Question 11 OR - A-Level Maths

CAIE FP2 2010 June — Question 11 OR

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Find or specify CDF
Difficulty	Challenging +1.2 This is a multi-part Further Maths question requiring integration to find the CDF, geometric distribution application, and transformation of random variables. While it involves several techniques, each part follows standard procedures: (i) is routine integration and evaluation, (ii) applies geometric distribution properties directly, and (iii) uses the standard Jacobian transformation method. The calculations are straightforward with no novel insights required, making it moderately above average difficulty for Further Maths.
Spec	5.02f Geometric distribution: conditions 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03e Find cdf: by integration 5.03g Cdf of transformed variables

The continuous random variable $T$ has probability density function given by $$\mathrm { f } ( t ) = \begin{cases} 0 & t < 2 \\ \frac { 2 } { ( t - 1 ) ^ { 3 } } & t \geqslant 2 \end{cases}$$

Find the distribution function of $T$, and find also $\mathrm { P } ( T > 5 )$.
Consecutive independent observations of $T$ are made until the first observation that exceeds 5 is obtained. The random variable $N$ is the total number of observations that have been made up to and including the observation exceeding 5. Find $\mathrm { P } ( N > \mathrm { E } ( N ) )$.
Find the probability density function of $Y$, where $Y = \frac { 1 } { T - 1 }$.

Show mark scheme Show mark scheme source

Question 11:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$I_{disc} = \frac{1}{2}\cdot 4ma^2 = [2ma^2$ or $18ma^2]$	B1	Find MI of disc about $O$ [or $A$]
$I_{ring} = m(2a)^2 = [4ma^2$ or $8ma^2]$	B1	Find MI of ring about $O$ [or $A$]
$I_{rod} = (4/3)ma^2\; [$ or $22ma^2/3]$	B1	Find MI of $AO$ about $O$ [or $BO$ about $A$]
$I_{wheel} = 10ma^2 + 8m(2a)^2 = 42ma^2$ A.G.	M1, A1	Find MI of wheel about $A$
$\frac{1}{2}I_{wheel}\omega^2 = 8mg \times 2a\sin 30°$; $\omega^2 = 8mga^2/21ma^2$	M1 A1	Find angular speed $\omega$ using energy
$\omega = \sqrt{8g/21a}$ (A.E.F.) or $1.95/\sqrt{a}$	A1
$I_{new} = 8ma^2 + 4m(2a)^2 = 24ma^2$	M1 A1	Find new MI about $A$
$\frac{1}{2}I_{new}\omega^2 = M_{new}g \times 2a\sin\theta$	M1	Find required angle $\theta$ using energy
$M_{new} = m + 3m = 4m$	A1	Find and use new mass
$(32/7)mga = 8mga\sin\theta$ (A.E.F.)	A1	Substitute for $I_{new},\; M_{new},\; \omega^2$
$\theta = \sin^{-1}(4/7) = 0.608\;\text{rad}$ or $34.8°$	A1	Solve for $\theta$

Total: 14 marks

Question 11 OR:

Part (i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Integrate to find $F(t)$ for $t \geq 2$: $F(t) = c - (t-1)^{-2}$	M1	$c$ needed
Use $F(2) = 0$ to find $c$: $F(t) = 1 - (t-1)^{-2}$	A1
Find $p = P(T > 5)$: $p = 1 - F(5) = 1 - (1 - 4^{-2}) = \frac{1}{16}$	B1

Part mark: 3

Part (ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
State or imply distribution: $P(N > n) = p(1-p)^{n-1}$ or geometric distribution with parameter $p$	M1
Find $P(N > E(N))$: $(1-p)^{1/p} = (15/16)^{16} = 0.356$	M1 A1

Part mark: 3

Part (iii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Relate distribution function $G(y)$ of $Y$ to $T$: $G(y) = P(Y < y) = P(1/(T-1) < y)$	M1
Rearrange: $= P(T > 1 + 1/y)$	A1
Relate to distribution function $F$: $= 1 - F(1 + 1/y)$	M1
Substitute expression for $F$: $= 1 - \{1 - (1 + 1/y - 1)^{-2}\}$	A1
Simplify: $= y^2$	A1
Differentiate to find probability density function: $g(y) = 2y$	M1 A1
Give complete statement of $g(y)$: $g(y) = 2y \ (0 \leq y \leq 1)$, $0$ otherwise	A1

Part mark: 8

Total: [14]

