QUESTION 1Classification is a key method of abstracting reality into simplified map. The reasons why the cartographers need to classify the data are technical constraints either in manual and digital. Manual means that cartographers need to work in hardcopy method while digital means the cartographers use the software to the classification and this digital method is easier to use compared with manual method. The second reason why need classification is for data accuracy. For this case, the classification is smooth out the data inaccuracy to give the effects of look of the map. This is important to make sure that the map is easier to understand for the map reader and the location on the map is exactly same with the reality world. The last reason why the cartographers need to classify the data is for perceptual demands. In this reason, the classification result must in clearer map image, enables selective perception of seeing groups and patterns, and lastly helpful to enhance insight in the data.

QUESTION 2

NATURAL BREAKS a method that captures the character of the data set. Classes are based on the natural grouping of the data values. If used the ArcMap software, it identifies break points by looking for grouping and patterns inherent in the data. The features are divided into classes whose boundaries are set where there are relatively big jumps in the data values.

EQUAL INTERVAL This method encloses equal amounts of the total data range into each class interval. This classification scheme divides the range of attribute values into equal sized subranges.

QUANTILES This method divides the data set into equal number of values in each class. This minimizes the importance of class boundaries, but it can be misleading because one class could have widely differing values.

STANDARD DEVIATION This method is used if a data set displays a normal frequency distribution. It measure for the spread of data around the mean, the mean is calculated and then the standard deviation using statistical mathematics. Usually no more than 6 classes are necessary to convey the information.

ARITHMETIC PROGREESION This classification method is used only when the shape of the data set approximates the shape of a typical arithmetic progression. The width of class increases with constant value. If number of class is known, the formula that will be used is Xmin+I+2I+3I+4I+..=Xmax.

JENKS OPTIMIZATION The Jenks method is used to generate intervals (ranges) within numeric series. The typical application is to automatically generated value ranges on map legends. It is based on the nature of the data and it groups data taking into account the inherent breaks in the data searching for points where that difference is maximized and using these as the limits of each class or interval. This method calculates the differences in values between individual statistics in ascending order. Then it places a limit to separate the groups where the differences in values are high.

If I is given a set of quantitative data, the classification that I will be choose is natural breaks because its allows to take into account observable discontinuities, does not allow direct comparisons and lastly this method (which is done automatically in GIS) can be adjusted based on a graph.

QUESTION 3

1. Classification methods to classify the cities in three groups.a. Equal intervali. Sorting data in ascending order.ii. Determine the interval

AVERAGE TEMPERATURES IN JANUARY AND JULY IN THE CAPITALS OF EU COUNTRIES (EQUAL INTERVAL)

JANUARY Where, X = variable of data valuesNumber of class, n = 3 Maximum value, max = 10Minimum value, min = -11 JULY Where, X = variable of data valuesNumber of class, n = 3Maximum value, max = 37Minimum value, min = 20 iii. Determine the class limit based on the calculated interval.

Average Temperatures in JANUARY and JULY in the Capitals of EU Countries (equal interval)CLASSJANUARYJULY

1= -11 + 7= -4= 20 + 5.67= 25.67

2= -4 + 7= 3= 25.67 + 5.67= 31. 34

3= 3 + 7= 10= 31.34 + 5.67= 37.01

The Class Limit of Temperatures in JANUARY and JULY in the Capitals of EU Countries (equal interval)JANUARYJULY

CLASSCLASS LIMITCLASSCLASS LIMIT

1-11 to -4120 to 25

2-3 to 3226 to 31

34 to 10332 to 37

Tabulation of data for Average Temperature in JANUARY and JULY in the capitals of the EU Countries (equal interval)NOCOUNTRYCAPITALTEMPTCLASS

