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• Level: GCSE
• Subject: Maths
• Word count: 2537

# MayfieldHigh School Statistics Project.

Extracts from this document...

Introduction

Samuel .J. Verlander                                                                                                        Statistics Coursework                           2003

Mayfield High School Statistics Project

For this year 11 GCSE mathematics coursework, I was given statistics on all of the children in Mayfield high school from year 7 to 11. The title Mayfield high school is fictional, but the statistics are from a real school in America. The vast amount of data that I was provided with is:

 Year Group Number of Boys Number of Girls Total 7 151 131 282 8 145 125 270 9 118 143 261 10 106 94 200 11 84 86 170 Total 604 579 1183

As the above table shows, there are 1183 students in the school, from year 7 to year 11. In addition to this data, more has been provided on each of the students, such as:

1. Name

2. Age

3. Year Group

4. IQ

5. Weight

6. Height

7. Hair colour

8. Eye colour

9. Distance from home to school

10. Usual method of travel to school

11. Number of brothers or sisters

12. Key Stage 2 results in English mathematics and science

There is a total of 27 different sets of data. This data is provided on every student from years 7 to 11.

From this the total number of datum points can be taken by multiplying the total number of students (1183) by the number of different data types on each of the students (27) giving the sum 1183 x 27 = 31941 datum points.

This project is about statistics, so I will be using this area of mathematics and this information combined to produce a project on the statistical values of some of the lines of enquiry of this data.

There is far too much data to analyze it all; it would take too much time and effort. Therefore in this project a sample of the whole sum of students will have to be used.

Middle

To obtain the 5 random numbers from the list of 13, two figure numbers between 01 and 13 would be needed. If the random starting location was the fourth figure in the second row, and reading across in groups of two because two figure numbers are needed, the numbers obtained are:

84,39,74,52,26,36,32,03,24,74,00,55.......

All duplicates and numbers larger than 13 must be discarded, this process continues until 5 numbers between 01 and 13 have been found. This method is too lengthy to be used; luckily, there is another, faster way in which the 5 digit can be obtained. This is to take the first five numbers and turn them into decimals. In doing this, the numbers obtained are:

0.84, 0.39, 0.74, 0.52, 0.26

They must then be multiplied by the total sum of cards that are being sampled. These random digits will then therefore become:

0.84 x 13 = 10.92 rounded = 11

0.39 x 13 = 5.07   rounded = 5

0.74 x 13 = 9.62   rounded = 10

0.52 x 13 = 6.76   rounded = 7

0.26 x 13 = 3.38   rounded = 3

As the numbers being dealt with are all whole, i.e. there are no half cards or bits of cards in a suit, the numbers are rounded, leaving the random samples that were being looked for at the beginning.

This can be a long and laborious process, and takes much time. However, thankfully there is a much shorter way of producing the results, by using a calculator. I will be using this process for the duration of the project. Most calculators now have a key that produces random numbers, by using this key five times; the desired number of random numbers is produced between 0 and 1. These are then multiplied by the total number of data, giving the random sample.

Conclusion

This prediction does not need a great deal of research to prove. For this prediction, I will need bar charts. The average of boys and girls in each class will be placed next to each other and the overall average at the end. The class average shall be taken from the samples boys and girls in the year, averaged, and placed onto the bar next to the opposite gender. The overall result will be taken by a) showing how many bars are higher for each gender and b) by taking the overall average of all the samples of each year.

Averages

Average height of boys in year 7 = (1.50 + 1.67 + 1.45 + 1.49 + 1.52 + 1.65 + 1.54 + 1.55)/8

=1.55m  sd: 0.072

Average height for Girls in year 7= (1.80 + 1.62 + 1.60 + 1.51 + 1.42 +1.56 + 1.54)/7

= 1.6m  sd: 0.11

Average height for boys in year 8 = (1.75 + 1.83 + 1.72 + 1.60 + 1.82 + 1.90 + 1.71)/7

= 1.56m  sd: 0.091

Average height for girls in year 8 = (1.74 + 1.69 + 1.72 + 1.54 + 1.52)/5

= 1.64m  sd: 0.093

Average height for boys in year 9 = (1.54 + 1.78 + 1.77 + 1.65 + 1.32 + 1.70)/6

= 1.63m  sd: 0.16

Average height for girls in year 9 = (1.6 + 1.75+ 1.76 + 1.65 + 1.65 + 1.59 + 1.62 + 1.58)/8

= 1.44m  sd: 2.58

Average height for boys in year10 = (1.72 + 1.57 + 1.75 +1.50 + 1.62 + 1.73)/6

= 1.65m  sd: 0.092

Average height for girls in year 10 = (1.55 + 1.55 + 1.53 + 1.55 + 1.72)/5

= 1.58  sd:

Average height for boys in year 11 = (1.8 + 1.67 + 1.61 + 1.8)/4

= 1.72m  sd: 0.082

Average height for girls in year 11 = (1.63 + 1.62 + 1.56 + 1.75)/4

=1.64m  sd: 0.069

As the graph shows, the boys in the 3rd to 5th quarter are more than in the 1st and 2nd.

Prediction 2-I predict that children in Mayfield who live further from the school will weigh less than those who live closer, due to the fact the journey gives them more exercise.

This prediction can be solved by taking the children from each year and plotting their weight against the distance they travel to school. In this scatter graph, a negative correlation will be needed to prove the theorem. This is because the further from the school that they get the lighter they will be.

This student written piece of work is one of many that can be found in our GCSE Height and Weight of Pupils and other Mayfield High School investigations section.

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