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• Level: GCSE
• Subject: Maths
• Word count: 1731

# Maths Mayfield High School Data Handling

Extracts from this document...

Introduction

I have been presented with data of secondary nature, about a school named Mayfield high school. This is table shows the number of pupils. 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 There 1183 students in this high school, and they have carried out several surveys, and put information on every single students into their own record on a database. The database contains several kinds of information, for example, name, age, year group, IQ, weight, height, eye colour, hair colour, test results, etc. The variations I have chosen to follow for my coursework are: 1. The relationship between height and weight Keeping in mind that there are 1183 students, I cannot provide these enquiries onto each pupil. I must take a suitable sample. Sampling helps to pick and choose some data needed to gain a result. Here are the methods available: Year group Total number of students Number students to be taken 7 282 282/1183 x 100 = 24 8 270 270/1183 x 100 = 23 9 261 261/1183 x 100 = 22 10 200 200/1183 x 100 = 17 11 170 170/1183 x 100 = 14 Total students = 100 The students taken must be taken at random. ...read more.

Middle

The equation is y = mx + c. To find the gradient this could also be shown as: y y1 m = x1 y1 x1 c > Minimizing bias - throughout the coursework, problems will arise regarding being bias. This means in some cases I may favour someone else or be unfair. This is not good because it affects my results. For example I may choose more girls than boys or more year 11 than year 7. This can be avoided by using different kinds of sampling, which suits the situation. This way my results will remain fair. > Mean of deviations from the mean - this is another measure of spread. This again provides evidence for comparison. From this you can work out the how far away a height is from the average. Height - mean height = deviation Although if you wish to work out the mean of the deviations, you must work all the deviations out then use this equation: ?|x - x| = Mean of deviations n When working this out you ignore the negative values a deviation may have. After the calculations are made you can compare various types of data found out. ...read more.

Conclusion

I will now work out the mode. From my results the mode numbers for height are 1.72m, 1.65m, 1.62m and 1.60m. The mode for weight is 50 kg. From my results the median height is: 192.52 + 1 = 96.76th position 2 Which is 1.70m From my results the median weight is: 6008 + 1 = 3004.5th position 2 Which is 85kg From my results the height average is: 192.52 Mean = 118 = 1.632m The average weight is: 6008 Mean = 118 = 50.92kg Therefore we learn the average person is 50.92kg in weight, and 1.632m in height. I will also work out the mean of the deviations, The mean deviation for height is: 1501.6 Mean deviation = 118 = 12.73cm The mean deviation for weight is: 1088.720 Mean deviation = 118 = 9.226kg Using the equation on my graph, I can work what height or weight a person will be based on the line of best fit. If a person weighs 50kg, y= 0.0073 x 50 + 1.2482 =1.6078m is their height (from the line of best fit) If a person weighs 40kg, y= 0.0073 x 40 + 1.2482 =1.5402m is their height (from the line of best fit) Therefore from all of these calculations I have proven my hypothesis to be correct. This is because I have shown as the weight increases so to does the height. ?? ?? ?? ?? ...read more.

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