# Math Internal Assessment Gold Medal Heights

Gold Medal Heights The heights achieved by gold medalists in the high jump have been recorded starting from the 1932 Olympics to the 1980 Olympics. The table below shows the Year in row 1 and the Height in centimeters in row 2 Year| 1932| 1936| 1948| 1952| 1956| 1960| 1964| 1968| 1972| 1976| 1980| Height (cm)| 197| 203| 198| 204| 212| 216| 218| 224| 223| 225| 236| They were recorded to show a pattern year after year and to reveal a trend. The data graph below plots the height on the y-axis and the year on the x-axis.

**Math Internal Assessment Gold Medal Heights**specifically for you

Data Graph 1 Height (cm) Height (cm) Year Year

In Data Graph 1 the data shown represents the height in cm achieved by gold medalists in accordance to the year in which the Olympic games were held. The Graph shows a gradual increase in height as the years increase. The parameters shown in this are the heights, which can be measured during each year to show the rise. The constraints of this task are finding a function to fit the data point shown in Data Graph 1. Some other constraints would be that there aren’t any outliers in the graph and it has been a pretty steady linear rise. The type of function that models the behavior of the function is linear.

This type of function models it because the points resemble a line rather than a curve. To represent the points plotted in Data Graph 1 a function is created. To start deciphering a function I started with the equation – Y = mx + b To show the slope of the line since the function is linear. For the first point the function would have to satisfy 197 = m (1932) + b In order for the line to be steep the b value or y intercept will have to be low to give it a more upward positive slope. Y = mx -1000 197 = m (1932) -1000 1197 = m (1932) m = 0. 619

The final linear equation to satisfy some points would be y = 0. 62x – 1000 The graph below shoes the model linear function and the original data points to show their relationship. Graph 2 Year Year Height (cm) Height (cm) The graph above shows the linear function y = 0. 62x – 1000 in relation to the data points plotted on Data Graph 1. The differences between the function and the points plotted is that the function does not full satisfy all the x and y values. The outliers in this case are from the years 1948, 1952, and 1980 which all of y values that do not meet the function closely.

Using regression the following function and graph is found. The function and line found using regression matches the one found by me. The linear function does not cross all points but shows the gradual shape in which the points plotted make. Another function that is used is a quadratic function Quadratic functions are set up as: Y = px2 + tx +b To make this function resemble the points plotted on the Graph 1 the p value will have to be very small to widen the shape of the quadratic The b value also has to be small to resemble the y intercept and to give the graph a more upward slope

I used the function: Y = 0. 0000512×2 + 0. 5171x – 1010 In order for this function to work it must satisfy the point of (1964, 218) Y = 0. 0000512 (1964)2 + 0. 5171 (1964) – 1010 Y = 0. 0000512 (3857296) + 1015. 58 – 1010 Y = 197. 49 + 1015. 58 – 1010 Y = 218 This graph of the function y = 0. 0000512×2 +0. 5171x – 1010 is shown in the following Graph 3 as it is against the points plotted in Data Graph 1 Graph 3 Height (cm) Height (cm) Year Year It is shown in Graph 3 that the quadratic function does resemble the shape of the line plotted by the points in Data Graph 1.

In Graph 4 both functions are shown against the original data points plotted in Data Graph 1. Graph 4 Height (cm) Height (cm) Year Year Had the games been held in 1940 and 1944 the winning heights would be estimated as: Y = 0. 62(1940) – 1000 Y = 1202. 8 – 1000 Y = 202. 8 When the x value of 1940 is plugged into the linear equation y = 0. 62x – 1000 the y value, or winning height in the year 1940, would be 202. 8. Y = 0. 0000512(1940)2 + 0. 5171(1940) – 1010 Y = 0. 0000512(3763600) + 1003. 17 – 1010 Y = 192. 69 + 1003. 17 – 1010 Y = 185. 86 When the x value 1940 is plugged into the quadratic equation y = 0. 000512×2 + 0. 5171x – 1010 the y value, or winning height for 1940, is 185. 86. In order to make a more accurate estimate for the winning height in 1940 the average of both terms, 202. 8 and 185. 86, are taken. 202. 8 + 185. 86 / 2 = 194. 33 The estimated winning height in 1940 would be 194. 33 For the estimated winning height in 1944: Y = 0. 62(1944) – 1000 Y = 1205. 28 – 1000 Y = 205. 28 When the x value of 1944 is plugged into the linear equation 0. 62x – 1000 the y value, or winning height in 1944, is 205. 8. Y = 0. 0000512(1944)2 + 0. 5171(1944) – 1010

Y = 0. 0000512(3779136) + 1005. 24 – 1010 Y = 193. 49 + 1005. 24 – 1010 Y = 188. 73 When the x value of 1944 is plugged into the quadratic equation of 0. 0000512×2 + 0. 5171x + 1010 the y value, or winning height in 1944, is 188. 73. To have a more accurate estimate of the winning height the average of the two y values of two functions is taken. 188. 73 + 205. 28 / 2 = 197. 01 The winning height for the 1944 Olympics would be estimated at 197. 01 Graph 5 Year Year Height (cm) Height (cm) Graph 5 shows all the data from the 1896 Olympics to the 2008 Olympics.

The red line is the quadratic function f(x) = 0. 0000512×2 + 0. 5171x – 1010 and the blue line represents the linear function f(x) = 0. 62x -1000. The function models that are represented in Graph 5 do not fully resemble the points plotted from the gold medal high jump heights but have an overall shape of the rise in height from year to year. Graph 6 Height (cm) Height (cm) Year Year The graph above shows all data point plotted on a graph. The overall trend or shape resemble that of a cubic. Graph 7 The shape of the cubic resembles more of the shape shown in the data points plotted on Graph 6.

The modifications that need to be made on my models are that the quadratic function needs to have a more curve to it than just a straight line. To do that I need to change the variables a, b, and c in the equation f(x) = ax2 + bx + c f(x) = 0. 0000512×2 + 0. 5171x -1010 Graph 8 Height (cm) Height (cm) Year Year In a zoomed out view the graph shows the overall shape of the quadratic shape of the red line and how it goes through the black data points and how the linear equation is completely straight. IB Mathematics SL Year 2 Internal Assessment Gold Medal Heights Date Due: February 11, 2013