## Question 11:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_{disc} = \frac{1}{2}\cdot 4ma^2 = [2ma^2$ or $18ma^2]$ | B1 | Find MI of disc about $O$ [or $A$] |
| $I_{ring} = m(2a)^2 = [4ma^2$ or $8ma^2]$ | B1 | Find MI of ring about $O$ [or $A$] |
| $I_{rod} = (4/3)ma^2\; [$ or $22ma^2/3]$ | B1 | Find MI of $AO$ about $O$ [or $BO$ about $A$] |
| $I_{wheel} = 10ma^2 + 8m(2a)^2 = 42ma^2$ **A.G.** | M1, A1 | Find MI of wheel about $A$ |
| $\frac{1}{2}I_{wheel}\omega^2 = 8mg \times 2a\sin 30°$; $\omega^2 = 8mga^2/21ma^2$ | M1 A1 | Find angular speed $\omega$ using energy |
| $\omega = \sqrt{8g/21a}$ (A.E.F.) or $1.95/\sqrt{a}$ | A1 | |
| $I_{new} = 8ma^2 + 4m(2a)^2 = 24ma^2$ | M1 A1 | Find new MI about $A$ |
| $\frac{1}{2}I_{new}\omega^2 = M_{new}g \times 2a\sin\theta$ | M1 | Find required angle $\theta$ using energy |
| $M_{new} = m + 3m = 4m$ | A1 | Find and use new mass |
| $(32/7)mga = 8mga\sin\theta$ (A.E.F.) | A1 | Substitute for $I_{new},\; M_{new},\; \omega^2$ |
| $\theta = \sin^{-1}(4/7) = 0.608\;\text{rad}$ or $34.8°$ | A1 | Solve for $\theta$ |

**Total: 14 marks**

## Question 11 OR:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate to find $F(t)$ for $t \geq 2$: $F(t) = c - (t-1)^{-2}$ | M1 | $c$ needed |
| Use $F(2) = 0$ to find $c$: $F(t) = 1 - (t-1)^{-2}$ | A1 | |
| Find $p = P(T > 5)$: $p = 1 - F(5) = 1 - (1 - 4^{-2}) = \frac{1}{16}$ | B1 | |

**Part mark: 3**

---

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply distribution: $P(N > n) = p(1-p)^{n-1}$ *or* geometric distribution with parameter $p$ | M1 | |
| Find $P(N > E(N))$: $(1-p)^{1/p} = (15/16)^{16} = 0.356$ | M1 A1 | |

**Part mark: 3**

---

### Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Relate distribution function $G(y)$ of $Y$ to $T$: $G(y) = P(Y < y) = P(1/(T-1) < y)$ | M1 | |
| Rearrange: $= P(T > 1 + 1/y)$ | A1 | |
| Relate to distribution function $F$: $= 1 - F(1 + 1/y)$ | M1 | |
| Substitute expression for $F$: $= 1 - \{1 - (1 + 1/y - 1)^{-2}\}$ | A1 | |
| Simplify: $= y^2$ | A1 | |
| Differentiate to find probability density function: $g(y) = 2y$ | M1 A1 | |
| Give complete statement of $g(y)$: $g(y) = 2y \ (0 \leq y \leq 1)$, $0$ otherwise | A1 | |

**Part mark: 8**

**Total: [14]**

Show LaTeX source

The continuous random variable $T$ has probability density function given by

$$\mathrm { f } ( t ) = \begin{cases} 0 & t < 2 \\ \frac { 2 } { ( t - 1 ) ^ { 3 } } & t \geqslant 2 \end{cases}$$

(i) Find the distribution function of $T$, and find also $\mathrm { P } ( T > 5 )$.\\
(ii) Consecutive independent observations of $T$ are made until the first observation that exceeds 5 is obtained. The random variable $N$ is the total number of observations that have been made up to and including the observation exceeding 5. Find $\mathrm { P } ( N > \mathrm { E } ( N ) )$.\\
(iii) Find the probability density function of $Y$, where $Y = \frac { 1 } { T - 1 }$.

\hfill \mbox{\textit{CAIE FP2 2010 Q11 OR}}

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Q1 5 Q2 7 Q3 9 Q4 9 Q5 13 Q6 5 Q7 7 Q8 9 Q9 9 Q10 13 Q11 EITHER Q11 OR

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(I_{disc} = \frac{1}{2}\cdot 4ma^2 = [2ma^2\) or \(18ma^2]\)	B1	Find MI of disc about \(O\) [or \(A\)]
\(I_{ring} = m(2a)^2 = [4ma^2\) or \(8ma^2]\)	B1	Find MI of ring about \(O\) [or \(A\)]
\(I_{rod} = (4/3)ma^2\; [\) or \(22ma^2/3]\)	B1	Find MI of \(AO\) about \(O\) [or \(BO\) about \(A\)]
\(I_{wheel} = 10ma^2 + 8m(2a)^2 = 42ma^2\) A.G.	M1, A1	Find MI of wheel about \(A\)
\(\frac{1}{2}I_{wheel}\omega^2 = 8mg \times 2a\sin 30°\); \(\omega^2 = 8mga^2/21ma^2\)	M1 A1	Find angular speed \(\omega\) using energy
\(\omega = \sqrt{8g/21a}\) (A.E.F.) or \(1.95/\sqrt{a}\)	A1
\(I_{new} = 8ma^2 + 4m(2a)^2 = 24ma^2\)	M1 A1	Find new MI about \(A\)
\(\frac{1}{2}I_{new}\omega^2 = M_{new}g \times 2a\sin\theta\)	M1	Find required angle \(\theta\) using energy
\(M_{new} = m + 3m = 4m\)	A1	Find and use new mass
\((32/7)mga = 8mga\sin\theta\) (A.E.F.)	A1	Substitute for \(I_{new},\; M_{new},\; \omega^2\)
\(\theta = \sin^{-1}(4/7) = 0.608\;\text{rad}\) or \(34.8°\)	A1	Solve for \(\theta\)