IBDP Maths SL 4.3 Measures of central tendency AA HL Paper 2- Exam Style Questions- New Syllabus

Step 1: Use the Mean formula
Total frequency \(\sum f = 12 + q + 8 = 20 + q\).
Mean \(\bar{x} = \frac{\sum f x}{\sum f} = 31\):
\(\frac{12(20) + 35q + 8p}{20 + q} = 31\)
\(240 + 35q + 8p = 31(20 + q) \implies 240 + 35q + 8p = 620 + 31q\)
\(4q + 8p = 380 \implies q + 2p = 95 \quad \text{—(Eq 1)}\)

Step 2: Use the Variance formula
Variance \(\sigma^2 = \frac{\sum f x^2}{\sum f} – \bar{x}^2 = 124\):
\(\frac{12(20^2) + q(35^2) + 8p^2}{20 + q} – 31^2 = 124\)
\(\frac{4800 + 1225q + 8p^2}{20 + q} – 961 = 124 \implies \frac{4800 + 1225q + 8p^2}{20 + q} = 1085 \quad \text{—(Eq 2)}\)

Step 3: Solve the simultaneous equations
From (Eq 1), \(q = 95 – 2p\). Substitute this into (Eq 2):
\(\frac{4800 + 1225(95 – 2p) + 8p^2}{20 + (95 – 2p)} = 1085\)
\(\frac{4800 + 116375 – 2450p + 8p^2}{115 – 2p} = 1085\)
\(121175 – 2450p + 8p^2 = 1085(115 – 2p)\)
\(121175 – 2450p + 8p^2 = 124775 – 2170p\)
\(8p^2 – 280p – 3600 = 0\)
Dividing by 8: \(p^2 – 35p – 450 = 0\)
Factorizing: \((p – 45)(p + 10) = 0\)
Since \(p\) must be a positive integer, \(p = 45\).
Then \(q = 95 – 2(45) = 5\).

Answer: \( \boxed{p = 45, \ q = 5} \)