| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Find quantiles from CDF |
| Difficulty | Challenging +1.2 Part (a) is straightforward quartile calculation from a given CDF. Part (b) requires finding the PDF by differentiation, then computing E(X²) and E(X⁴) using integration, which is moderately demanding. Part (c) involves the transformation technique for continuous random variables. While this covers multiple techniques, each step follows standard procedures taught in Further Statistics, making it above average but not exceptionally challenging. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(U) = 0.75\), so \(1 - \frac{1}{144}(12-U)^2 = \frac{3}{4}\) | M1 | |
| \((12-U)^2 = 36\), \(12-U = \pm 6\), \(U = 6\) only | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([f(x) =]\ \frac{1}{72}(12-x)\) | B1 | Differentiate given PDF |
| \(E(X^2) = \int_0^{12} \frac{1}{72}(12x^2 - x^3)\,dx = \frac{1}{72}\left(4x^3 - \frac{1}{4}x^4\right) [= 24]\) | M1 | Integrate \(x^2 f(x)\) with their \(f(x)\) |
| \(E(X^4) = \int_0^{12} \frac{1}{72}(12x^4 - x^5)\,dx = \frac{1}{72}\left(\frac{12}{5}x^5 - \frac{1}{6}x^6\right) [= 1382.4]\) | M1 | Integrate \(x^4 f(x)\) with their \(f(x)\) |
| \(\text{Var}(X^2) = E(X^4) - E^2(X^2)\ [= 1382.4 - 24^2]\) | M1 | Using correct formula with numerical values possibly unsimplified |
| \(1382.4 - 576 = 806(.4)\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(y) = \begin{cases} 0 & y < 0 \\ 1 - \frac{1}{144}(12-y^2)^2 & 0 \leq y \leq \sqrt{12} \\ 1 & y > \sqrt{12} \end{cases}\) | M1 | Change the variable: \(1 - \frac{1}{144}(12-y^2)^2\). Allow domain unchanged at this stage |
| \(\frac{1}{36}(12y - y^3)\) | M1 | Differentiate their \(G(y)\) |
| \(g(y) = \begin{cases} \frac{1}{36}(12y-y^3) & 0 \leq y \leq \sqrt{12} \\ 0 & \text{otherwise} \end{cases}\) | A1 | Fully correct |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(U) = 0.75$, so $1 - \frac{1}{144}(12-U)^2 = \frac{3}{4}$ | M1 | |
| $(12-U)^2 = 36$, $12-U = \pm 6$, $U = 6$ only | A1 | |
| | **2** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[f(x) =]\ \frac{1}{72}(12-x)$ | B1 | Differentiate given PDF |
| $E(X^2) = \int_0^{12} \frac{1}{72}(12x^2 - x^3)\,dx = \frac{1}{72}\left(4x^3 - \frac{1}{4}x^4\right) [= 24]$ | M1 | Integrate $x^2 f(x)$ with their $f(x)$ |
| $E(X^4) = \int_0^{12} \frac{1}{72}(12x^4 - x^5)\,dx = \frac{1}{72}\left(\frac{12}{5}x^5 - \frac{1}{6}x^6\right) [= 1382.4]$ | M1 | Integrate $x^4 f(x)$ with their $f(x)$ |
| $\text{Var}(X^2) = E(X^4) - E^2(X^2)\ [= 1382.4 - 24^2]$ | M1 | Using correct formula with numerical values possibly unsimplified |
| $1382.4 - 576 = 806(.4)$ | A1 | |
| | **5** | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(y) = \begin{cases} 0 & y < 0 \\ 1 - \frac{1}{144}(12-y^2)^2 & 0 \leq y \leq \sqrt{12} \\ 1 & y > \sqrt{12} \end{cases}$ | M1 | Change the variable: $1 - \frac{1}{144}(12-y^2)^2$. Allow domain unchanged at this stage |
| $\frac{1}{36}(12y - y^3)$ | M1 | Differentiate their $G(y)$ |
| $g(y) = \begin{cases} \frac{1}{36}(12y-y^3) & 0 \leq y \leq \sqrt{12} \\ 0 & \text{otherwise} \end{cases}$ | A1 | Fully correct |5 The continuous random variable $X$ has cumulative distribution function F given by
$$F ( x ) = \begin{cases} 0 & x < 0 \\ 1 - \frac { 1 } { 144 } ( 12 - x ) ^ { 2 } & 0 \leqslant x \leqslant 12 \\ 1 & x > 12 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the upper quartile of $X$.
\item Find $\operatorname { Var } \left( X ^ { 2 } \right)$.\\
The random variable $Y$ is given by $Y = \sqrt { X }$.
\item Find the probability density function of $Y$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q5 [10]}}