependentlywithoutdrawinghistogram.Forthis,werequirethemidpointsoftheclassintervalofthedata.

DoThese1.Constructthefrequencypolygonsofthefollowingfrequencydistributions.

(i)Runsscoredbystudentsofaclassinacricketfriendlymatch.Runsscored10–2020–3030–4040–5050–60Noofstudents35842

(ii)Saleofticketsfordramainanauditorium.Rateofticket1015202530Noofticketssold5030603020

7.5.2.3CharacteristicsofaFrequencyPolygon:

1.Frequencypolygonisagraphicalrepresentationofafrequencydistribution(discrete/continuous)2.ClassmarksorMidvaluesofthesuccessiveclassesaretakenonX-axisandthecorrespondingfrequenciesonthe

Y-axis.3.Areaoffrequencypolygonandhistogramdrawnforthesamedataareequal.

Think,DiscussandWrite

1.Histogramrepresentsfrequencyoveraclassinterval.Canitrepresentthefrequencyataparticularpointvalue?2.Canafrequencypolygongiveanideaoffrequencyofobservationsataparticularpoint?

7.5.2.4ConstructionofaFrequencyPolygonforagroupedfrequencydistributionwithoutusinghistogram:Inastudyofdiabeticpatients,thefollowingdatawereobtained.Ages10–2020–3030–4040–5050–60Noofpatients5916113

Letusconstructfrequencypolygonforitwithoutusingthehistogram.Step1:Findtheclassmarksofdifferentclasses.Step2:Selectthescale:

X-axis1cm=1classintervalY-axis1cm=2marks

Step3:If‘x’denotestheclassmarkandfdenotesthecorrespondingfrequencyofaparticularclass,thenplot(‘x’,f)onthegraph.

Step4:Jointheconsecutivepointsinorderbylinesegments.Step5:Imaginetwomoreclasses,onebeforethefirstclassandtheotherafterthelastclasseachhavingzerofrequency.

Marktheirmidvaluesonthegraph.Step6:Completethepolygon.ClassIntervalNoofClassPoints(Ages)PatientsMark0–1005(5,0)10–20515(15,5)20–30925(25,9)30–401635(35,16)40–501145(45,11)50–60355(55,3)60–70065(65,0)

7.5.3FrequencyCurveforagroupedfrequencydistribution

Itisanotherwayofrepresentationofthedatabyafreehandcurve.Letusconstructfrequencycurvefortheabovedatawithoutusingthehistogram.Step1:Findtheclassmarksofdifferentclasses.Step2:Selectthescale:

X-axis1cm=1classintervalY-axis1cm=2marks

Step3:If‘x’denotestheclassmarkandfdenotesthecorrespondingfrequencyofaparticularclass,thenplot(x,f)onthegraph.

Step4:Jointheconsecutivepointssuccessivelybyafreehandcurve.ClassNoofClassPointsIntervalPatientsMark(Ages)0–1005(5,0)10–20515(15,5)20–30925(25,9)30–401635(35,16)40–501145(45,11)50–60355(55,3)60–70065(65,0)

7.5.4GraphofaCumulativeFrequencyDistribution

Agraph representing thecumulative frequenciesof agrouped frequencydistributionagainst thecorresponding lower /upperboundariesofrespectiveclassintervalsiscalledCumulativeFrequencyCurveorOgiveCurve.Thesecurvesareusefulinunderstandingtheaccumulationoroutstandingnumberofobservationsateveryparticularlevelofcontinuousseries.7.5.4.1LessthanCummulativefrequencycurveConsider thegrouped frequencydistributionofnumberof tenders receivedby adepartment from the contractors for acivilworkinacourseoftime.

CI(days)0–44-88-1212–1616–20Nooftenders2512103

Step1:Ifthegivenfrequencydistributionisininclusiveform,thenconvertitintoanexclusiveform.Step2:Constructthelessthancumulativefrequencytable.

ClassNoofUBL.Cu.FrIntervalTenders(Days)0–42424–85878–1212121912–1610162916–2032032

Step3:Mark theupperboundariesof theclass intervalsalongX -axisand theircorrespondingcumulative frequenciesalongY-axisSelectthescale:X-axis1cm=1classintervalY-axis1cm=4tenders

Step4:Also,plotthelowerboundaryofthefirstclass(upperboundaryoftheclassprevioustofirstclass)intervalwithcumulativefrequency0.

Step5:Jointhesepointsbyafreehandcurvetoobtaintherequiredogive.

