Question 11 OR - A-Level Maths

CAIE FP2 2013 November — Question 11 OR

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Composite/applied transformation
Difficulty	Challenging +1.2 This question requires understanding the relationship T = 60 - X, finding the CDF through transformation and integration of a cubic polynomial, then computing median and mean from the derived distribution. While it involves multiple steps and careful algebraic manipulation, the techniques are standard for Further Pure probability: variable transformation, polynomial integration, and using CDF properties. The cubic makes arithmetic slightly tedious but conceptually this is a straightforward application of learned methods rather than requiring novel insight.
Spec	5.03a Continuous random variables: pdf and cdf 5.03e Find cdf: by integration

Guided tours of a museum begin every 60 minutes. A randomly chosen tourist arrives $X$ minutes after the start of a tour. The continuous random variable $X$ has probability density function f given by $$f ( x ) = \begin{cases} \frac { ( x - 20 ) ^ { 2 } } { 24000 } & 0 < x < 60 \\ 0 & \text { otherwise } \end{cases}$$ The random variable $T$ is the time that the tourist has to wait for the next tour to begin. Show that the distribution function G of $T$ is given by $$\mathrm { G } ( t ) = \begin{cases} 0 & t \leqslant 0 \\ \frac { 8 } { 9 } - \frac { ( 40 - t ) ^ { 3 } } { 72000 } & 0 < t < 60 \\ 1 & t \geqslant 60 \end{cases}$$ Find the median and the mean of $T$.

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Question 11(a)(i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$I_{AB} = \frac{1}{3}\ 4m(3a)^2 + 4m(7a)^2 = 208ma^2$	M1 A1	State or find MI of rod $AB$ about $C$
$I_{disc} = \frac{1}{2}\ 8m(2a)^2 + 8m(2a)^2 = 48ma^2$	M1 A1	Find MI of disc about $C$
$I = I_{AB} + I_{disc} = 256ma^2$	A1	Find MI of body about $C$
$I\ \frac{d^2\theta}{dt^2} = [-](4 \times 7 + 8 \times 2)mga\sin\theta$	M1 A1	Use equation of circular motion to find $\frac{d^2\theta}{dt^2}$
$\frac{d^2\theta}{dt^2} = -\frac{11g}{64a}\ \theta$	A1	Approximate $\sin\theta$ by $\theta$ and substitute for $I$
$T = 16\pi\sqrt{\frac{a}{11g}}$ A.G.	B1	Find period $T = \frac{2\pi}{\omega}$ with $\omega = \sqrt{\frac{11g}{64a}}$

Question 11(a)(ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\frac{1}{2}I\omega^2 = 4mg\ 7a(1 - \cos\theta) + 8mg\ 2a(1-\cos\theta)$	M1 A1	Use energy to find max. angular velocity $\omega$
$\omega^2 = \frac{2(11mga)0.4}{256ma^2}$		Substitute for $I$ and $\cos\theta$ and simplify
$= \frac{11g}{80a}$	A1
$v_A = 10a\omega = \sqrt{\frac{55ga}{4}}$	M1 A1	Find maximum speed of $A$ (A.E.F.)

Question 11(b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$F(x) = \frac{1}{3}(x-20)^3/24000 + \frac{1}{3}\ 20^3/24000$	M1	Find $F(x)$ for $0 < x < 60$ by integration
$= (x-20)^3/72000 + \frac{1}{9}$	A1
$F(x) = 0\ (x \leq 0),\ 1\ (x \geq 60)$	B1	State $F(x)$ for other values of $x$
$G(t) = P(T < t) = P(60 - X < t) = P(X > 60 - t)$		Find $G(t)$ for $0 < t < 60$ from $X + T = 60$
$= 1 - F(60-t)$
$= \frac{8}{9} - \frac{(40-t)^3}{72000}$ A.G.	M1 A1
$\frac{8}{9} - \frac{(40-m)^3}{72000} = \frac{1}{2}$	M1	Formulate equation for median $m$ of $T$
$(40-m)^3 = (\frac{8}{9} - \frac{1}{2})72000 = 28000$	M1	Find value of $m$
$m = 40 - 28000^{1/3} = 9.63$	A1
$g(t) = (40-t)^2/24000$	M1 A1	Find $g(t)$ for $0 < t < 60$
$E(T) = \int t \cdot g(t)\, dt = \int(40^2t - 80t^2 + t^3)\, dt/24000$	M1	Find $E(T)$ from $\int t\, g(t)\, dt$
$= \left[\frac{1}{2}40^2t^2 - \frac{1}{3}80t^3 + \frac{1}{4}t^4\right]_0^{60}/24000$	A1
$= 120 - 240 + 135 = 15$	A1

## Question 11(a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{AB} = \frac{1}{3}\ 4m(3a)^2 + 4m(7a)^2 = 208ma^2$ | M1 A1 | State or find MI of rod $AB$ about $C$ |
| $I_{disc} = \frac{1}{2}\ 8m(2a)^2 + 8m(2a)^2 = 48ma^2$ | M1 A1 | Find MI of disc about $C$ |
| $I = I_{AB} + I_{disc} = 256ma^2$ | A1 | Find MI of body about $C$ |
| $I\ \frac{d^2\theta}{dt^2} = [-](4 \times 7 + 8 \times 2)mga\sin\theta$ | M1 A1 | Use equation of circular motion to find $\frac{d^2\theta}{dt^2}$ |
| $\frac{d^2\theta}{dt^2} = -\frac{11g}{64a}\ \theta$ | A1 | Approximate $\sin\theta$ by $\theta$ and substitute for $I$ |
| $T = 16\pi\sqrt{\frac{a}{11g}}$ **A.G.** | B1 | Find period $T = \frac{2\pi}{\omega}$ with $\omega = \sqrt{\frac{11g}{64a}}$ |

