| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths statistics question involving a uniform distribution. Finding the CDF requires simple integration of a constant, the transformation Y=X³ uses standard change of variable formula (differentiating the inverse), and E(Y) and Var(Y) involve routine integration. All techniques are mechanical applications of standard methods with no conceptual challenges. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integrate to find \(F(x)\) for \(1 \leq x \leq 3\): \(F(x) = \frac{1}{2}(x-1)\) | B1 | |
| State \(F(x)\) for other intervals: \(0\,(x<1)\), \(1\,(x>3)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Relate dist. fn. \(G(y)\) of \(Y\) to \(X\): \(G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3}) = \frac{1}{2}(y^{1/3} - 1)\) | M1 A1 | working may be omitted |
| Differentiate to find \(g(y)\): \(g(y) = y^{-2/3}/6\;(1 \leq y \leq 27)\), \([= 0 \text{ otherwise}]\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(E(Y)\): \(E(Y) = \int_1^{27} y\,(y^{-2/3}/6)\,dy = [y^{4/3}/8]_1^{27}\) or \([x^4/8]_1^3 = (81-1)/8 = 10\) | B1 | |
| Find \(\text{Var}(Y)\): \(E(Y^2) = \int_1^{27} y^2(y^{-2/3}/6)\,dy = [y^{7/3}/14]_1^{27}\) or \([x^7/14]_1^3 = (2187-1)/14 = 1093/7\) | ||
| \(\text{Var}(Y) = E(Y^2) - 10^2 = 393/7\) or \(56.1[4]\) | M1 A1 | |
| Subtotal: 3 marks | Total: 8 marks |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrate to find $F(x)$ for $1 \leq x \leq 3$: $F(x) = \frac{1}{2}(x-1)$ | B1 | |
| State $F(x)$ for other intervals: $0\,(x<1)$, $1\,(x>3)$ | B1 | |
**Subtotal: 2 marks**
## Question 6(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Relate dist. fn. $G(y)$ of $Y$ to $X$: $G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3}) = \frac{1}{2}(y^{1/3} - 1)$ | M1 A1 | working may be omitted |
| Differentiate to find $g(y)$: $g(y) = y^{-2/3}/6\;(1 \leq y \leq 27)$, $[= 0 \text{ otherwise}]$ | B1 | |
**Subtotal: 3 marks**
## Question 6(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $E(Y)$: $E(Y) = \int_1^{27} y\,(y^{-2/3}/6)\,dy = [y^{4/3}/8]_1^{27}$ or $[x^4/8]_1^3 = (81-1)/8 = 10$ | B1 | |
| Find $\text{Var}(Y)$: $E(Y^2) = \int_1^{27} y^2(y^{-2/3}/6)\,dy = [y^{7/3}/14]_1^{27}$ or $[x^7/14]_1^3 = (2187-1)/14 = 1093/7$ | | |
| $\text{Var}(Y) = E(Y^2) - 10^2 = 393/7$ or $56.1[4]$ | M1 A1 | |
**Subtotal: 3 marks | Total: 8 marks**
---6 The continuous random variable $X$ has probability density function f given by
$$\mathrm { f } ( x ) = \begin{cases} 0 & x < 1 \\ \frac { 1 } { 2 } & 1 \leqslant x \leqslant 3 \\ 0 & x > 3 \end{cases}$$
Find the distribution function of $X$.
The random variable $Y$ is defined by $Y = X ^ { 3 }$. Find\\
(i) the probability density function of $Y$,\\
(ii) the expected value and variance of $Y$.
\hfill \mbox{\textit{CAIE FP2 2011 Q6 [8]}}