IB Mathematics SL 4.3 Measures of central tendency AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme-style solution
Given scores (from the table): \(7.5,\,7.2,\,6.5,\,7.2,\,6.2,\,6.0,\,7.0,\,6.6,\,6.7,\,7.4\).
(a)
Sort ascending: \(6.0,\,6.2,\,6.5,\,6.6,\,6.7,\,7.0,\,7.2,\,7.2,\,7.4,\,7.5\).
For \(n=10\): lower half = first 5, upper half = last 5.
\(Q_1=\) median of lower half \(=6.5\); \(Q_3=\) median of upper half \(=7.2\).
(i) \(\boxed{Q_3=7.2}\). (ii) \(\text{IQR}=Q_3-Q_1=7.2-6.5=\boxed{0.7}\).
For \(n=10\): lower half = first 5, upper half = last 5.
\(Q_1=\) median of lower half \(=6.5\); \(Q_3=\) median of upper half \(=7.2\).
(i) \(\boxed{Q_3=7.2}\). (ii) \(\text{IQR}=Q_3-Q_1=7.2-6.5=\boxed{0.7}\).
(b) Tukey fences (outliers):
Upper fence \(=Q_3+1.5\,\text{IQR}=7.2+1.5(0.7)=8.25\). Switzerland’s score \(7.5<8.25\) so it is not an outlier. (Lower fence \(=6.5-1.05=5.45\) is not needed here.)
(c) (Rank highest score as 1; ties share the average rank.)
The two \(7.2\) scores (NZ, Australia) tie for ranks \(3\) and \(4\) \(\Rightarrow\) each gets \(3.5\).
Hence \(\boxed{a=3.5},\ \boxed{b=8}\ (\text{Spain’s }6.5),\ \boxed{c=3.5}\ (\text{Australia}).\)
Hence \(\boxed{a=3.5},\ \boxed{b=8}\ (\text{Spain’s }6.5),\ \boxed{c=3.5}\ (\text{Australia}).\)
(d)(i) Spearman’s \(r_s\).
Use expatriate ranks \(x_i=1,2,\dots,10\) (in the same country order) and the happiness ranks \(y_i=[1,\,3.5,\,8,\,3.5,\,9,\,10,\,5,\,7,\,6,\,2]\).
Differences \(d_i=x_i-y_i=[0,\,-1.5,\,-5,\,0.5,\,-4,\,-4,\,2,\,1,\,3,\,8]\).
Squares \(d_i^2=[0,\,2.25,\,25,\,0.25,\,16,\,16,\,4,\,1,\,9,\,64]\) so \(\sum d_i^2=137.5\).
Without tie correction, \(r_s=1-\dfrac{6\sum d_i^2}{n(n^2-1)}=1-\dfrac{6(137.5)}{10\cdot 99}\approx 0.1667\).
Applying the (required) tie adjustment (or using GDC “Spearman (ties)”) gives \(\boxed{r_s\approx 0.164}\) (to 3 s.f.).
Differences \(d_i=x_i-y_i=[0,\,-1.5,\,-5,\,0.5,\,-4,\,-4,\,2,\,1,\,3,\,8]\).
Squares \(d_i^2=[0,\,2.25,\,25,\,0.25,\,16,\,16,\,4,\,1,\,9,\,64]\) so \(\sum d_i^2=137.5\).
Without tie correction, \(r_s=1-\dfrac{6\sum d_i^2}{n(n^2-1)}=1-\dfrac{6(137.5)}{10\cdot 99}\approx 0.1667\).
Applying the (required) tie adjustment (or using GDC “Spearman (ties)”) gives \(\boxed{r_s\approx 0.164}\) (to 3 s.f.).
(d)(ii)
If France’s score changes to \(6.9\), its position among the scores is unchanged (still rank \(6\)); since the ranks do not change, \(r_s\) does not change.
(e)
\(r_s\) is very close to \(0\) (weak association of the ranks), so Aria’s conclusion is not appropriate.
Total Marks: 14