IB Mathematics SL 4.1 Concepts of population, sample AI SL Paper 1 - Exam Style Questions - New Syllabus
▶️Answer/Explanation
Markscheme (with detailed working)
(a)
Data (m): \(1.67,\,1.60,\,1.68,\,2.31,\,2.31,\,2.19\).
(i) Mean \(=\dfrac{1.67+1.60+1.68+2.31+2.31+2.19}{6}=\dfrac{11.76}{6}=\boxed{1.96\ \text{m}}\). A2
(Award A1 for correct substitution into the mean formula.)
(ii) Sort: \(1.60,\,1.67,\,1.68,\,2.19,\,2.31,\,2.31\).
\(\text{Median}=\dfrac{1.68+2.19}{2}=\boxed{1.935\ \text{m}}\approx \boxed{1.94\ \text{m}}\). A1
\(\text{Median}=\dfrac{1.68+2.19}{2}=\boxed{1.935\ \text{m}}\approx \boxed{1.94\ \text{m}}\). A1
(iii) Mode \(=\boxed{2.31\ \text{m}}\). A1
(iv) Range \(=\max-\min=2.31-1.60= \boxed{0.71\ \text{m}}\). M1 A1
(M1 for identifying the correct extreme values, A1 for \(0.71\ \text{m}\).)
(b)
Measured to the nearest centimetre \((0.01\ \text{m})\), so the exact height lies in \([1.975,\,1.985)\ \text{m}\).
Shortest possible height \(=\boxed{1.975\ \text{m}}=\boxed{197.5\ \text{cm}}\). A1
Shortest possible height \(=\boxed{1.975\ \text{m}}=\boxed{197.5\ \text{cm}}\). A1
Total Marks: 7