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  • Level: GCSE
  • Subject: Maths
  • Word count: 13867

Used car prices.

Extracts from this document...

Introduction

Used car prices

Introduction

During this coursework I will be working from the form the data that has been given to me. It is established on figuring out and representing the data in different forms. The table below is the given data that is to be interpreted:

Car

Make

Model

Price when new

Second hand price

Age (Years)

Mileage

Engine Size

1

Ford

Orion

16000

7999

1

7000

1.8

2

Mercedes

A140 Classic

14425

10999

1

14000

1.4

3

Vauxhall

Vectra

18580

7999

2

20000

2.5

4

Vauxhall

Astra

14325

6595

4

30000

1.6

5

Nissan

Micra

7995

3999

3

37000

1

6

Renault

Megane

13610

4999

4

33000

1.6

7

Mitsubishi

Carisma GDI

14875

5999

2

24000

1.8

8

Rover

623 Gsi

22980

6999

4

30000

2.3

9

Renault

Megane

13175

6999

3

41000

1.6

10

Vauxhall

Tigra

13510

7499

4

27000

1.4

11

Fiat

Bravo

10351

3495

5

51000

1.4

12

Vauxhall

Vectra

18140

6499

4

49000

2.5

13

BMW

525i SE

28210

5995

8

55000

2.5

14

Vauxhall

Corsa

8900

4995

2

24000

1.6

15

Fiat

Punto

8601

3995

4

31000

1.2

16

Rover

820 SLi

21586

3795

6

51000

2

17

Mitsubishi

Carisma

15800

5999

2

33000

1.8

18

Fiat

Cinquecento

6009

1995

6

20000

0.9

19

Rover

416i

13586

3795

6

49000

1.6

20

Nissan

Micra

6295

1795

8

47000

1.2

21

Daewoo

Lanos

11225

5999

3

42000

1.6

22

Rover

114 Sli

8595

2495

6

33000

1.4

23

Ford

Escort

8785

1595

7

68000

1.3

24

Fiat

Uno

6864

1495

8

51000

1

25

Rover

Metro

6645

895

7

43000

1.1

26

Vauxhall

Nova

5599

1000

10

75000

1.4

27

Toyota

Corrolla

13800

7495

2

25000

2

28

Vauxhall

Cavalier

10150

850

10

73000

1.6

29

Volkswagen

Golf

400

15

1.4

30

Volkswagen

Golf

9524

3695

7

49000

1.4

31

Seat

Ibiza

5995

795

7

45000

0.9

32

Rover

214i

9565

1700

8

55000

1.4

33

Ford

Fiesta

7310

1050

8

90000

1.1

34

Fiat

Tempra

10423

1295

6

81000

1.6

35

Ford

Fiesta

7875

1495

11

74000

1.8

36

Hyundai

Sonnata

11598

1195

9

65000

2

37

Renault

Clio

6795

1995

8

47000

1.2

38

Citroen

Debut

5715

1495

7

50000

0.95

39

Renault

Clio

7403

1495

9

98000

1.2

40

Fiat

Tipo

8272

1500

7

32000

1.4

41

BMW

316i

13650

6995

6

71000

1.8

42

Citroen

AX Diesel

7680

1295

6

1.5

43

Ford

Fiesta LX

8748

1995

7

60000

1.1

44

Nissan

Micra

5340

1595

9

40000

1.1

45

Ford

Escort Duet

9105

2300

7

64000

1.4

46

Nissan

Sunny

7799

2595

7

41000

1.4

47

Vauxhall

Astra

13740

2900

6

58000

1.6

48

Hyundai

Accent

6899

2800

6

49000

1.3

49

Daewoo

Nubira

13850

6895

4

14730

2

50

Daewoo

Lanos

9525

4395

3

32400

1.4

51

Volkswagen

Golf

12999

3595

6

58000

1.8

52

Ford

Escort

12125

4295

3

29000

1.4

53

Ford

Escort

11800

4700

5

34000

1.8

54

Bentley

TurboR

170841

37995

8

55000

2

55

Fiat

Punto

7864

4500

3

13000

1.2

56

Lexus

LS400

39728

6250

7

4

57

Ford

Fiesta

8680

3200

4

27000

1.8

58

Nissan

Almera

12590

4300

4

17000

1.6

59

Rover

623GSi

24086

2975

5

96000

2.3

60

Rover

620Si

17795

3400

5

66000

2

61

Fiat

Bravo

10954

6795

1

3000

1.2

62

Peugot

406LX

13975

5795

3

53000

1.8

63

Volkswagen

Golf GTi

16139

6995

5

35000

2

64

Ford

Focus

14505

8800

2

7200

1.6

65

Ford

Puma

13230

8250

3

34000

1.4

66

Peugot

206

9125

7500

1

18000

1.1

67

Peugot

406

...read more.

