# Math Lab Sl 11

Pages: 7 (1671 words) Published: December 20, 2012
Cindy Hwang
IB Math 11 SL 1-3
IB Math Portfolio
Gold medal heights
Aim:
The aim of this task is to consider the winning height for the men’s high jump in the Olympic Games. Introduction:
The Olympic Games which are held in every four years have an event called Men’s High Jump and usually an athlete tries to jump over a bar which is set up in a certain range from1 meter to 3 meters. The table 1 shows the record of the gold medalists during the Olympic Games held during 1932 to 1980. Note: The Olympic Games were not held in 1940 and 1944. Table 1

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| Variables:
Independent Variables:
The heights that are achieved by the gold medalists
Dependent Variables:
The years of Olympic Games
Figure 1:

From the figure 1, the independent variables are the x axis which shows the years of the Olympic Games, and y axis is the dependent variables which represents the heights that are achieved by the gold medalists. Also it shows that it is not constant.

Linear Regression
To create a certain equation, you draw the best fit line on the graph. The difference between the red graph and the linear function is that the red does not have a predictable pattern. When the best fit is drawn it is possible to find the equation of this graph. Though the equation that is made by the best fit and three points on the graph is actually on the line, there is a limit. Such as the y axis which represents the height that the gold medalist achieved cannot be under zero because it is impossible for a person to jump under the ground. Also as you see in the graph, after the War World Ⅱ, it decreased a bit then it started to increase little by little. However, on linear regression, there isn’t any sign that it is decreasing. Since we can’t guarantee that it will increase constantly, it is not a best idea to state this is a best regression for this data.

Figure 2:

In the figure 3, the graph is connected with the linear function. The black dotes are the points that fits well on the best fit line. In order to find out the equation, we use the points (1932, 197) and (1972, 223)

figure3:

Equation
y= mx+b

m=
Substituting:
(1932, 197) (1972, 223)
233 = 1972m + b
197= 1932m + b
b= y intercept when x=0

Then in order to find the value of m which represents slope we insert slope equation with our points.

m=

(1932, 197) (1972, 223)
1972-1932

223-197
m=

m= 40/26

∴m= 20/13

When the value of m is found, b can be identified. Now the equation is this:

y= 20/13x+b

Chose one point among those two and insert in the equation.
(1932, 197)
y= 20/13x+b
197= 20/13*1932 + b
197= 898.60 + b
b= -701.6

As the result the equation for the linear function is:
y= 20/13x – 701.6

Figure 4: Figure 5: Equation for the quadratic regression.

Since the linear regression not might be the best function to determine, I tried quadratic Regression. For the quadratic regression, it looks more realistic than linear regression though it might look similar. As it said on the graph, the curve shows that the function is gradually increasing which is more accurate than the linear function. Because the heights are increasing little by little, it is more efficient to use a quadratic function if the graph shows that the data is increasing very diminutively. However, same as the linear function, we cannot assume that the height will increase gradually in the future and there is no chance that the graph itself is going to decrease because it is a quadratic function.

Comparing Linear Regression and Quadratic Regression
Figure 5
In Figure 5, both linear and quadratic regressions are shown. However, there isn’t a huge difference between them except that linear function increases dramatically...