IB Mathematics SL 4.2 Presentation of data AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
From the bar graph provided in the context, the frequency for \(3\) goals is identified as \(5\).
\(\boxed{5}\)

(b)
(i) Mean \(= \frac{\text{Total Goals}}{\text{Total Matches}}\).
Total Matches \(= 3 + 11 + 7 + 5 + p + 1 = p + 27\).
Total Goals \(= (0 \times 3) + (1 \times 11) + (2 \times 7) + (3 \times 5) + (4 \times p) + (5 \times 1) = 45 + 4p\).
Given Mean \(= 2.2\), the equation is: \( \boxed{45 + 4p = 2.2(p + 27)} \)

(ii) Solving for \(p\):
\(45 + 4p = 2.2p + 59.4\)
\(1.8p = 14.4\)
\(p = 8\).
\(\boxed{8}\)

(c)
From the \(2024\) data: Total matches \(= 35\). Median is at the \(18^{th}\) position \(= 2\). \(Q_1\) is at the \(9^{th}\) position \(= 1\). \(Q_3\) is at the \(27^{th}\) position \(= 4\). \(\text{IQR} = 4 – 1 = 3\). Range \(= 5 – 0 = 5\).
From the \(2025\) box plot: Median \(= 2\), \(Q_1 = 1\), \(Q_3 = 4\), \(\text{IQR} = 3\), Range \(= 5\).
Consistent observations:
1. The median number of goals is \(2\) for both years.
2. The interquartile range (\(\text{IQR}\)) is \(3\) goals for both years.

(d)
Event \(F\) is “scoring \(0\) or \(1\) goal”. The complement \(F’\) is “not scoring \(0\) or \(1\) goal”, which is equivalent to “scoring at least \(2\) goals”.
Equivalent events from the list: \(B\) and \(C\).
\(\boxed{B, C}\)

(e)
Total matches in \(2024 = 35\). Initially, there are \(11\) matches with \(1\) goal.
After watching one match with \(1\) goal, \(34\) matches remain, of which \(10\) contain \(1\) goal.
Probability \(= \frac{10}{34} = \frac{5}{17}\).
\(\boxed{\frac{5}{17}}\)

(f)
Probability of first match having \(5\) goals \(= \frac{1}{35}\).
Probability of second match having \(0\) goals (given the first was different) \(= \frac{3}{34}\).
Total Probability \(= \frac{1}{35} \times \frac{3}{34} = \frac{3}{1190}\).
\(\boxed{\frac{3}{1190}}\)