IBDP Maths AHL 4.14 Variance of random variable AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
\(X \sim \text{B}(5, p)\). \({\text{P}}(X = 4) = \binom{5}{4} p^4 (1 – p) = 5p^4(1 – p)\).
\(\boxed{5p^4(1 – p)}\)
(i) Maximize \({\text{P}}(X = 4) = 5p^4(1 – p)\). Derivative: \(\frac{d}{dp}[5p^4(1 – p)] = 5[4p^3(1 – p) + p^4(-1)] = 5(4p^3 – 5p^4)\).
Set to zero: \(20p^3 – 25p^4 = 0\). Factor: \(5p^3(4 – 5p) = 0\). Solutions: \(p = 0\) or \(p = \frac{4}{5}\). Maximum at \(p = \frac{4}{5}\).
(ii) \({\text{E}}(X) = np = 5 \times \frac{4}{5} = 4\).
\(\boxed{\frac{4}{5}, 4}\)
\(X \sim \text{B}(5, p)\) M1
\({\text{P}}(X = 4) = \binom{5}{4} p^4 (1 – p)\) A1
[2 marks]
(i) \(\frac{d}{dp}(5p^4 – 5p^5) = 20p^3 – 25p^4\) M1A1
\(5p^3(4 – 5p) = 0 \Rightarrow p = \frac{4}{5}\) M1A1
Note: No \(p = 0\) in final answer for A1.
(ii) \({\text{E}}(X) = np = 5 \left(\frac{4}{5}\right)\) M1
\(= 4\) A1
[6 marks]
▶️ Answer/Explanation
Graph \(y = \frac{t|\sin 2t|}{\pi}\) from 0 to \(\pi\), with \(y = 0\) outside. Two peaks, one near \(t = \frac{\pi}{2}\), other near \(t = \frac{3\pi}{2}\), symmetric about \(t = \frac{\pi}{2}\).
(Note: Sketch not generated, but described for reference.)
From sketch, mode is where \(f(t)\) peaks, approximately \(t = 2.46\) (near \(\frac{3\pi}{4}\)).
\(\boxed{2.46}\)
\({\text{E}}(T) = \frac{1}{\pi} \int_0^\pi t \cdot \frac{t|\sin 2t|}{\pi} \, dt = \frac{1}{\pi^2} \int_0^\pi t^2 |\sin 2t| \, dt\).
Approximate value: 2.04.
\(\boxed{2.04}\)
\({\text{Var}}(T) = \int_0^\pi t^2 \cdot \frac{t|\sin 2t|}{\pi} \, dt – (2.04)^2\).
Approximate value: 0.516.
\(\boxed{0.516}\)
Probability = \(\frac{1}{\pi} \int_{2.04}^{2.46} t |\sin 2t| \, dt \approx 0.285\).
\(\boxed{0.285}\)
(i) \(\int_0^{\pi/2} \frac{t|\sin 2t|}{\pi} \, dt\). Use integration by parts: \(u = t\), \(dv = |\sin 2t| \, dt\), \(v = -\frac{1}{2} \cos 2t\).
\(\frac{1}{\pi} \left[ – \frac{t}{2} \cos 2t \right]_0^{\pi/2} + \frac{1}{\pi} \int_0^{\pi/2} \frac{1}{2} \cos 2t \, dt = \frac{1}{4}\).
(ii) Since \(\int_0^{\pi/2} f(t) \, dt = \frac{1}{4}\), lower quartile is \(\frac{\pi}{2}\).
\(\boxed{\frac{1}{4}, \frac{\pi}{2}}\)
—Markscheme—
Two enclosed regions (0 to \(\frac{\pi}{2}\) and \(\frac{\pi}{2}\) to \(\pi\)) bounded by curve and t-axis A1
Correct non-symmetrical shape, mode between \(\frac{\pi}{2}\) and \(\pi\) A1
[2 marks]
Mode = 2.46 A1
[1 mark]
\({\text{E}}(T) = \frac{1}{\pi} \int_0^\pi t \cdot \frac{t|\sin 2t|}{\pi} \, dt\) M1
\(= 2.04\) A1
[2 marks]
Either: \({\text{Var}}(T) = \int_0^\pi (t – 2.03788\ldots)^2 \cdot \frac{t|\sin 2t|}{\pi} \, dt\) M1A1
Or: \({\text{Var}}(T) = \int_0^\pi t^2 \cdot \frac{t|\sin 2t|}{\pi} \, dt – (2.03788\ldots)^2\) M1A1
\({\text{Var}}(T) = 0.516\) A1
[3 marks]
\(\frac{1}{\pi} \int_{2.03788\ldots}^{2.456590\ldots} t |\sin 2t| \, dt = 0.285\) M1A1
[2 marks]
(i) Integration by parts attempt M1
\(u = t, dv = \sin 2t \, dt, v = -\frac{1}{2} \cos 2t\) A1
\(\frac{1}{\pi} \left[ t \left( -\frac{1}{2} \cos 2t \right) \right]_0^\pi – \frac{1}{\pi} \int_0^\pi \left( -\frac{1}{2} \cos 2t \right) \, dt\) A1
\(=\frac{\sin 2t}{4\pi} – \frac{t \cos 2t}{2\pi}\) A1
(ii) \(\frac{\sin \pi}{4\pi} – \frac{\frac{\pi}{2} \cos \pi}{2\pi} = \frac{1}{4}\) A1
Since \(\text{P}(0 \leqslant T \leqslant \frac{\pi}{2}) = \frac{1}{4}\), lower quartile is \(\frac{\pi}{2}\) R1AG
[5 marks]