Home / IB Mathematics SL 4.4 Linear correlation of bivariate data AI SL Paper 1- Exam Style Questions

IB Mathematics SL 4.4 Linear correlation of bivariate data AI SL Paper 1- Exam Style Questions- New Syllabus

Question

James is a keen hiker who keeps a record of his performance. The following table shows the time, in minutes, it takes him to walk one kilometre up slopes with different inclines. The incline of each slope is constant.
Incline G (%)04101520
Time T (min.)6.858.4211.2014.4917.88
(a) (i) Find the equation of the regression line of T on G .
     (ii) Describe the correlation between T and G with reference to the value of r , the Pearson’s product-moment correlation coefficient. [4]
On Sunday, James intends to walk up a slope with an incline of 13 %.
(b) Estimate the time it will take James to walk one kilometre up the slope. [2]
This morning, James walked one kilometre up a slope, and it took 22 minutes.
(c) Explain why it would be inappropriate to use the equation found in part (a) to estimate the incline of this slope. [1]
▶️Answer/Explanation
Markscheme (with detailed working)

(a)

Set up linear model \(T = aG + b\). Use least squares (technology acceptable). From the data:

\(\displaystyle a = 0.552139\ldots,\quad b = 6.35703\ldots\) A1 A1
So the regression line (rounded to 3 s.f.) is \(\boxed{T = 0.552\,G + 6.36}\).
Pearson correlation \(r \approx 0.993910\ldots \approx \boxed{0.994}\). This indicates a (very) strong positive linear correlation between \(T\) and \(G\). A1 R1

(b)

Substitute \(G=13\) into the regression line of \(T\) on \(G\):
\(T(13) = 0.552139\ldots \times 13 + 6.35703\ldots = 13.5348\ldots \Rightarrow \boxed{13.5\ \text{min}}\) (to 3 s.f.). M1 A1

(c)

The computed line is the regression of \(T\) on \(G\); using it to predict \(G\) from a given \(T\) is inappropriate. (Equivalently: it models “Time on Incline”, not “Incline on Time”.) R1
Total Marks: 7
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