1LTVILNIUS-111

2LVRIGA-10

3ESTTALLINN-10

4FINHELSINKI-9

5PLWARSAW-6

6SPRAGUE-5

7CZSTOCKHOLM-5

8AVIENNA-4

9HBUDAPEST-4

10SLOLJUBLJANA-4

11SKBERLIN-32

12DBRATISLAVA-3

13DKCOPENHAGEN-2

14BBRUSSELS-1

15LLUXEMBOURG-1

16NLAMSTERDAM-1

17FINPARIS1

18IRLDUBLIN1

19GBMADRID2

20ESTLONDON2

21CYNICOSIA53

22IRLROME5

23GRATHENS6

24PLLISBON8

25MVALLETTA10

`NOCOUNTRYCAPITALTEMPTCLASS

1ESTTALLIN201

2IRLDUBLIN20

3DKCOPENHAGEN22

4FINHELSINKI22

5GBLONDON22

6LVRIGA22

7NLAMSTERDAM22

8SSTOCKHOLM22

9BBRUSSELS23

10CZPRAGUE23

11LLUXEMBOURG23

12LTVILNIUS23

13DBERLIN 24

14PLWARSAW24

15AVIENNA25

16FPARIS25

17SKBRATISLAVA262

18SLOLJUBLJANA27

19PLISBON27

20HBUDAPEST28

21MVALLETTA29

22IROME30

23EMADRID31

24GRATHENS333

25CYNICOSIA37

iv. Construct the graph

b. Quantilesi. Sort data in ascending order.ii. Calculate the category ranges.

iii. Tabulate the data with respect to category ranges.NOCOUNTRYCAPITALTEMPTCLASS

1LTVILNIUS-111

2LVRIGA-10

3ESTTALLINN-10

4FINHELSINKI-9

5PLWARSAW-6

6SPRAGUE-5

7CZSTOCKHOLM-5

8AVIENNA-4

9HBUDAPEST-42

10SLOLJUBLJANA-4

11SKBERLIN-3

12DBRATISLAVA-3

13DKCOPENHAGEN-2

14BBRUSSELS-1

15LLUXEMBOURG-1

16NLAMSTERDAM-1

17FINPARIS13

18IRLDUBLIN1

19GBMADRID2

20ESTLONDON2

21CYNICOSIA5

22IRLROME5

23GRATHENS6

24PLLISBON8

25MVALLETTA10

NOCOUNTRYCAPITALTEMPTCLASS

1ESTTALLIN201

2IRLDUBLIN20

3DKCOPENHAGEN22

4FINHELSINKI22

5GBLONDON22

6LVRIGA22

7NLAMSTERDAM22

8SSTOCKHOLM22

9BBRUSSELS232

10CZPRAGUE23

11LLUXEMBOURG23

12LTVILNIUS23

13DBERLIN24

14PLWARSAW24

15AVIENNA25

16FPARIS25

17SKBRATISLAVA263

18SLOLJUBLJANA27

19PLISBON27

20HBUDAPEST28

21MVALLETTA29

22IROME30

23EMADRID31

24GRATHENS33

25CYNICOSIA37

iv. Construct the graph.

c. Natural breaksi. Sorting the data in ascending orderii. Determine the class interval.

Class interval of the average temperatures in january and july in the capitals of the eu countries(natural breaks)

CLASSJANUARYJULY

1-11 to -920 to 21

2-8 to 222 to 31

33 to 1032 to 37

iii. Determine the differences and select the three largest class limit.TABULATION OF DATA FOR AVERAGE TEMPERATURE IN JANUARY AND JULY IN THE CAPITALS OF THE EU COUNTRIES (NATURAL BREAKS)NOCOUNTRYCAPITALTEMPTCLASS

1LTVILNIUS-111

2LVRIGA-10

3ESTTALLINN-10

4FINHELSINKI-9

5PLWARSAW-62

6SPRAGUE-5

7CZSTOCKHOLM-5

8AVIENNA-4

9HBUDAPEST-4

10SLOLJUBLJANA-4

11SKBERLIN-3

12DBRATISLAVA-3

13DKCOPENHAGEN-2

14BBRUSSELS-1

15LLUXEMBOURG-1

16NLAMSTERDAM-1

17FINPARIS1

18IRLDUBLIN1

19GBMADRID2

20ESTLONDON2

21CYNICOSIA53

22IRLROME5

23GRATHENS6

24PLLISBON8

25MVALLETTA10

NOCOUNTRYCAPITALTEMPTCLASS

1ESTTALLIN201

2IRLDUBLIN20

3COPENHAGEN222

4FINHELSINKI22

5GBLONDON22

6LVRIGA22

7NLAMSTERDAM22

8SSTOCKHOLM22

9BBRUSSELS23

10CZPRAGUE23

11LLUXEMBOURG23

12LTVILNIUS23

13DBERLIN24

14PLWARSAW24

15AVIENNA25

16FPARIS25

17SKBRATISLAVA26

18SLOLJUBLJANA27

19PLISBON27

20HBUDAPEST28

21MVALLETTA29

22IROME30

23EMADRID31

24GRATHENS333

25CYNICOSIA37

iv. Contruct the graph.

