Home / IB Mathematics AHL 2.10 Scaling very large or small numbers AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 2.10 Scaling very large or small numbers AI HL Paper 1- Exam Style Questions- New Syllabus

IB DP AA Maths HL Paper 1 -All Topics

Question

A veterinarian is researching the correlation between the age, measured in days (\(d\)), and the body mass, measured in kilograms (\(w\)), for a specific dog breed. By analyzing a comprehensive dataset, he constructs two different regression models. The statistical parameters for these models are presented in the following table.
 Horizontal axisVertical axisGradientIntercept\(R^2\)
Graph 1\(d\)\(\ln w\)\(0.00571\)\(1.54\)\(0.72\)
Graph 2\(\ln d\)\(\ln w\)\(0.302\)\(0.693\)\(0.95\)
Using the information provided, identify which graph represents the more accurate relationship between \(w\) and \(d\). Derive the equation for this relationship in the form \(w = f(d)\), ensuring the expression is simplified. Provide a justification for your choice.

Most-appropriate topic codes:

• AHL 4.13: Non-linear regression; coefficient of determination (\(R^2\)) as a measure of fit
• AHL 2.10: Linearizing data using logarithms to determine exponential or power relationships
▶️ Answer/Explanation
Detailed solution

Justification:
Graph 2 represents the superior model as it possesses a significantly higher coefficient of determination (\(R^2 = 0.95\)) compared to Graph 1 (\(R^2 = 0.72\)), indicating a much stronger fit to the data.

Derivation:
The linear equation from Graph 2 is given by:
\(\ln w = 0.302(\ln d) + 0.693\)
Applying the exponential function to both sides:
\(w = e^{0.302 \ln d + 0.693}\)
Using exponent laws:
\(w = e^{0.693} \cdot e^{\ln(d^{0.302})}\)
Since \(e^{0.693} \approx 2.00\):
\(w = 2.00 \cdot d^{0.302}\)
Expression: \(w = 2d^{0.302}\).

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