Middle

Mid-interval (m)

f*m

4 ≤ x < 6

-

0

0

5

0

6 ≤ x < 8

III

3

3

7

21

8 ≤ x < 10

IIII

5

8

9

45

10 ≤ x < 12

I

1

9

11

11

12 ≤ x < 14

III

3

12

13

39

14 ≤ x < 16

II

2

14

15

30

16 ≤ x < 18

II

2

16

17

34

I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be easily seen that 5 is the highest frequency. Therefore mode is equal to £8k ≤ x < £10k. This shows me that this is the most frequent price. It is the central tendency from all the Ford cars.  

Secondly I shall calculate the range. This will show me the price between the lowest and the highest price of this car make. The range can be calculated as the following:

Range = Highest Price – Lowest Price

Range = 18 – 4 = £14k

The mean can be found by dividing the number of cars by means of the number of makes.

This can be written as the following:

Mean = ∑ Frequency × Mid-interval

∑ Frequency

Mean = 21 + 45 + 11 + 39 + 30 + 34 ÷ 16

Mean = £11.25k

The median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median:

Median = n + 1 ÷ 2

Where n is equal to highest cumulative frequency

Median = n + 1 ÷ 2

Median = 16 + 1 ÷ 2

Median = 17 ÷ 2

Median = 8.5

Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 8.5 lies between 8 and 9. It can now be said that the median is equal to 10 ≤ x < 12.

Therefore, Median = 10 ≤ x < 12

Lower Quartile, Upper Quartile and Inter Quartile Range – Ford  

To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range.  

Median = n + 1 ÷ 2

Median = 16 + 1 ÷ 2

Median = 8.5

I now looked at 8.5 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the x-axis. This has been shown on the next page.

Hence, Median is equal to £10.5k

To find the lower quartile (LQ) I will use the following formula:

¼ (n + 1)

Where ‘n’ again is known as the highest cumulative frequency.

¼ (n + 1)

¼ (16 + 1)

¼ (17)

17 ÷ 4 = 4.25

I now looked at 4.25 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the x-axis. This has been shown on the next page

It is now known that the lower quartile is equal to £8.4k. This is the value of one-quarter way into the division.  

The same steps are to be considered while finding out the upper quartile, but this time the following formula was used:

Upper Quartile = ¾ (n + 1)

¾ (16 + 1)

¾ (17)

17 х 3 ÷ 4

51 ÷ 4

12.75

The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the x-axis is £15k.

UQ = £14.5k

Now with the known upper and lower quartile I can determine calculating the inter-quartile range. The following formula shows how to calculate the inter-quartile range:

Inter- quartile range (IQR) = Upper quartile – Lower quartile

IQR = 14.5 – 8.4

I.Q.R = £6.1k

image49.pngimage12.pngimage13.pngimage11.pngimage11.pngimage01.png
image14.pngimage15.png
Mean, mode, median and range – Vauxhall  

I now have shown the mean, mode, median and range of a specific make of car. I have taken Vauxhall into consideration. The mean, mode and median will be found by using the prices of Vauxhall cars when new. From the following table the mean, mode, median and range can be found:

Price of new cars (£k)

Tally

Frequency (f)

Cumulative Frequency

Mid-interval (m)

f*m

4 ≤ x < 6

I

1

1

5

5

6 ≤ x < 8

II

2

3

7

14

8 ≤ x < 10

II

2

5

9

18

10 ≤ x < 12

I

1

6

11

11

12 ≤ x < 14

III

3

9

13

39

14 ≤ x < 16

I

1

10

15

15

16 ≤ x < 18

-

0

10

17

0

18 ≤ x < 20

III

3

13

19

57

I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be seen that 3 is the highest frequency because there are two numbers that are 3. This is known as a bi-modal. This basically means that there are two highest frequencies in the Vauxhall model. Therefore mode is equal to £12k ≤ x < £14k and £18k ≤ x < £20k. This shows me that this is the most frequent price. It is the central tendency from all the Vauxhall cars.  

