| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Square root transformation (Y = √X) |
| Difficulty | Challenging +1.2 This is a two-part question requiring calculation of expectation with a piecewise PDF (straightforward integration) and finding the PDF of a transformed variable using the standard Jacobian method. While the piecewise nature adds some computational work, the transformation Y = √X is a standard textbook technique for Further Maths students, requiring no novel insight—just careful application of the formula g(y) = f(x)|dx/dy|. |
| Spec | 5.03c Calculate mean/variance: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \int_0^1 \frac{3}{16}x(2-\sqrt{x})\,dx + \int_1^9 \frac{\frac{3}{16}x}{\sqrt{x}}\,dx\) | M1 | |
| \(\frac{3}{16}\left[x^2 - \frac{2}{5}x^{\frac{5}{2}}\right] + \frac{3}{16}\left[\frac{2}{3}x^{\frac{3}{2}}\right]\) | A1 | |
| \(\dfrac{269}{80}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(x) = \begin{cases} \frac{3}{8}\!\left(x - \frac{1}{3}x^{3/2}\right), & 0 \leq x < 1 \\ \frac{3}{8}x^{1/2} - \frac{1}{8}, & 1 \leq x \leq 9 \end{cases}\) | M1A1 | |
| \(G(y) = \begin{cases} \frac{3}{8}\!\left(y^2 - \frac{1}{3}y^3\right), & 0 \leq y < 1 \\ \frac{3}{8}y - \frac{1}{8}, & 1 \leq y \leq 3 \end{cases}\) | M1A1 | |
| \(G(y) = 0\) for \(y \leq 0\) and \(= 1\) for \(y \geq 3\) | A1 | |
| \(g(y) = \begin{cases} \frac{3}{8}(2y - y^2), & 0 \leq y < 1 \\ \frac{3}{8}, & 1 \leq y \leq 3 \\ 0 & \text{otherwise} \end{cases}\) |
## Question 3:
### Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \int_0^1 \frac{3}{16}x(2-\sqrt{x})\,dx + \int_1^9 \frac{\frac{3}{16}x}{\sqrt{x}}\,dx$ | M1 | |
| $\frac{3}{16}\left[x^2 - \frac{2}{5}x^{\frac{5}{2}}\right] + \frac{3}{16}\left[\frac{2}{3}x^{\frac{3}{2}}\right]$ | A1 | |
| $\dfrac{269}{80}$ | A1 | |
### Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = \begin{cases} \frac{3}{8}\!\left(x - \frac{1}{3}x^{3/2}\right), & 0 \leq x < 1 \\ \frac{3}{8}x^{1/2} - \frac{1}{8}, & 1 \leq x \leq 9 \end{cases}$ | M1A1 | |
| $G(y) = \begin{cases} \frac{3}{8}\!\left(y^2 - \frac{1}{3}y^3\right), & 0 \leq y < 1 \\ \frac{3}{8}y - \frac{1}{8}, & 1 \leq y \leq 3 \end{cases}$ | M1A1 | |
| $G(y) = 0$ for $y \leq 0$ and $= 1$ for $y \geq 3$ | A1 | |
| $g(y) = \begin{cases} \frac{3}{8}(2y - y^2), & 0 \leq y < 1 \\ \frac{3}{8}, & 1 \leq y \leq 3 \\ 0 & \text{otherwise} \end{cases}$ | | |
---3 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} \frac { 3 } { 16 } ( 2 - \sqrt { x } ) & 0 \leqslant x < 1 \\ \frac { 3 } { 16 \sqrt { x } } & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( X )$.\\
The random variable $Y$ is such that $Y = \sqrt { X }$.
\item Find the probability density function of $Y$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q3 [8]}}