#### "Males in the 11-18years age range will guess the angles and lengths better than females in the 30+years age range."

INTRODUCTION: We have been given a task to work with the statistical side of maths. We were told that a random sample of one hundred and fifty people were asked to estimate the degree of an angle which looked like this: Also each person was asked to estimate the length of a line which looked something like this: From this we were asked to think about what sort of conclusions we could give for the analysed results. I thought about this and came up with the following hypothesis: "Males in the 11-18years age range will guess the angles and lengths better than females in the 30+years age range." I have chosen this hypothesis because it allows me to look at both age and gender of the surveyed people. (The actual measurement of the angle was 36.5degrees and the actual measurement of the line was 4.35cm) The data I have received from the survey is on the following page... PLAN: To investigate this hypothesis I may need to use a number of methods to calculate the data I will receive. I may need to use many statistical tools which I already know; such as Averages, Standard Deviation, Frequency Polygons, Histograms, Scatter Diagrams, Cumulative frequency and maybe other methods which I will have to research. As the sample I am being given is random, it has a large range of varied aged males and varied aged females. I need to divide the groups up into the groups I have to

#### I chose to investigate the relationship between the height and weight of year 7 students. I think that as the height increases, so will the weight.

Jenny Hodgson 11A Maths Statistics Coursework Hypothesis I chose to investigate the relationship between the height and weight of year 7 students. I think that as the height increases, so will the weight. This means there would be a positive correlation. I think that the correlation coefficient will be between 0.5 and 1 because that is strong positive correlation. -1 0 1 negative no positive correlation correlation correlation Method To investigate my hypothesis, I used given data from Mayfield High School. It contained information such as name, age, height, weight, IQ, eye colour, hair colour etc. Using a computer, I opened the data using Microsoft Excel. As I was only investigating height and weight of year 7 students, I deleted all the irrelevant columns, such as year 8, 9, 10, 11 students, IQ, hair colour, eye colour etc. I then sorted the data, again using the computer. I discarded my outliers in the data. I discarded data which either had very tall or very heavy people on and this may be false data, unreliable and affect my final result. I needed a sample of 50 students and I had 200 year 7 students. To get my sample of 50 students, I used

#### I am going to estimate the population of woodlice in parts of the school ground, using Lincoln Index.

ESTIMATING THE POPULATION OF WOODLICE IN PARTS OF THE SCHOOL GROUND Aim: - I am going to estimate the population of woodlice in parts of the school ground, using Lincoln Index. Materials White trays, gloves, nail polish and spoon. This investigation would be performed, using the mark-release recapture technique. Method A certain area is marked out and then by lifting up stones and pieces of decaying wood, woodlice found are picked up by using a spoon and placed in a white tray. The amount of woodlice caught is counted and then each woodlouse is marked with the nail polish in such a way that will not harm them or attract predators. The woodlice are then released into their habitat and allowed to mix with the rest of the population. After two days, woodlice are then caught from the same location and the amount caught is recorded, along with the number of the woodlice that have already been marked from the first sample. Result Number of Woodlice Original sample, S1 35 Second sample, S2 39 Number marked in second sample, m 2 Lincoln Index Total population = (S1 x S2)/ m Total population of woodlice = (35 x 39)/ 2 = 682.5 Approximately 683 woodlice are present in this area. This method is reliable if I assume that * The marked woodlouse mix evenly with the unmarked; * Being caught the first time does not increase or decrease the likelihood of the