QUESTION 4The calculation of size of proportional circles using:

I. SQUARE ROOT METHOD

CITYPOPULATION

A5,000,000

B2,000,000

C100,000

D400,000

FORMULA: = radius

Population of city A:= 2236.068

Population of city B: = 1414.214

Population of city C: = 316.228

Population of city D: = 632.456

CITYRADIUS (r)RADIUS SIZE (cm)

A2236.0681.0

B1414.2140.6

C316.2280.14

D 632.4560.3

Radius size of city A:For beginning is started with 1.0 cm

Radius size of city B:Formula Radius B Radius A 1414.214/ 2236.068 = 0.6cm

Radius size of city C:Formula Radius C Radius A 316.228/ 2236.068= 0.14cm

Radius size of city D:Formula Radius D Radius A 632.456/ 2236.068= 0.3 cm

II. FLANNERY METHODProcedure using this method(a) Determine the logarithm of the data(b) Multiply these by 0.57(c) Determine the antilogarithms(d) Divide these values by the chosen unit value for the radii of the circlesPopulation of city ALog 5,000,000 = 6.6986.698X 0.57 = 3.817Anti logarithm 3.817 = 6561.453 Population of city BLog 2,000,000 = 6.3016.301 X 0.57 =3.591 Anti logarithm 3.591 = 3899.420

Population of city CLog 100,000 = 55 X 0.57 = 2.850Anti logarithm 2.850 = 707.946Population of city DLog 400,000 = 5.6025.602 X 0.57 = 3.193Anti logarithm 3.193 = 1559.553

CITYRADIUS (r)RADIUS SIZE (cm)

A6561.4531.000

B3899.4200.6

C707.9460.1

D1559.5530.3

Radius size of city A:For beginning is started with 1.000 cm

Radius size of city B:Formula Radius B Radius A 3899.420/ 6561.453= 0.594 cm

Radius size of city C:Formula Radius C Radius A 707.946/ 6561.453= 0.109 cm

Radius size of city D:Formula Radius D Radius A 1559.553/ 6561.453= 0.238cm

Map

ADDBNUMBER OF POPULATIONC

100,000400,0002,000,0005,000,000

POPULATION IN SERI KEMBANGAN IN YEAR 2010 (FLANNERY METHOD)

ADDBNUMBER OF POPULATIONC

100,000400,0002,000,0005,000,000

QUSTION 5For this question the classification of data values need to be done first. The classification method that used is quartiles method because each class is approximately equally represented on the final map. This method can be minimizes the importance of class boundaries but it can be misleading because one class could have widely differing values. Moreover, quartiles are very useful for ordinal data, since the class assignment of quartiles is based on the ranked data. So, the procedure for quartiles are shown below: (i) Sort the original data from lowest to biggest value based on the total value

DistrictsNo. MarriagesNo. DivorcesTotal

119005002400

214502851735

375095845

49001201020

520005462546

675070820

721040250

842065485

915063301836

Sort Process

Sort process

Sort NumberDistrictsNo. MarriagesNo. DivorcesTotal

1721040250

2842065485

3675070820

4375095845

549001201020

6214502851735

7915063301836

8119005002400

9520005462546

(ii) Decide the number of classes from nine country and calculate the mean of each classes

Number of Observations per Class =

Therefore,Number of Observations per class = 9 / 3

= 3 classes

Mean 1 =

Mean 1 = (250 + 485 + 820 / 3 = 518.3

Mean 2 =

Mean 2 = (845+ 1020+ 1735) / 3 = 1200

Mean 3 =

Mean 3 = (1836+ 2400 + 2546) / 3 = 2260.67

(iii) Calculation of the proportional circle size

MEANMEAN VALUECALCULATION RADIUS SIZERADIUS SIZE (cm)