## Question 11(a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}I\omega^2 = 4mg\ 7a(1 - \cos\theta) + 8mg\ 2a(1-\cos\theta)$ | M1 A1 | Use energy to find max. angular velocity $\omega$ |
| $\omega^2 = \frac{2(11mga)0.4}{256ma^2}$ | | Substitute for $I$ and $\cos\theta$ and simplify |
| $= \frac{11g}{80a}$ | A1 | |
| $v_A = 10a\omega = \sqrt{\frac{55ga}{4}}$ | M1 A1 | Find maximum speed of $A$ (A.E.F.) |

---

## Question 11(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \frac{1}{3}(x-20)^3/24000 + \frac{1}{3}\ 20^3/24000$ | M1 | Find $F(x)$ for $0 < x < 60$ by integration |
| $= (x-20)^3/72000 + \frac{1}{9}$ | A1 | |
| $F(x) = 0\ (x \leq 0),\ 1\ (x \geq 60)$ | B1 | State $F(x)$ for other values of $x$ |
| $G(t) = P(T < t) = P(60 - X < t) = P(X > 60 - t)$ | | Find $G(t)$ for $0 < t < 60$ from $X + T = 60$ |
| $= 1 - F(60-t)$ | | |
| $= \frac{8}{9} - \frac{(40-t)^3}{72000}$ **A.G.** | M1 A1 | |
| $\frac{8}{9} - \frac{(40-m)^3}{72000} = \frac{1}{2}$ | M1 | Formulate equation for median $m$ of $T$ |
| $(40-m)^3 = (\frac{8}{9} - \frac{1}{2})72000 = 28000$ | M1 | Find value of $m$ |
| $m = 40 - 28000^{1/3} = 9.63$ | A1 | |
| $g(t) = (40-t)^2/24000$ | M1 A1 | Find $g(t)$ for $0 < t < 60$ |
| $E(T) = \int t \cdot g(t)\, dt = \int(40^2t - 80t^2 + t^3)\, dt/24000$ | M1 | Find $E(T)$ from $\int t\, g(t)\, dt$ |
| $= \left[\frac{1}{2}40^2t^2 - \frac{1}{3}80t^3 + \frac{1}{4}t^4\right]_0^{60}/24000$ | A1 | |
| $= 120 - 240 + 135 = 15$ | A1 | |

Show LaTeX source

Guided tours of a museum begin every 60 minutes. A randomly chosen tourist arrives $X$ minutes after the start of a tour. The continuous random variable $X$ has probability density function f given by

$$f ( x ) = \begin{cases} \frac { ( x - 20 ) ^ { 2 } } { 24000 } & 0 < x < 60 \\ 0 & \text { otherwise } \end{cases}$$

The random variable $T$ is the time that the tourist has to wait for the next tour to begin. Show that the distribution function G of $T$ is given by

$$\mathrm { G } ( t ) = \begin{cases} 0 & t \leqslant 0 \\ \frac { 8 } { 9 } - \frac { ( 40 - t ) ^ { 3 } } { 72000 } & 0 < t < 60 \\ 1 & t \geqslant 60 \end{cases}$$

Find the median and the mean of $T$.

\hfill \mbox{\textit{CAIE FP2 2013 Q11 OR}}

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Q2 9 Q3 9 Q4 10 Q5 12 Q6 5 Q7 7 Q8 10 Q9 10 Q10 11 Q11 EITHER Q11 OR

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(F(x) = \frac{1}{3}(x-20)^3/24000 + \frac{1}{3}\ 20^3/24000\)	M1	Find \(F(x)\) for \(0 < x < 60\) by integration
\(= (x-20)^3/72000 + \frac{1}{9}\)	A1
\(F(x) = 0\ (x \leq 0),\ 1\ (x \geq 60)\)	B1	State \(F(x)\) for other values of \(x\)
\(G(t) = P(T < t) = P(60 - X < t) = P(X > 60 - t)\)		Find \(G(t)\) for \(0 < t < 60\) from \(X + T = 60\)
\(= 1 - F(60-t)\)
\(= \frac{8}{9} - \frac{(40-t)^3}{72000}\) A.G.	M1 A1
\(\frac{8}{9} - \frac{(40-m)^3}{72000} = \frac{1}{2}\)	M1	Formulate equation for median \(m\) of \(T\)
\((40-m)^3 = (\frac{8}{9} - \frac{1}{2})72000 = 28000\)	M1	Find value of \(m\)
\(m = 40 - 28000^{1/3} = 9.63\)	A1
\(g(t) = (40-t)^2/24000\)	M1 A1	Find \(g(t)\) for \(0 < t < 60\)
\(E(T) = \int t \cdot g(t)\, dt = \int(40^2t - 80t^2 + t^3)\, dt/24000\)	M1	Find \(E(T)\) from \(\int t\, g(t)\, dt\)
\(= \left[\frac{1}{2}40^2t^2 - \frac{1}{3}80t^3 + \frac{1}{4}t^4\right]_0^{60}/24000\)	A1
\(= 120 - 240 + 135 = 15\)	A1