#### Factors that most affect the prices of second hand cars

Maths Coursework Hypothesis For my maths project I will hope to find out the factors that most affect the prices of second hand cars. From my preliminary work I have chosen five factors that could possibly affect the prices the most. The five that I have chosen are: . Mileage 2. Age 3. Owners 4. Insurance Group 5. MPG (Miles Per Gallon) Evidently these factors will all be compared to the percentage decrease in price from new to second hand. By the end of the project I will hopefully have found the biggest factor in percentage decrease in price. In this project I will be using several different graphs including correlation coefficient, scatter graphs and histograms. For my scatter graphs I will use the mileage and the miles per gallon compared to the percentage decrease in price. For my correlation coefficient I will be using the age compared to the percentage decrease in price Prediction I predict that the mileage will affect the price of second hand car the most. This is because the mileage determines how much the car has been used and this leads to the quality of the engine. I also think that the number of owners and the age will affect the prices of second hand cars tremendously. I believe that the MPG will effect the prices of second hand cars significantly as the miles per gallon can show the cost of running the cars and if the MPG is high then that might

#### C3 Mei - Numerical Methods to solve equations

C3 Coursework In this coursework, I will use numerical methods to solve the following equation, as I cannot solve it algebraically. I can only obtain an approximation of the solution as it is impossible or hard to find the exact value of the function. Decimal Search The graph below is the function I will use decimal search in order to find an approximation of one of the roots. The table below shows decimal search. Each boundary is tested for sign change which indicates that a root exists between them. The x where the sign change occurs in now the new boundaries and tested for sign change again. This method is repeated until an approximation of the root is found to a suitable number of decimal places. x f(x) 0 2 0.1 .9501 0.2 .8016 0.3 .5581 0.4 .2256 0.5 0.8125 0.6 0.3296 0.7 -0.2099 0.8 -0.7904 0.9 -1.3939 -2 x f(x) 0.66 0.011747 0.661 0.006295 0.662 0.000838 0.663 -0.00462 0.664 -0.01009 0.665 -0.01556 0.666 -0.02104 0.667 -0.02652 0.668 -0.032 0.669 -0.03749 0.67 -0.04299 x f(x) 0.662 0.000838 0.6621 0.000292 0.6622 -0.00025 0.6623 -0.0008 0.6624 -0.00135 0.6625 -0.00189 0.6626 -0.00244 0.6627 -0.00299 0.6628 -0.00353 0.6629 -0.00408 0.663 -0.00462 x f(x) 0.6 0.3296 0.61 0.277958 0.62 0.225763 0.63 0.17303 0.64 0.119772 0.65 0.066006 0.66 0.011747 0.67 -0.04299 0.68 -0.09819

#### Solving Cubic equations or polynomials of greater order

Solving Cubic equations or polynomials of greater order Quadratic I firstly solved our quadratic equations using factorising, this worked well until I found equations that have factors that are not whole. I used the formula for these equations, although the formula is very hard to remember and if it's written incorrectly the answer will be wrong, so I tried using graphs this was ok although the graphs are not very accurate, they are only around 1 decimal place. Factorise X² - 6x + 5 (x - 5) (x - 1) = 0 x = 5 or x = 1 Solve using formula - b +/- V b² - 4ac 2a + 6 +/- V 36 - 4 x 1 x -5 2 x 1 + 6 + V 56 or + 6 - V 56 2 2 Cubic Here I first tried solving the equations by factorising, as with the quadratic equations this worked well until I came across equations with factors including decimal places. For these I tried as before to use the formula but I firstly need to find one of the factors, and if all factors include decimal places this can be difficult. So I lastly tried to solve the equation using graphs, this was extremely hard as the cubic graph needs more detail and I found that the graph as before was only correct to one decimal place. This is not accurate enough for a cubic graph. Now, I have decided to go and look for other ways in which it would be possible to find all the factors of the equations which is accurate. I have now found a new