Similarlywecanconstruct ‘Greater thancumulative frequencycurve’by takinggreater thancumulativeonY-axis andcorresponding‘LowerBoundaries’ontheX-axis.

Exercise-7.3

1.The following tablegives thedistributionof45studentsacross thedifferent levelsof IntelligentQuotient.Drawthehistogramforthedata.IQ60-7070-8080-9090-100100-110110-120120-130Noofstudents25610985

2.Constructahistogramforthemarksobtainedby600studentsintheVIIclassannualexaminations.Marks360400440480520560Noofstudents100125140958060

3.Weeklywages of 250workers in a factory are given in the following table.Construct the histogramand frequencypolygononthesamegraphforthedatagiven.Weeklywage500-550550-600600-650650-700700-750750-800Noofworkers304250554528

4.Agesof60teachersinprimaryschoolsofaMandalaregiveninthefollowingfrequencydistributiontable.ConstructtheFrequencypolygonandfrequencycurveforthedatawithoutusingthehistogram.(Useseparategraphsheets)Ages24–2828–3232–3636–4040–4444–48Noofteachers121015986

5.Constructclassintervalsandfrequenciesforthefollowingdistributiontable.Alsodrawtheogivecurvesforthesame.MarksobtainedLessthan5Lessthan10Lessthan15Lessthan20Lessthan25Noofstudents28182735

Whatwehavediscussed

•Arithmeticmeanoftheungroupeddata= or = (shortrepresentation)where representsthesumofallxiswhere‘i’takesthevaluesfrom1ton

•Arithmeticmean=Estimatedmean+Averageofdeviations

Or =A+•Meanisusedintheanalysisofnumericaldatarepresentedbyuniquevalue.•Medianrepresentsthemiddlevalueofthedistributionarrangedinorder.

•Themedianisusedtoanalysethenumericaldata,particularlyusefulwhenthereareafewobservationsthatareunlikemean,itisnotaffectedbyextremevalues.

•Modeisusedtoanalysebothnumericalandverbaldata.•Modeisthemostfrequentobservationofthegivendata.Theremaybemorethanonemodeforthegivendata.•Representationofclassifieddistinctobservationsofthedatawithfrequenciesiscalled‘FrequencyDistribution’

or‘DistributionTable’.•Differencebetweenupperandlowerboundariesofaclassiscalledlengthoftheclassdenotedby‘C’.•Inaaclasstheinitialvalueandendvalueofeachclassiscalledthelowerlimitandupperlimitrespectivelyof

thatclass.•Theaverageofupperlimitofaclassandlowerlimitofsuccessiveclassiscalledupperboundaryofthatclass.•Theaverageofthelowerlimitofaclassanduperlimitofpreceedingclassiscalledthelowerboundaryofthe

class.•Theprogressivetotaloffrequenciesfromthelastclassofthetabletothelowerboundaryofparticularclassis

calledGreaterthanCumulativeFrequency(G.C.F).•TheprogressivetotaloffrequenciesfromfirstclasstotheupperboundaryofparticularclassiscalledLessthan

CumulativeFrequency(L.C.F.).•Histogramisagraphicalrepresentationoffrequencydistributionofexclusiveclassintervals.•When the class intervals in a grouped frequency distribution are varying we need to construct rectangles in

histogramonthebasisoffrequencydensity.

Frequencydensity= Leastclasslengthinthedata•Frequencypolygonisagraphicalrepresentationofafrequencydistribution(discrete/continuous)•Infrequencypolygonorfrequencycurve,classmarksormidvaluesoftheclassesaretakenonX-axisandthe

correspondingfrequenciesontheY-axis.•Areaoffrequencypolygonandhistogramdrawnforthesamedataareequal.•Agraphrepresenting thecumulativefrequenciesofagroupedfrequencydistributionagainst thecorresponding

lower / upper boundaries of respective class intervals is called Cumulative Frequency Curve or“OgiveCurve”.

ThinkingCriticallyTheabilityofsomegraphsandcharts todistortdatadependsonperceptionof individuals tofigures.Considerthesediagramsandanswereachquestionbothbeforeandafterchecking.

(a)Whichislonger,theverticalorhorizontalline?(b)Arelineslandmstraightandparallel?(c)Whichlinesegmentislonger: or(d)Howmanysidesdoesthepolygonhave?Isitasquare?(e)Stare at the diagrambelow.Canyou see four large posts rising upout of the paper?State

someandseefoursmallposts.

(a) (b) (c) (d) (e)