| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Standard +0.3 This is a standard piecewise PDF question requiring integration to find k, then quartiles and variance. The piecewise structure is simple (constant and linear), and all parts follow routine procedures taught in Further Stats. Slightly above average difficulty due to the piecewise nature and multiple calculations, but no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 k\,dx + \int_1^2 kx\,dx = [kx] + \left[\frac{k}{2}x^2\right],\quad k + \frac{3}{2}k = 1,\quad k = \frac{2}{5}\) | B1 | Integration and \(k + 3k/2 = 1\) oe seen |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(x) = \begin{cases} kx & 0 \leq x < 1 \\ \frac{k}{2}x^2 + \frac{1}{5} & 1 \leq x \leq 2 \end{cases}\) | M1 | CDF or integration with \(\frac{1}{5}x^2 + \frac{1}{5}\) seen |
| UQ: \(\frac{k}{2}x^2 + \frac{1}{5} = \frac{3}{4}\) | M1 | Method for UQ |
| \(\left[x^2 + 1 = 5 \times \frac{3}{4},\quad x^2 = \frac{11}{4}\right]\quad x = \frac{1}{2}\sqrt{11}\) | A1 | UQ |
| LQ: \(kx = \frac{1}{4},\quad x = \frac{5}{8}\) | B1 | LQ |
| \(\text{IQR} = \frac{1}{2}\sqrt{11} - \frac{5}{8} = 1.03(3)\) | A1 | CAO |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \int_0^1 kx\,dx + \int_1^2 kx^2\,dx = \left[\frac{k}{2}x^2\right] + \left[\frac{k}{3}x^3\right] = \frac{17}{15}\) | M1 | With correct limits |
| \(E(X^2) = \int_0^1 kx^2\,dx + \int_1^2 kx^3\,dx = \left[\frac{k}{3}x^3\right] + \left[\frac{k}{4}x^4\right] = \frac{49}{30}\) | M1 | With correct limits |
| \(\text{Var}(X) = \frac{49}{30} - \left(\frac{17}{15}\right)^2\) | M1 | Using correct formula with numerical values |
| \(\frac{157}{450}\ (= 0.349)\) | A1 | |
| 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 k\,dx + \int_1^2 kx\,dx = [kx] + \left[\frac{k}{2}x^2\right],\quad k + \frac{3}{2}k = 1,\quad k = \frac{2}{5}$ | B1 | Integration and $k + 3k/2 = 1$ oe seen |
| | **1** | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = \begin{cases} kx & 0 \leq x < 1 \\ \frac{k}{2}x^2 + \frac{1}{5} & 1 \leq x \leq 2 \end{cases}$ | M1 | CDF or integration with $\frac{1}{5}x^2 + \frac{1}{5}$ seen |
| UQ: $\frac{k}{2}x^2 + \frac{1}{5} = \frac{3}{4}$ | M1 | Method for UQ |
| $\left[x^2 + 1 = 5 \times \frac{3}{4},\quad x^2 = \frac{11}{4}\right]\quad x = \frac{1}{2}\sqrt{11}$ | A1 | UQ |
| LQ: $kx = \frac{1}{4},\quad x = \frac{5}{8}$ | B1 | LQ |
| $\text{IQR} = \frac{1}{2}\sqrt{11} - \frac{5}{8} = 1.03(3)$ | A1 | CAO |
| | **5** | |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \int_0^1 kx\,dx + \int_1^2 kx^2\,dx = \left[\frac{k}{2}x^2\right] + \left[\frac{k}{3}x^3\right] = \frac{17}{15}$ | M1 | With correct limits |
| $E(X^2) = \int_0^1 kx^2\,dx + \int_1^2 kx^3\,dx = \left[\frac{k}{3}x^3\right] + \left[\frac{k}{4}x^4\right] = \frac{49}{30}$ | M1 | With correct limits |
| $\text{Var}(X) = \frac{49}{30} - \left(\frac{17}{15}\right)^2$ | M1 | Using correct formula with numerical values |
| $\frac{157}{450}\ (= 0.349)$ | A1 | |
| | **4** | |
---4 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} k & 0 \leqslant x < 1 \\ k x & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 2 } { 5 }$.
\item Find the interquartile range of $X$.
\item Find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q4 [10]}}