IB Mathematics SL 4.4 Linear correlation of bivariate data AI SL Paper 2 - Exam Style Questions - New Syllabus

(a)
\( m = \frac{9.45 – 8.73}{1958 – 1708} = \frac{0.72}{250} \).
\(\boxed{0.00288}\)

(b)
(i) The gradient represents the average increase in temperature per year.
(ii) Units: \(\boxed{^\circ\text{C}/\text{year}}\)

(c)
Using \( y – y_1 = m(x – x_1) \):
\( y – 8.73 = 0.00288(x – 1708) \)
\( y = 0.00288x – 4.91904 + 8.73 \)
\(\boxed{y = 0.00288x + 3.81}\)

(d)
\( y = 0.00288(2000) + 3.81 = 5.76 + 3.81 \).
\(\boxed{9.57 ^\circ\text{C}}\)

(e)
Using a GDC for linear regression on all 7 data points:
(i) \(\boxed{y = 0.00256x + 4.46}\)
(ii) \(\boxed{r \approx 0.861}\)

(f)
\( y = 0.00256(2000) + 4.46 = 5.12 + 4.46 \).
\(\boxed{9.58 ^\circ\text{C}}\)

(g)
1. Extrapolation: \( 15 ^\circ\text{C} \) is well outside the range of observed data; linear trends may not continue indefinitely.
2. Regression Direction: The line \( y \) on \( x \) is used to predict temperature from years. To predict years from temperature, an \( x \) on \( y \) regression line should ideally be used.