Math Internal Assessment Gold Medal Heights

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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.

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 = 0.0000512x2 + 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 = 0.0000512x2 +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 –...
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