#### Design, make and test a Sundial.

Sundial Coursework Statement of Task: To design, make and test a Sundial. Introduction: The sundial is the oldest know device for measuring time. The first confirmed uses of it where by the Babylonians in around 2000BC. However it is safe to say that shadows have been used as a rough measure of time ever since primitive man discovered that as the day progresses the shadows of trees and rocks get shorter and then longer again. The sundials used by the Babylonians were hollow half spheres, set with edges flat and with a small bead at the centre. As the day wore on the shadow of the bead followed a circular arc, which is divided into 12 "temporary hours" (they changed through the seasons). A modern sundial consists of a plane (dial face) and the gnomon (style). The dial face is divided into hours and the gnomon is the flat piece of metal, or stick in the dials centre, it points towards the North Pole (in the northern hemisphere) or South Pole (in the southern hemisphere). The gnomon must tilt at the angle of the latitude at the location. Although a sundial seems like a simple device for measuring time, it is not. One cannot simply look at the shadow and find the time. Firstly the sundial will show local time, not GMT. Secondly we must allow for the 'equation of time.' Because the Earth's orbit is not circular, its velocity changes at the perihelion. To allow for this

#### Statistic: Is reading age a predictor for future attainment?

Statistic: Is reading age a predictor for future attainment? In this coursework I am going to investigate if reading age affects future results in KS3 Maths. I am going to consider gender differences. I believe that girls will make more improvement than boys because girls have a higher reading age than boys; I also predict that reading age will not be a predictor for future attainments. In order to collect my data fairly I am going to take a stratified sample because it will randomly select boys and girls, and won't give me a bias selection. I am going to collect a random sample by using my calculator and the random button. I pressed in the number of girls and boys (96 for girls, 143 for boys) I then pressed the random button which gave me numbers between 0 and the number of boys and girls which I then take down and draw up a table from it. I am going to present my data using the following statistical techniques a cumulative frequency diagram so it can show both discrete and continues data. Scatter diagrams, I can use that to show whether two sets of data are related. Box & Whisker plot of reading ages, mode, mean, median, and range, IQR to show me how the data is distributed. I will then do the same for the boys and girls. This will tell me which genders achieving well depending on their reading age. To summarise the KS3 Maths data I will use Bar charts, Pie charts mean,

#### Numerical solution of equations

Pure Mathematic 2 Coursework Numerical solution of equations By Michael Pang I am going to show 3 of the numerical methods for solving the equation which cannot be solved algebraically. They are interval estimation, fixed point estimation and Newton-Raphson method. Those of these numerical methods are used when algebraic ones are not available. When you found any equations which cannot be solved algebraically, probably you will draw the graph and see where the roots are. However, the other problem is found that you cannot get all roots accuracy. Actually, the numerical methods cannot find the exactly root but the answers are more accuracy than sketching the graphs. Therefore, we usually provide the answer to 5 or 6 decimal places depending on what the questioner needs. Finally, we check the answer by setting the lower bound and upper bound to see whether it has sign change or not. Even if the three numerical can solve the equation non-algebraically, they have its advantages and disadvantages. And now I am going to show how these methods works and their problems by using the equation F(x) = x³-9x+3. Interval estimation Assume that the roots of the equation F(x) = x³-9x+3, and I am looking for the roots which F(x) = 0. The roots of the equation are the values of x which the graph of y = x³-9x+3 crosses the x axis. By using the computer, I recognise that there are 3

#### Mathematics portfolio - Translations.

Man Ju Y12D Mathematics - portfolio Translations . 2. is the effect of translation vector of . It moves up 2 units. is the effect of translation vector of . It moves up 4 units. is the effect of translation vector of . It moves down 3 units. 3. + 3 is the effect of translation vector of . It moves up 3 units. -1 is the effect of translation vector of . It moves down 1 unit. 4. Curves move either up or down vertically. The units they move is according to the number after x2 in the equation . If the number is positive, the curve will be pulled upwards. If the number is negative, the curve will be pulled downwards. 5. f(x) = sinx -2 is the effect of translation vector of f(x) = sinx. It moves down 2 units. This has the same effect with the previous examples. The unit it moves is according to the number after sinx in the equation f(x) = sinx c. If the number is positive, the curve will be pulled upwards. If the number is negative, the curve will be pulled downwards. Therefore the generalization extends to any function f(x). 6. 7. is the effect of translation vector of . It moves right 2 units. is the effect of translation vector of . It moves right 3 units. is the effect of translation vector of . It moves left 1 unit. is the