IBDP Maths AHL 4.14 Variance of random variable AA HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)
\[ \int_{2.25}^{4.5} \frac{4}{21} \left( 1 – \cos \left( \frac{4\pi}{9}(t – 2.25) \right) \right) dt = \frac{4}{21} \left[ t – \frac{9}{4\pi} \sin \left( \frac{4\pi}{9}(t – 2.25) \right) \right]_{2.25}^{4.5} \]
Evaluating at the limits gives \( \frac{4}{21} \left( 4.5 – 2.25 \right) = \frac{4}{21} \times 2.25 = \frac{3}{7} \).
Answer: \( \boxed{0.429} \)

(a)(ii)
The mode is the \( t \)-value where the pdf is at its maximum. From the definition and graph:
Answer: \( \boxed{4.5} \)

(a)(iii)
Since \( P(T < 4.5) = \frac{3}{7} \approx 0.429 \), and the median \( m \) satisfies \( P(T < m) = 0.5 \), the median must be greater than 4.5 because more area is needed to reach 0.5.
Answer: \( \boxed{\text{Median is greater than Mode}} \)

(b)
Elite runners \( T \leq 3.5 \): \( \int_{2.25}^{3.5} f(t) dt \approx 0.1037 \).
Answer: \( \boxed{0.104} \)

(c)
\( P(T \leq 3 \mid T \leq 3.5) = \frac{P(T \leq 3)}{P(T \leq 3.5)} \approx \frac{0.0247}{0.1037} \approx 0.238 \).
Answer: \( \boxed{0.238} \)

(d)
Lower quartile \( q_1 \) satisfies \( \int_{2.25}^{q_1} f(t) dt = 0.25 \). Since \( 0.25 < 0.429 \), \( q_1 \) is in the first interval.
Solving \( \frac{4}{21} \left( q_1 – 2.25 – \frac{9}{4\pi} \sin \left( \frac{4\pi}{9}(q_1 – 2.25) \right) \right) = 0.25 \) numerically:
Answer: \( \boxed{4.01} \)

(e)
1. \( E(P) = a – b E(T) \implies a – 4.723b = 100 \)
2. Max score at min time \( t=2.25 \): \( a – 2.25b = 150 \)
Subtracting the two equations: \( (4.723 – 2.25)b = 50 \implies 2.473b = 50 \implies b \approx 20.218 \).
Substituting back: \( a = 150 + 2.25(20.218) \approx 195.49 \).
Answer: \( \boxed{a = 195, \ b = 20} \)

(f)
Using the property \( \text{Var}(aX+b) = a^2 \text{Var}(X) \):
\( \text{Var}(P) = (-b)^2 \text{Var}(T) = (20.218)^2 \times 0.906 \approx 370.35 \).
Answer: \( \boxed{370} \)