| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Power transformation (Y = X^n, n≥2) |
| Difficulty | Standard +0.8 This is a Further Pure Mathematics 2 question requiring transformation of continuous random variables using the Jacobian method (Y = X³), finding cumulative distribution functions, and solving probability equations. While the calculus is straightforward, the transformation technique and multi-step nature elevate it above standard A-level, placing it moderately above average difficulty for Further Maths content. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = \int f(x)\,dx = x^2/12 + c\) | M1 A1 | Find or state distribution function \(F(x)\) for \(2 \leq x \leq 4\) |
| \(F(x) = x^2/12 - \frac{1}{3}\ [2 \leq x \leq 4],\ 0\ (x<2),\ 1\ (x>4)\) | A1 | Use \(F(2)=0\) or \(F(4)=1\) to find \(F(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})\) | M1 A1 | Find or state \(G(y)\) from \(Y=X^3\) for \(2 \leq x \leq 4\); allow \(<\) or \(\leq\) throughout |
| \(= y^{2/3}/12 - \frac{1}{3}\) | A0 if \(G(y)\) incorrect | |
| \(g(y) = (1/18)\,y^{-1/3}\ [8 \leq y \leq 64,\ 0 \text{ otherwise}]\) | M1 A1 | Find \(g(y)\) by differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (AEF) \(7/12 = 1 - G(k) = 1 - k^{2/3}/12 + \frac{1}{3}\) | M1 A1 | Formulate condition for \(k\); M0 for \(7/12 = G(k)\) |
| \(k^{2/3} = 9,\ k = 27\) | A1 | Find \(k\) |
# Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = \int f(x)\,dx = x^2/12 + c$ | M1 A1 | Find or state distribution function $F(x)$ for $2 \leq x \leq 4$ |
| $F(x) = x^2/12 - \frac{1}{3}\ [2 \leq x \leq 4],\ 0\ (x<2),\ 1\ (x>4)$ | A1 | Use $F(2)=0$ or $F(4)=1$ to find $F(x)$ |
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# Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})$ | M1 A1 | Find or state $G(y)$ from $Y=X^3$ for $2 \leq x \leq 4$; allow $<$ or $\leq$ throughout |
| $= y^{2/3}/12 - \frac{1}{3}$ | | A0 if $G(y)$ incorrect |
| $g(y) = (1/18)\,y^{-1/3}\ [8 \leq y \leq 64,\ 0 \text{ otherwise}]$ | M1 A1 | Find $g(y)$ by differentiation |
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# Question 7(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (AEF) $7/12 = 1 - G(k) = 1 - k^{2/3}/12 + \frac{1}{3}$ | M1 A1 | Formulate condition for $k$; M0 for $7/12 = G(k)$ |
| $k^{2/3} = 9,\ k = 27$ | A1 | Find $k$ |
---7 The random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(i) Find the distribution function of $X$.
The random variable $Y$ is defined by $Y = X ^ { 3 }$. Find\\
(ii) the probability density function of $Y$,\\
(iii) the value of $k$ for which $\mathrm { P } ( Y \geqslant k ) = \frac { 7 } { 12 }$.
\hfill \mbox{\textit{CAIE FP2 2016 Q7 [8]}}