Secondly I shall calculate the range. The range can be calculated as the following:

Range = Highest Price – Lowest Price

Range = 20 –4 = £16k

The mean can be found by dividing the number of cars by means of the number of makes. This can be written as the following:

Mean = ∑ Frequency × Mid-interval

∑ Frequency

Mean = 5 + 14 + 18 + 11 + 39 + 15 + 0 + 57 ÷ 19

Mean = £8.37k

The median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median:

Median = n + 1 ÷ 2

Where n is equal to highest cumulative frequency

Median = n + 1 ÷ 2

Median = 13 + 1 ÷ 2

Median = 14 ÷ 2

Median = 7

Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 7 lies between 6 and 18. It can now be said that the median is equal to 12 ≤ x < 14.

Therefore, Median = £12k ≤ x < £14k

Lower Quartile, Upper Quartile and Inter Quartile Range – Vauxhall  

To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range.

Median = n + 1 ÷ 2

Median = 13 + 1 ÷ 2

Median = 7

I now looked at 7 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the x-axis. This has been shown on the next page.

Hence, Median is equal to £13k

To find the lower quartile (LQ) I will use the following formula:

¼ (n + 1)

Where ‘n’ again is known as the highest cumulative frequency.

¼ (n + 1)

¼ (13 + 1)

¼ (14)

14 ÷ 4 = 3.5

I now looked at 3.5 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the x-axis. This has been shown on the next page

It is now known that the lower quartile is equal to £8.5k. This is the value of one-quarter way into the division.  

The same steps are to be considered while finding out the upper quartile, but this time the following formula was used:

Upper Quartile = ¾ (n + 1)

¾ (13 + 1)

¾ (14)

14 х 3 ÷ 4

42 ÷ 4

10.5

The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the x-axis is £18.5k.

UQ = £18.5k

Now with the known upper and lower quartile I can determine calculating the inter-quartile range. The following formula shows how to calculate the inter-quartile range:

Inter- quartile range (IQR) = Upper quartile – Lower quartile

IQR = 18.5 – 8.5

I.Q.R = £10k

image50.pngimage16.png

Mean, mode, median and range – Fiat  

The third make of car I shall investigate is Fiat. I have shown a table below, with all the necessary details that I will be using during the calculations of discovering the mean, mode, median and range.

Price of new cars (£k)

Tally

Frequency (f)

Cumulative Frequency

Mid-interval (m)

f*m

4 ≤ x < 6

-

0

0

5

0

6 ≤ x < 8

IIII

4

4

7

28

8 ≤ x < 10

II

2

6

9

18

10 ≤ x < 12

IIII

4

10

11

44

I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be seen that 4 is the highest frequency because there are two numbers that are 3. This is known as a bi-modal. This basically means that there are two highest frequencies in the Fiat model. Therefore mode is equal to £6k ≤ x < £8k and £10k ≤ x < £12k. This shows me that this is the most frequent price. It is the central tendency from all the Fiat cars.  

Secondly I shall calculate the range. This will show me the price between the lowest and the highest price of this car make. The range can be calculated as the following:

Range = Highest Price – Lowest Price

Range = 12 – 4 = £8k

The mean can be found by dividing the number of cars by means of the number of makes.

This can be written as the following:

Mean = ∑ Frequency × Mid-interval

∑ Frequency

Mean = 0 + 28 + 18 + 44  ÷ 10

Mean = £9k

The median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median:

Median = n + 1 ÷ 2

Where n is equal to highest cumulative frequency. The added one is done so that the figure can be rounded off to give a most accurate result that could be obtained.  

Median = n + 1 ÷ 2

Median = 10 + 1 ÷ 2

Median = 11 ÷ 2

Median = 5.5

Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 5.5 lies between 4 and 6. It can now be said that the median is equal to 8 ≤ x < 10.

Therefore, Median = 8 ≤ x < 10

Lower Quartile, Upper Quartile and Inter Quartile Range – Fiat  

To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range.  

Median = n + 1 ÷ 2

Median = 10 + 1 ÷ 2

Median = 5.5

I now looked at 5.5 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the x-axis. This has been shown on the next page.

Hence, Median is equal to £8.4k

To find the lower quartile (LQ) I will use the following formula:

¼ (n + 1)

Where ‘n’ again is known as the highest cumulative frequency.

