• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# IB Math Methods SL: Internal Assessment on Gold Medal Heights

Extracts from this document...

Introduction

International Baccalaureate

IB Mathematics SL Y2

Internal Assessment

Task: Gold Medal Heights (Type 2)

Introduction

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

 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

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.

Middle

th year-elapsed point, with a difference of 12 centimeters. While the linear function meets at the last data point at 48 years with 236 cm; it also has a wide difference at the starting point – the very first year with a difference of 10cm. The linear model also faces issues that it does not account for a sinusoidal deviation from the equation line; for example, the 4th year and the 16th year had very different ranges from the function; and were extreme in range from each other; as the linear line goes in a straight fashion, it cannot account for these outlier points. Another weakness in the model is that the function will continue to increase over value in time; however this is not possible in this real life situation as there will be a limit to how high a human can jump due to gravitational forces; which the linear function cannot account for. On the other hand; the equation keeps decreasing in value when going back; and does not account for that time cannot be negative; nor that a high jump cannot go “down” below the ground (negative value) or remain on the ground (zero). However, for a linear model; this equation appears to be the best fit possible; as a rather precise method - Least Squares Approach, which involved all data points, was used rather than conducting a simple slope formula of two points to discover a gradient.

As discussed above, while a linear function works well as a analytical approach; using technology to determine a best-fit would be better approach to discover a better fit than the original linear equation. Let us explore an alternative method of finding a function; this time using regression.

Conclusion

In conclusion, let us conduct one example of a modification; linear regression - on Autograph to modify our current linear function to improve the best-fit. After inputting data (see Appendix, Figure 5) Let’s conduct linear regression with the programme; which then gives us a best-fit function. Let’s see the next page for a revised linear function.

Graph 10: Revised Linear Function As is evident from the graph; the revised linear function corresponds with many more points than the original linear function, and in general fits the data points much better than the original function. While improving the linear function still does not address for the constant increase over time; the function works very well in this instance as the gradient of the revised function allows more data points to accurately correspond with the function.

In conclusion, we can find from the data that records do not quickly increase over time, but gradually increase and may sometimes decrease due to extraneous events such as the Second World War. It was also beneficial to realize the natural limits behind real-life examples such as how high a human can naturally jump. This project also demonstrates the limitation in technology; while technology is very accurate within ranges of data that is already given; it cannot accurately predict futures within the graph and certain anomalies such as outliers in the data.

Colophon

The technology used in this assessment includes but is not limited to: TI-84 calculator, AutoGraph Windows Version 3.1, Mac OS Grapher, Windows Publisher, and Microsoft Excel.

Appendix

Figure 1 – Data input onto Autograph Figure 2 – Calculator display of values

Calculator Screen 1                        Calculator Screen 2  Figure 3: Editing data sets on Autograph Figure 4: Broad view of quintic and linear function Figure 5: Revissed linear regression data plotting Black and white: 1-7, 9, 14, 15, 17-19

Colour: Pages 8. 10. 11, 12, 13, 16

Page  of

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Tide Modeling

???????? ??????????-???????????????? ??????????|-???????????????? ??????????.�100 it is only %3.64 showing that the model is accurate. Points 0,12 1,13 2,14 3,15 4,16 5,17 6,18 7,19 8,20 9,21 10,22 11,23 Model 6.475 9.313 11.39 12.15 11.39 9.313 6.475 3.638 1.56 0.8 1.16 3.638 Average 7.2 9.95 11.7 12.15 11.25 9.4 6.8 4.05 1.85 0.8 1.3 3.2

2. ## Math IB SL BMI Portfolio

In conclusion, this study has shown that for girls living in urban conditions, BMI can be modeled by the sinusoidal and logistic mathematical functions.

1. ## A logistic model

2 ?10 a. Consider an initial population u1=2.5x104 fish. The equation could be formulated as follows: ?5 2 3 un?1 ? (?1? 10 )(un?1 ) ? 1.6(un?1 ) ? 8 ?10 {19} Table 9.1. The population of fish in a lake over a time range of 30 years estimated using the logistic function model {14}.

2. ## Maths IA Type 2 Modelling a Functional Building. The independent variable in ...

The equation to finding the volume of multiple cuboids is Method 9. Using Excel: To now further validate this statement through the use of Microsoft Excel to construct the follow table below. Column C was found by applying the equation, , to Column C, where is the value directly left in Column B.

1. ## IB math SL type 2 project

the operating cost should be considered with the efficiency of the container; although the capital cost is very reasonable, if the operating cost is too expensive, McNewton is likely to buy another container that costs reasonably. Conclusion: The Newton's Law of Cooling can be practically used in the real world

2. ## Math SL Circle Portfolio. The aim of this task is to investigate positions ...

However, there is no P', for and do not intersect. In the following graph, 2, r =5. and does not intersect, so there is no Point A. cannot be created without A as its centre point. Therefore, there is no point P', and the length of ' cannot be determined.

1. ## Crows Dropping Nuts

? ? ? ? ? ? Ex: x=3 y=21.37241705 x (.83584326053) =21.37241705 x .5839483675 12.48 After subbing in each x, to find the y-value, the data table became: Height of drop (m) 3 4 5 6 7 8 10 15 Number of Drops 12.48 10.43 8.72 7.29 4.4 4.1

2. ## MATH IB SL INT ASS1 - Pascal's Triangle

If we have a closer look at d1, for example, we can easily determine this pattern: The first number is 1, also can be written as and the denominator is therefore 1. The second fraction is and the denominator is therefore 2. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 