- Level: AS and A Level
- Subject: Maths
- Word count: 2232
Maths Statistics Investigation
Extracts from this document...
Introduction
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 |
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
Conclusion
- 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
- 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
This student written piece of work is one of many that can be found in our AS and A Level Probability & Statistics section.
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