| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| 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 the constant using ∫f(x)dx=1, calculating E(X²) by integration, and constructing the CDF by integrating each piece. While it involves multiple parts and careful handling of the piecewise structure, these are routine techniques for Further Maths students with no novel problem-solving required—slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Total prob \(= 1\), so \(\int_0^1\left(a+\frac{1}{5}x\right)dx + \int_1^2\left(2a-\frac{1}{5}x\right)dx = 1\) | *M1 | Correct expression integrated and equated to 1. Powers of \(x\) correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(3a - \frac{1}{5} = 1\) | DM1 | Limits used and attempt to solve. |
| Solve to give \(a = \frac{2}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 \frac{1}{5}(2+x)x^2\,dx + \int_1^2 \frac{1}{5}(4-x)x^2\,dx\) | M1 | Correct expressions integrated, FT their \(a\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{13}{10}\) | A1 | \(\frac{11}{60}+\frac{67}{60}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{5}\left(2x+\frac{1}{2}x^2\right) & 0 \leq x < 1 \\ \frac{1}{5}\left(4x-\frac{1}{2}x^2-1\right) & 1 \leq x \leq 2 \\ 1 & x > 2 \end{cases}\] | M1 | Integration of their PDF. |
| Middle 2 parts correct | A1 | |
| \(= 0\ (x<0)\) and \(= 1\ (x>2)\) | A1 | First and last parts correct and all domains correct, with no gaps. |
## Question 3(a):
Total prob $= 1$, so $\int_0^1\left(a+\frac{1}{5}x\right)dx + \int_1^2\left(2a-\frac{1}{5}x\right)dx = 1$ | *M1 | Correct expression integrated and equated to 1. Powers of $x$ correct.
$\left[ax+\frac{1}{10}x^2\right] + \left[2ax-\frac{1}{10}x^2\right] = 1$
$3a - \frac{1}{5} = 1$ | DM1 | Limits used and attempt to solve.
Solve to give $a = \frac{2}{5}$ | A1 |
---
## Question 3(b):
$\int_0^1 \frac{1}{5}(2+x)x^2\,dx + \int_1^2 \frac{1}{5}(4-x)x^2\,dx$ | M1 | Correct expressions integrated, FT their $a$.
$\frac{1}{5}\left[\frac{2}{3}x^3+\frac{1}{4}x^4\right] + \frac{1}{5}\left[\frac{4}{3}x^3-\frac{1}{4}x^4\right]$
$\frac{13}{10}$ | A1 | $\frac{11}{60}+\frac{67}{60}$
---
## Question 3(c):
$$F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{5}\left(2x+\frac{1}{2}x^2\right) & 0 \leq x < 1 \\ \frac{1}{5}\left(4x-\frac{1}{2}x^2-1\right) & 1 \leq x \leq 2 \\ 1 & x > 2 \end{cases}$$ | M1 | Integration of their PDF.
Middle 2 parts correct | A1 |
$= 0\ (x<0)$ and $= 1\ (x>2)$ | A1 | First and last parts correct and all domains correct, with no gaps.
---3 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} a + \frac { 1 } { 5 } x & 0 \leqslant x < 1 \\ 2 a - \frac { 1 } { 5 } x & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item Find $\mathrm { E } \left( X ^ { 2 } \right)$.
\item Find the cumulative distribution function of $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q3 [8]}}