¼ (n + 1)

¼ (10 + 1)

¼ (11)

17 ÷ 4 = 2.75

I now looked at 2.75 on the y-axis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the x-axis. This has been shown on the next page. It is now known that the lower quartile is equal to £7.4k. This is the value of one-quarter way into the division.  

The same steps are to be considered while finding out the upper quartile, but this time the following formula was used:

Upper Quartile = ¾ (n + 1)

¾ (10 + 1)

¾ (11)

11 х 3 ÷ 4

33 ÷ 4

8.25

The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the x-axis is £11.4k.

UQ = £11.4k

Now with the known upper and lower quartile I can determine calculating the inter-quartile range. The following formula shows how to calculate the inter-quartile range:

Inter- quartile range (IQR) = Upper quartile – Lower quartile

IQR = 11.4 – 7.4

I.Q.R = £4k

image51.pngimage03.pngimage17.pngimage04.pngimage02.png
image05.png
image07.pngimage06.png

Box whisker diagrams

In this part pf my coursework I shall produce box whisker diagrams for each of these makes that I have calculated the mean, mode, median, range, upper quartile, lower quartile and inter quartile range. After making the box whisker diagrams I shall compare each of the diagrams produced. This will also tell me how much tendency there is from the central point. The box whisker diagram shall look like the diagram below:

image08.png

I will be producing all the box whisker diagrams on the same scale this will allow me to compare them easily. I have shown the box whisker diagrams for three makes below:

image09.png

image10.png

Conclusion on box whisker diagrams

By looking at the previous page, which shows the three box whisker diagrams for Ford, Vauxhall and Fiat, it can be seen that the Vauxhall is the widest drawn diagram and Fiat is the smallest.

Once again the median shows the middle price of the make of car. By visual observation id can be seen that Vauxhall once again has the largest median and the Fiat has the smallest median. This tells me that the central tendency of Vauxhall is higher than any of the other two makes. If the middle price of the Vauxhall make of cars is taken into consideration that would be more expensive than the other makes, this will not be suitable for a normal working class person. I would prefer buying the Fiat make of cars, which has a not so expensive central tendency. The Ford is the average car between these two makes.

It can also be seen that Vauxhall has cars that are of a higher range than the other two makes that I have been investigating. This is a good feature, as the person that will buy the vat will obviously buy a car, which is reliable to his/her pocket, and this may vary to very rich and not so rich. There I mean to say that there is a wide choice of choosing a very expensive car or which will be reasonable and affordable, by working class people.

I have also notices that comparison between the medians to the upper quartile also varies. Vauxhall again has the more dispersion. This is due to the contributing factors that raise the price of the second hand car. There can be many contributing factors that raise the used car price. Some of them have been mentioned below:

  • The engine size: the more the engine sizes then more the cost of the car.
  • The number of doors: the more the doors can also raise the price.
  • The style: this is defining a contributing factor because sport cars tend to have a more expensive price than the hatch or estate cars.
  • Central locking: this can hoist the price, as it is expensive to have in a car.  
  • Number of seats: this can raise the price.
  • Gearbox: this tends to have an effect on the price as well, as the automatic cars verge to have a highest price tan the manual cars.
  • Air conditions.

I would prefer a sales man that has to travel up and down different states most probably drives the Vauxhall make of. A normal working class person that travels to and from work every weekday would most probably drive the Fiat. And the Ford is a mixture of both.

Second hand price and new price

I know shall begin the second part of the coursework. In this part of the coursework I will be comparing the current price (second hand price) and the original cost (price when new) of the each make of car. I am investigating this, as I will be able to judge which make of car decrease in value the most and which preserves most value. I will be scrutinizing five make of cars. The following are the make of cars that I shall be investigating:

  • Peugeot
  • Fiat
  • Rover
  • Ford
  • Vauxhall

Peugeot

I shall begin with Peugeots; the data for this make of car has been shown on the table below:

Car Number

Price when new (£) (n)

Second hand price (£) (s)

Depreciation (s/n)

1

7600

2497

0.42

2

9125

7500

0.82

3

12350

3995

0.43

4

13875

5795

0.33

5

17490

7500

0.32

The above data has been represented on the following bar chart. This bar chart has especially been created as it clearly shows the comparisons between the cost of each car when new and the second hand price for a certain years of age.

image52.png

By looking at the bar chart it is obviously known that the price when new is higher than the second hand price. It clearly represents a comparison between them both. The age is a contributing factor to the depreciation of the price. Many other things can also act as a contributing factor such as door number and engine size.

