BMI Internal Assessment

By: Jason Chau

Year 11 SL Mathematics - 2009-08-09


Introduction

This piece of work is based on the data of the median BMI for females of different ages in the US in the year 2000. I will be modelling the data using different types of functions, as it is a modelling task.

The Body Mass Index or BMI is a ratio of a person’s high in relation to their weight. It is calculated by dividing the person’s weight (kg) by the square of their height (m). This ration is usually used to determine a person’s health.

  1. Using technology, plot the data points on a graph. Define all variables used and state any parameters clearly.

I put the age in the x column because it is the constant variable. I called this column the ‘Age’ column

Next, I put the BMI values in the y column as it is the dependent variable and called it the ‘BMI’ column

The graph is plotted using these data points in the table with the all points unconnected, as this makes the graph more accurate. The independent variable is the Age, which is the x axis on the graph while the dependent variable is the BMI, which is the y axis on the graph. The parameter is a constant in the equation of a curve that can be varied to yield a family of similar curves.

  1. What type of function models the behaviour of the graph? Explain why you chose this function. Create an equation (a model) that fits the graph.

I chose the polynomial, quadratic function because the graph looks similar to a parabola. A quadratic function is:

y=ax2 + bx + c; we need to find the values for a, b and c.

In this equation, there are three unknown variables (a, b and c) so therefore, I would need to find three equations. I will be using simultaneous equations to find the values of the unknown variables.

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  1. x=5 is the line of symmetry as

x=  = 5

Using data from the graph, I let the x value be 3 and therefore, the y value is 15.7,

  1. 15.7 = a(3)2 + b(3)2 + c

15.7 = 9a + 3b + c

Using another point on the graph, I let the x value be 17 and therefore, the y value is 20.85

  1. 20.85 = a(17)2 + b(17)2 + c

20.85 = 289a + 17b + c

Now that I have 3 equations, I can use simultaneous equations to solve the values of a, ...

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