| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| 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 k, then E(X²) for variance, and solving for the median. All techniques are routine for Further Maths students, though the piecewise nature and algebraic manipulation add modest complexity beyond basic continuous distributions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k2x^2 - \frac{1}{3}x^3 + k\left[6x - \frac{1}{2}x^2\right] = 1\); \(k\left[8 - \frac{8}{3} + 36 - 18 - 12 + 2\right] = \frac{40}{3}k = 1\); \(k = \frac{3}{40}\) | B1 | Integrate and equate to 1, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X^2) = k\int_0^2\left(4x^3 - x^4\right)dx + k\int_2^6\left(6x^2 - x^3\right)dx = k\left[x^4 - \frac{1}{5}x^5\right] + k\left[2x^3 - \frac{1}{4}x^4\right] = k[9.6 + 96] = 7.92\) | M1* | Attempt at \(E(X^2)\), correct limits, integrated |
| \(\text{Var}(X) = E(X^2) - E^2(X) = \text{their } (7.92) - 2.5^2\) | depM1 | Use of correct formula, with their \(E(X^2)\) |
| \(1.67\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^2 kx(4-x)dx + \int_2^m k(6-x)dx = \frac{1}{2}\) | M1* | Correct expression and attempt to integrate |
| \(\int_0^2 kx(4-x)dx = \frac{2}{5}\) | B1 | May be implied |
| \(\frac{2}{5} + k\left[6x - \frac{1}{2}x^2\right]_2^m = \frac{1}{2}\); \(6m - \frac{1}{2}m^2 - 12 + 2 = \frac{1}{10k}\); \(m^2 - 12m + \frac{68}{3} = 0\) | depM1 | First integral and \(\frac{1}{2}\) may be seen combined as \(\frac{1}{10}\). Obtain quadratic equation |
| \(m = 6 \pm \frac{2}{3}\sqrt{30}\); \(m = 6 - \frac{2}{3}\sqrt{30}\ (= 2.35)\) | A1 | Single answer |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k2x^2 - \frac{1}{3}x^3 + k\left[6x - \frac{1}{2}x^2\right] = 1$; $k\left[8 - \frac{8}{3} + 36 - 18 - 12 + 2\right] = \frac{40}{3}k = 1$; $k = \frac{3}{40}$ | B1 | Integrate and equate to 1, no errors seen |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X^2) = k\int_0^2\left(4x^3 - x^4\right)dx + k\int_2^6\left(6x^2 - x^3\right)dx = k\left[x^4 - \frac{1}{5}x^5\right] + k\left[2x^3 - \frac{1}{4}x^4\right] = k[9.6 + 96] = 7.92$ | M1* | Attempt at $E(X^2)$, correct limits, integrated |
| $\text{Var}(X) = E(X^2) - E^2(X) = \text{their } (7.92) - 2.5^2$ | depM1 | Use of correct formula, with their $E(X^2)$ |
| $1.67$ | A1 | |
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 kx(4-x)dx + \int_2^m k(6-x)dx = \frac{1}{2}$ | M1* | Correct expression and attempt to integrate |
| $\int_0^2 kx(4-x)dx = \frac{2}{5}$ | B1 | May be implied |
| $\frac{2}{5} + k\left[6x - \frac{1}{2}x^2\right]_2^m = \frac{1}{2}$; $6m - \frac{1}{2}m^2 - 12 + 2 = \frac{1}{10k}$; $m^2 - 12m + \frac{68}{3} = 0$ | depM1 | First integral and $\frac{1}{2}$ may be seen combined as $\frac{1}{10}$. Obtain quadratic equation |
| $m = 6 \pm \frac{2}{3}\sqrt{30}$; $m = 6 - \frac{2}{3}\sqrt{30}\ (= 2.35)$ | A1 | Single answer |3 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} k x ( 4 - x ) & 0 \leqslant x < 2 \\ k ( 6 - x ) & 2 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 3 } { 40 }$.
\item Given that $\mathrm { E } ( X ) = 2.5$, find $\operatorname { Var } ( X )$.
\item Find the median value of $X$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q3 [8]}}