Fiat

The next make of car that I will be investigating will be Fiat. The same steps will be taken into consideration as for the Peugeot.      

The data for this make of car has been shown on the table below:

Car Number

Price when new (£) (n)

Second hand price (£) (s)

Depreciation (s/n)

Age (Yrs)

1

6009

1995

0.33

6

2

6864

1495

0.22

8

3

7518

3769

0.50

4

4

7864

4500

0.57

3

5

8272

1500

0.18

7

6

8601

3995

0.46

4

7

10351

3495

0.34

5

8

10423

1295

0.12

6

9

10810

4995

0.46

2

10

10954

6795

0.62

1

Many factors can also be a contributing factor here that shows more depreciation in the second hand price. These can be for instance the age of the car as the more older the car is the more depreciation will be caused or the larger the engine size the more the price will be when new and then it would reduce in price much more faster as the person that would buy it would not buy a car with a rather large engine as this will cost him/her more money on fuel, therefore the owner would have to bring the price down so that it could be sold. Therefore depreciation can be caused by many contributing factors.    

The following bar chart shows that the cost of buying a car brand new is always more than the current price of the car:

image20.png

Rover

The data for this make of car has been shown on the table below:

Car Number

Price when new (£) (n)

Second hand price (£) (s)

Age (Yrs)

Depreciation (s/n)

8

19530

14999

1

0.77

10

22980

6999

4

0.30

7

17795

3400

5

0.19

3

8595

2495

6

0.29

5

13586

3795

6

0.28

6

14486

3685

6

0.25

9

21586

3795

6

0.18

11

24086

2975

6

0.12

1

5495

1995

7

0.36

2

6645

895

7

0.13

4

9565

1700

8

0.18

The above data has been epitomized in the bar graph below:

image21.png

Looking at the graph it is easily noticed that the decrease in the original cost of the car has reduces rapidly.  For example Car number 7 has decreased from its original price to a very low second hand price. I have also noticed that car number 4 is an anomaly, this is because its second hand price is more than car number 2 which is 7 years old whereas car number 4 is 8 years old.  Ignoring car number 4 that has been taken as an anomaly, I have noticed that the more the age of the car the more the car depreciates.  

Ford

The same steps are taken into deliberation. The data for this make of car has been shown on the table below:

Car Number

Price when new (£) (n)

Second hand price (£) (s)

Age (Yrs)

Depreciation (s/n)

1

6590

1664

10

0.25

2

7310

1050

8

0.14

3

7875

1495

11

0.19

4

8680

3200

4

0.37

5

8748

1995

7

0.23

6

8785

1595

7

0.18

7

9105

2300

7

0.25

8

9995

2995

6

0.30

9

11800

4700

5

0.40

10

12125

4295

3

0.35

11

13183

3495

7

0.27

12

13230

8250

3

0.62

13

14505

8800

2

0.61

14

15405 ...read more.

Conclusion

Conclusion

I have accomplished this coursework and I have come to some conclusions. These conclusions were regarding the relationships of all the investigations I have taken into consideration.

Firstly in the price and cost investigation the original cost of all the cars was always more that the second hand price. This is obvious as first hand price is always more than the second hand price. This can is seen clearly by looking at the bar charts. By looking at the graphs it can be noticed that Peugeot retains most value and depreciates the slightest from all other makes of cars. It can also be noticed that Rover depreciates the most and retains its original value the least.

The investigation that was concentrating on the age and price of all the five makes I took into consideration were all negative correlations. This shows that as the age will increase the price value would decrease. This can also be noticed by looking at the gradient that was calculated for all the make of cars. It can be noticed that Rover, depreciates the most out of all the makes. It can also be seen that Peugeots depreciate the least and retain the most value out of all the makes.

The positive correlations were the ‘age and mileage’ and the ‘cost and engine size’. This means that as the age of the car increases the mileage increases alongside it and as the engine size of the car increases the cost of the car also increases.

Lastly by calculating the standard deviation I was able to become aware of the average price of all the five makes I had taken into consideration. This told me when car was the most expensive and which was the least expensive. By taking standard deviation into consideration I have found out that Rover is the most expensive car to purchase, whereas Fiats are the cheapest cars to purchase.

Anjum Kohli         Used Car Prices         GCSE Maths Coursework

...read more.

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