| 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 | PDF from CDF |
| Difficulty | Standard +0.3 This is a straightforward Further Maths statistics question requiring differentiation of a piecewise CDF to find the PDF, then standard probability and expectation calculations. While it involves multiple parts and careful handling of piecewise functions, each step follows routine procedures without requiring novel insight or particularly challenging integration. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \begin{cases} 1+x & -1 \leq x \leq 0 \\ 1-x & 0 < x \leq 1 \\ 0 & \text{otherwise} \end{cases}\) | M1 | Differentiation attempted |
| All correct, including 0 otherwise | A1 | All correct, including 0 otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\left(-\frac{1}{2} \leq X \leq \frac{1}{2}\right) = F\left(\frac{1}{2}\right) - F\left(-\frac{1}{2}\right) = \frac{7}{8} - \frac{1}{8}\) | M1 | Can do by integration with correct limits |
| \(\frac{3}{4}\) | A1 |
| Answer | Marks |
|---|---|
| \(\int_{-1}^{0} x^2(1+x)dx + \int_{0}^{1} x^2(1-x)dx\) | M1 |
| \(\left[\frac{1}{3}x^3 + \frac{1}{4}x^4\right] + \left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]\) | |
| \(\frac{1}{12} + \frac{1}{12} = \frac{1}{6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^{0} x^4(1+x)dx + \int_{0}^{1} x^4(1-x)dx - \left(\frac{1}{6}\right)^2\) | M1 | Complete method with their mean squared explicit |
| \(\left[\frac{1}{5}x^5 + \frac{1}{6}x^6\right] + \left[\frac{1}{5}x^5 - \frac{1}{6}x^6\right] - \frac{1}{36}\) | ||
| \(\frac{1}{15} - \frac{1}{36} = \frac{7}{180}\) | A1 |
## Question 2(a):
$f(x) = \begin{cases} 1+x & -1 \leq x \leq 0 \\ 1-x & 0 < x \leq 1 \\ 0 & \text{otherwise} \end{cases}$ | M1 | Differentiation attempted |
| All correct, including 0 otherwise | A1 | All correct, including 0 otherwise |
## Question 2(b):
$P\left(-\frac{1}{2} \leq X \leq \frac{1}{2}\right) = F\left(\frac{1}{2}\right) - F\left(-\frac{1}{2}\right) = \frac{7}{8} - \frac{1}{8}$ | M1 | Can do by integration with correct limits |
$\frac{3}{4}$ | A1 | |
## Question 2(c):
$\int_{-1}^{0} x^2(1+x)dx + \int_{0}^{1} x^2(1-x)dx$ | M1 | |
$\left[\frac{1}{3}x^3 + \frac{1}{4}x^4\right] + \left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]$ | | |
$\frac{1}{12} + \frac{1}{12} = \frac{1}{6}$ | A1 | |
## Question 2(d):
$\int_{-1}^{0} x^4(1+x)dx + \int_{0}^{1} x^4(1-x)dx - \left(\frac{1}{6}\right)^2$ | M1 | Complete method with their mean squared explicit |
$\left[\frac{1}{5}x^5 + \frac{1}{6}x^6\right] + \left[\frac{1}{5}x^5 - \frac{1}{6}x^6\right] - \frac{1}{36}$ | | |
$\frac{1}{15} - \frac{1}{36} = \frac{7}{180}$ | A1 | |2 The continuous random variable $X$ has cumulative distribution function F given by
$$F ( x ) = \left\{ \begin{array} { l c }
0 & x < - 1 \\
\frac { 1 } { 2 } ( 1 + x ) ^ { 2 } & - 1 \leqslant x \leqslant 0 \\
1 - \frac { 1 } { 2 } ( 1 - x ) ^ { 2 } & 0 < x \leqslant 1 \\
1 & x > 1
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability density function of $X$.
\item Find $\mathrm { P } \left( - \frac { 1 } { 2 } \leqslant X \leqslant \frac { 1 } { 2 } \right)$.
\item Find $\mathrm { E } \left( X ^ { 2 } \right)$.
\item Find $\operatorname { Var } \left( X ^ { 2 } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q2 [8]}}