IB Math Methods SL: Internal Assessment on Gold Medal Heights

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International Baccalaureate

IB Mathematics SL Y2

Internal Assessment

Task: Gold Medal Heights (Type 2)


The aim of this investigation is to consider the winning height for the men’s high jump in the Olympic Games.

Firstly, we are given a table that lists the record height achieved by gold medalists in each competition from 1932 onto 1980.

Given Information

N.B. The Olympic games were not held in 1940 and 1944.

Let us first plot the points on a graphing application. After plotting the above points in the graphing programme (see Figure 1 in Appendix), we arrive at the graph below.

Graph 1: Height achieved by gold medalists in various Olympic Games

        Domain: 1920  t  1990

        Range: 195  t 240        

This graph demonstrates the record height changing over time through every Olympic game. The h-axis represents the height of the record jump (in centimeters). The t-axis represents the years in which the jumps took place. It should be noted that there are certain limitations to this data set; primarily that there was no men’s high jump taking place in 1940 and 1944; none in 1940 due to the revocation of Tokyo as the host venue for the Games due to the Sino-Japanese war; and none in 1944 due to the outbreak of the Second World War. The absence of two games was likely to cause a difference in the pattern of data; most likely due to many governments’ attention on the Second World War rather than a “sporting competition.” Hence, it is causable that this led to a drop in the record height in 1948 (when the Games restarted) due to the lack of competition or motivation.

Some parameters of this data set are limited in that the value “h” cannot be lesser than or equal to zero, as a jump cannot be negative or remain on the ground. A constraint in this task is that the “t” values cannot go before 1896; as the Olympic Games Men’s High Jump did not exist until 1896. Furthermore, there will always be a limit to how high the record will be (h-value) due to the gravitational force of the Earth.

Let’s now analyse Graph 1 in order to determine what kind of function best fits the given behaviour of the graph. However, before we proceed, let’s simplify the data points; instead of actual years like 1938 and 1980, which would cause big constant values in the equations we will do later on, let’s simplify it so that the t-axis shows how many years have elapsed since 1932, which will simplify our equations without sacrificing accuracy. Hence, we arrive at Graph 2, shown below.

Graph 2: Height achieved by gold medalists in various Olympic Games

        Domain: 0  t  60

        Range: 190  t  240

Now we can proceed onto attempting to find which function fits the given data. When looking at the data points on this graph; it appears for the most part that the dependent variable (record height) increases along with the independent variable (years elapsed); the exception being between 1936 and 1948 along with 1968 and 1972. The values for “h” also appear to be increasing for the most part at a steady pace; hence, it appears that the graph demonstrates linear growth and is a linear function.

Let us now find a linear function analytically that best fits the data points on Graph 2; we will use the Least Squares approach to deduce the linear equation. Let’s start by recalling our given data (reproduced below):

 Given Information

N.B. The Olympic games were not held in 1940 and 1944.

Because we changed our approach from literal years to years elapsed, let us reflect that with a new table of data points.

Information Table 1

Now we can begin with our Least Squares approach.

The Least Squares approach involves finding the best-fit equation using sums and arithmetic of the data. It enables us to analytically to find an equation that models the data of our table above. Let us begin. A calculator is used to derive the answers to the sigmas.

Sum of t values

Sum of t values squared

Product of the sums of t and h

Now, let us find the average of the values of t and h, respectively.

= , and =

Now let’s combine the various values we have derived.

b1 is the regression slope of our linear function.

Finding b0 enables us to find the constant value in our linear equation (regression incercept).

After simplifying to three significant figures, we then arrive at 187. 187 is the constant value in our linear equation. Hence, so far we now have:. We now can find the slope; as our slope is the value of  found above. Now, we finally have , which simplifies to .  Now, let’s plug that equation into our graphing programme and see what the results are.

Join now!

Graph 3: Plotting of Linear Equation

Domain: 0  t  50

Range: 185  h  235

Let us now construct a table of values from the linear equation (y = 1.02x + 187).  A data table was constructed using a calculator, and can be seen below in the Appendix (Figures 2 and 3).

Upon converting that data to a table and simplifying it to three significant figures; we arrive at this table.

Information Table 2 ...

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