IB Mathematics SL 4.4 Linear correlation of bivariate data AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Gradient through \((x_1,y_1)=(1708,8.73)\) and \((x_2,y_2)=(1958,9.45)\): \[ m=\frac{9.45-8.73}{1958-1708}=\frac{0.72}{250}=0.00288. \] M1 A1
(b)
(i) The gradient is the estimated mean yearly increase in Earth’s mean annual temperature. A1
(ii) Units: \(^{\circ}\text{C per year}\). A1
(ii) Units: \(^{\circ}\text{C per year}\). A1
(c)
Using \(y=mx+c\) with \(m=0.00288\) and point \((1708,8.73)\): \[ 8.73=0.00288(1708)+c \Rightarrow c=8.73-4.91904=3.81096\ldots \] Hence \(\boxed{y=0.00288\,x+3.81096\,(~\approx~0.00288x+3.81~)}\). M1 A1
(d)
Substitute \(x=2000\) into Emily’s line: \[ y=0.00288(2000)+3.81096=5.76+3.81096=9.57096\ldots\approx \boxed{9.57^\circ\text{C}}. \] M1 A1
(e)
Using technology (linear regression \(y\) on \(x\)):
(i) \( \displaystyle \boxed{y=0.00255714\ldots\,x+4.46454\ldots\ \ (\text{accept }y\approx 0.00256x+4.46)} \). (M1) A1
(ii) \( \displaystyle \boxed{r\approx 0.861\ (0.861333\ldots)} \). A1
(ii) \( \displaystyle \boxed{r\approx 0.861\ (0.861333\ldots)} \). A1
(f)
Estimate at \(x=2000\) using Daniel’s regression line: \[ y=0.00255714\ldots(2000)+4.46454\ldots = 5.11428\ldots + 4.46454\ldots =9.57882\ldots \approx \boxed{9.58^\circ\text{C}}. \] (Accept \(9.57^\circ\text{C}\) if \(4.46\) used.) M1 A1
(g)
Any two valid reasons, for example:
- The regression is \(y\) on \(x\); using it to predict a year from a temperature (i.e. \(x\) from \(y\)) is not generally reliable.
- Extrapolation far beyond the observed data range may be invalid (the relationship need not remain linear).
- Small dataset and other influencing factors (model limitations, correlation \(\ne\) causation).
A1 A1
Total Marks: 15