IB Mathematics SL 4.3 Measures of central tendency AI SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Total beetles = sum of all frequencies:
\(42 + 25 + 19 + 61 + 27 + 12 = 186\).
\(\boxed{186}\)
(b)
Use midpoint of each category:
Small: midpoint \(= 11\) mm, frequency \(= 42 + 61 = 103\)
Medium: midpoint \(= 14\) mm, frequency \(= 25 + 27 = 52\)
Large: midpoint \(= 17\) mm, frequency \(= 19 + 12 = 31\)
Estimated mean:
\( \bar{x} = \frac{11 \times 103 + 14 \times 52 + 17 \times 31}{186} \approx \frac{1133 + 728 + 527}{186} = \frac{2388}{186} \approx 12.8 \)
\(\boxed{12.8 \ \text{mm}}\)
(c)
Total female beetles = \(42 + 25 + 19 = 86\).
Probability one female is small = \(\frac{42}{86} = \frac{21}{43}\).
With replacement:
\( P(\text{both small}) = \left(\frac{21}{43}\right)^2 = \frac{441}{1849} \approx 0.239 \).
\(\boxed{0.239}\)
(d)
• Hypotheses: \(H_0\): length category and sex are independent; \(H_1\): not independent.
• Use \(\chi^2\) test for independence.
• \(p\)-value ≈ 0.127 (from calculator).
• Since \(p\text{-value} > 0.05\), do not reject \(H_0\).
• Conclusion: No significant evidence that length category and sex are dependent.
\(\boxed{p\approx 0.127,\ \text{do not reject } H_0}\)
(e)
• \(H_0: \phi = 0.45\), \(H_1: \phi > 0.45\).
• Use binomial test: \(X \sim \text{B}(186, 0.45)\).
• Observed males = \(61 + 27 + 12 = 100\).
• \(p\text{-value} = P(X \geq 100) \approx 0.0101\).
• Since \(p\text{-value} < 0.05\), reject \(H_0\).
• Conclusion: Significant evidence that more than 45% are male.
\(\boxed{p\approx 0.0101,\ \text{reject } H_0}\)