1518.3518.3/2260.670.2

212001200 / 2260.670.5

32260.672260.671.0

(iv) Calculation of percentage New and Repeat for each city for pie chartSort Number

Districts

No. Marriages

No. Divorces

Total

Percentage (%)

MARRIAGESDIVORCES

17210402508416

284206548586.613.4

367507082091.58.5

437509584588.611.4

54900120102088.211.8

621450285173583.616.4

791506330183682.018.0

811900500240079.220.8

952000546254678.621.4

Total86.513.5

PERCENTAGE new = (Marriage value /Total) / 100%PERCENTAGE repeat = (Divorces value / Total) 100%

THE NUMBER OF REGISTERED MARRIAGES AND DIVORCES OF MUSLIM IN PAHANG IN YEAR 2005

The number of cases :91543

Marriages1200 22661

Divorces518

QUESTION 6Use map examples to outline the cartographic methods and you would employ for the most effective depiction for Ech of the following details

i) The population density by districts in johor for year 2010for this requirement the method that used is proportional circle size because this method is effective to show the size of the population for the Johor By this method, we can see the differences of the population for each of the year which is ,The population density data is used.

ii) The population migration from the different states in Malaysia into the federal Territory Kuala Lumpur in year 2000 and 2010for this requirement the method that used is size visual variable of the line. The size of the line that to show is by the thickness of the line where if the thickness of the line is ticker that means the migration from neighboring to Kuala Lumpur is higher and if the line of the thickness is thinner that means the migration from neighboring is lower.

iii) The average annual rainfall for Selangor in 2010for this requirement the method that used is colour visual variable where the colour that used is value colour. For this case, the area that have darkness value have the highest rainfall and for the lightness area have the lowest rainfall. Therefore, the rainfall data for Selangor 2010 are needed to show the value visual variable concept.

iv) The number of working population in the agricultural, industrial, and tertiary sectors for the different states in Malaysia.for this requirement the method that used is proportional circle (points) and colour. Proportional circle use for the number of working population which is we can see the difference size proportional circle based on the number of working population and we use the colour to differentiate which one the agricultural, industrial, and tertiary sectors .

THE POPULATION OF JOHOR DISTRICTS IN YEAR 2010

ADDBNUMBER OF POPULATIONC

100,000400,0002,000,0005,000,000

THE POPULATION OF MIGRATION TO KUALA LUMPUR IN YEAR 201030 00050 000100 000The Population of Migration (people) :SELANGORJOHORPAHANGKUALA LUMPUR

THE POPULATION OF MIGRATION TO KUALA LUMPUR IN YEAR 200015 00030 00050 000The Population of Migration (people) :SELANGORJOHORPAHANGKUALA LUMPUR

THE AVERAGE ANNUAL RAINFALL AT SELANGOR IN YEAR 2010

E C

A

B C

DC

Percipitation (mm)

0 - 250

250 - 500

500- 750

THE NUMBERS OF WORKING POPULATION AT MALAYSIA IN YEAR 2015

JOHORDK.LUMPURDSELANGORPAHANGTYPE OF SECTORS :ADBNUMBER OF POPULATION

Tertiary Industrial Agricultural100,000400,0002,000,0005,000,000

QUESTION 7

Dasymetric mapping is method to improve of the choropleth mapping technique for phenomena that have an uneven distribution where it is used other geographical factors to determine the cause of uneven distribution. Local knowledge of the area is necessary. Example and calculations:

J.K.WRIGHT METHOD

Formula Dn = (D / 1-Am) [ (Dm x Am) / 1-Am) ] Where,Dn = Density in area nDm = Estimated density in area mD = Density over the whole area (m+n)Am = The fraction of m of the total area

D = 60 people/kmn = landm = water assume no inhabitant, Dm = 0Am = 0.5Dn = ??

Dn = (60 / 1-0.5) [(0 x 0.5 / 1-0.5)] = 60 / 0.5 = 120

m = 0.3 (WATER)n = 0.7 (LAND)