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Maths Statistics Investigation

Extracts from this document...

Introduction

image00.png

Introduction

I’m going to investigate how the condition of a car e.g. bodywork, mileage, age and any special features, affects the price.  In my opinion the age of the car will have the biggest effect on its price, as with most cars, the older it becomes the more its price will depreciate. I’m going to draw a scatter graph to find the relationship between the price and the age.

I will investigate the first 30 cars on my datasheet except number 15 because it doesn’t have a price number. I have a secondary data sheet showing different information which may help me in my investigation.

Car no.

Make

Model

    Price used

Age

1

Citroen

Saxo

1180

10

2

Mercedes

E-Class 2000

11395

6

3

Peugeot

206

2780

8

4

Rover

25

2970

7

5

Ford

Probe

1055

11

6

Suzuki

Liana

4340

4

7

Volvo

S80

9040

5

8

Mercedes

SLK

25995

4

9

Rover

75

4980

7

10

Mercedes

SLK

19955

6

11

Nissan

Micra

860

11

12

Fiat

Bravo

1885

8

13

Renault

Laguna

6630

5

14

BMW

525i SE

3480

11

16

Ford

Ka

2090

9

17

Vauxhall

Agila

2595

6

18

Rover

Mini

1190

11

19

Volvo

440

1155

11

20

Ford

Fiesta 95-99

755

11

21

Mercedes

A-Class

4635

8

22

Audi

A2

7100

6

23

Honda

Prelude

1810

11

24

BMW

3-Series 91-99

12825

5

25

Nissan

Terrano

3435

10

26

Ford

Focus

8500

5

27

Citroen

AX

1080

11

28

Ford

Fusion

6020

4

29

Mitsubishi

Colt

2665

7

30

Mercedes

E-Class 2000

24435

4

31

Skoda

Fabia

3585

6

image06.png

image01.png

...read more.

Middle

13660

1320

90.3

47

Subaru

Justy

64000

4

8995

3610

59.9

54

Ford

Escort

29000

11

13310

900

93.2

57

Honda

Jazz

13000

4

11300

8260

26.9

59

Audi

Coupe 88-96

27000

9

22295

6695

70

62

Fiat

Cinquecento

66000

11

6450

665

89.7

67

Seat

Ibiza 2003

7200

3

9030

6315

30.1

68

Ford

Mondeo 93-96

34000

11

12255

690

94.4

73

Mercedes

Cab 93-97

18500

9

51825

14225

72.2

78

Audi

80 Cabriolet

96000

9

19430

4125

78.8

79

Subaru

Forester

50000

11

16945

4550

73.1

82

Fiat

Seicento

5000

5

5980

1915

68

83

Land Rover

Discovery

43000

7

30805

8715

71.7

88

Daewoo

Tacuma

55000

6

12495

4675

62.6

89

Mercedes

CLK 97-03

37000

9

28770

10075

65

90

BMW

5-series 88-96

49000

11

52655

5160

90.2

100

Nissan

100 NX

43000

11

13500

1005

92.6

101

Fiat

Punto

52000

4

8625

4550

47.2

120

Nissan

Almera

90000

2

13075

9075

30.6

122

Land Rover

Discovery

60000

3

29995

19345

35.5

132

Mercedes

CL 91-99

21000

11

101975

15105

85.2

134

Ford

Galaxy

10000

11

18745

3885

79.3

137

Seat

Toledo

20000

9

14025

1410

89.9

144

Suzuki

Baleno

60000

10

10170

900

91.2

149

Ford

Galaxy 2000

10000

6

25597

8990

64.8

150

Ford

fiesta

10000

4

9450

5050

46.6

154

Alfa Romeo

Spider

12000

3

28282

16435

41.9

157

...read more.

Conclusion

X = 0.58 and Y = 0.42
  1. Car No – 89 is a Mercedes CLK 97 – 03 aged 9, its mileage 3700 and its % depreciation is 65.

% depreciation = X ((9.5x9) +8) + Y ((0.0005x37000) +24)

  • When X= 1
  • And Y = 0

1((9.5x9) +8) + 0((0.0005x37000) +24)

1(85.5+8)           0(18.5+24)

   =93.5            = 0

93.5+0=93.5 TOO BIG

  • When X= 0.7
  • And Y = 0.3

0.7((9.5x9) +8) + 0.3((0.0005x37000) +24)

0.7(85.5+8)           0.3(18.5+24)

=65.45                  =12.75

65.45+12.75=78.2 TOO BIG

  • When X= 0.5
  • And Y = 0.5

0.5((9.5x9) +8) + 0.5((0.0005x37000) +24)

0.5(85.5+8)           0.5(18.5+24)

=46.75                     =21.25

46.75+21.25=68 TOO BIG

  • When X= 0.45
  • And Y = 0.55

0.45((9.5x9) +8) + 0.55((0.0005x37000) +24)

0.45(85.5+8)           0.55(18.5+24)

=42.075                      =23.375

42.075+23.375=65.45 TOO BIG

  • When X= 0.445
  • And Y = 0.555

0.445((9.5x9) +8) + 0.555((0.0005x37000) +24)

0.445(85.5+8)           0.555(18.5+24)

=41.6075                    =23.5875

41.6075+23.5875=65.195

65.195~65

Therefore I have concluded that X = 0.445 and Y = 0.555

  1. Car No – 173is a Mazda Premacy aged 7, it’s mileage 18000and its % depreciation is 69.

% depreciation = X ((9.5x7) +8) + Y ((0.0005x18000) +24)

  • When X = 1
  • And Y= 0

1((9.5x7) +8)   + 0((0.0005x18000) +24)

1(66.5+8)            0(9+24)

=74.5                    =0

74.5+0=74.5 TOO BIG

  • When X = 0.8
  • And Y= 0.2

0.8((9.5x7) +8)   + 0.2((0.0005x18000) +24)

0.8(66.5+8)             0.2(9+24)

=59.6        =6.6

59.6+6.6=66.2 too small

  • When X = 0.9
  • And Y= 0.1

0.9((9.5x7) +8)   + 0.1((0.0005x18000) +24)

0.9(66.5+8)            0.1(9+24)

=67.05                       =3.3

67.05+3.3=70.35 TOO BIG

  • When X = 0.88
  • And Y= 0.12

0.88((9.5x7) +8)   + 0.12((0.0005x18000) +24)

0.88(66.5+8)           0.12(9+24)

=65.56                        =3.96

65.56+3.96=69.52 TOO BIG

  • When X = 0.87
  • And Y= 0.13

0.87((9.5x7) +8)   + 0.13((0.0005x18000) +24)

0.87(66.5+8)           0.13(9+24)

=64.815                        =4.29

64.815+4.29=69.105

69.105~69

Therefore I have concluded that X = 0.87and Y= 0.13

I will now work out the average value of X and Y by:

  • Adding all the Xs together and dividing by 5
  • Adding all the Ys together and dividing by 5
  • X = 0.67and Y = 0.33
  • X = 0.965 and Y = 0.035
  • X = 0.58 and Y = 0.42
  • X = 0.445 and Y = 0.555
  • X = 0.87and Y= 0.13
  • 0.67 + 0.965+ 0.58 + 0.445 + 0.87 = 3.53

3.53/5 = 0.706 ~ 0.71

So X= 0.71

  • 0.33 + 0.035 + 0.42 + 0.555 + 0.13= 1.47

1.47/5 = 0.294~ 0.29

So Y=0.29

...read more.

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