| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with multiple regions |
| Difficulty | Standard +0.3 This is a standard piecewise PDF question requiring routine integration techniques across different regions. While it involves multiple parts (CDF, median, expectation, probability), each step follows textbook procedures without requiring novel insight or complex problem-solving. The piecewise nature adds mild complexity but remains a typical Further Stats exercise, placing it slightly above average difficulty. |
| 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 |
|---|---|
| \(F(x) = \begin{cases} \frac{x^2}{10}, & 0 \leq x < 2 \\ \frac{1}{15}(10x - x^2) - \frac{2}{3}, & 2 \leq x \leq 5 \end{cases}\) | M1A1 |
| \(= 0\) for \(x < 0\) and \(= 1\) for \(x > 5\) | A1 |
| Answer | Marks |
|---|---|
| \(F(m) = \frac{1}{2}\) so \(2m^2 - 20m + 35 = 0\) | M1 |
| \(m = 5 - \frac{1}{2}\sqrt{30} = 2.26\) | A1 |
| Answer | Marks |
|---|---|
| \(E(X^2) = \int_0^2 x^2 \cdot \frac{x}{5}\,dx + \int_2^5 \frac{2}{15}(5-x) \cdot x^2\,dx = \left[\frac{x^4}{20}\right] + \left[\frac{2}{15}\left(\frac{5x^3}{3} - \frac{x^4}{4}\right)\right]\) | M1 |
| \(= 6.5\) | A1 |
| Answer | Marks |
|---|---|
| \(F(3) - F(1)\) | M1 |
| \(= \frac{11}{15} - \frac{1}{10} = \frac{19}{30}\) | A1 |
| Answer | Marks |
|---|---|
| \(\int_1^2 \frac{x}{5}\,dx + \int_2^3 \frac{2}{15}(5-x)\,dx = \left[\frac{x^2}{10}\right] + \left[\frac{2}{15}\left(5x - \frac{x^2}{2}\right)\right]\) | M1 |
| \(= \frac{3}{10} + \frac{1}{3} = \frac{19}{30}\) | A1 |
## Question 3:
**Part 3(a):**
$F(x) = \begin{cases} \frac{x^2}{10}, & 0 \leq x < 2 \\ \frac{1}{15}(10x - x^2) - \frac{2}{3}, & 2 \leq x \leq 5 \end{cases}$ | M1A1 |
$= 0$ for $x < 0$ and $= 1$ for $x > 5$ | A1 |
**Part 3(b):**
$F(m) = \frac{1}{2}$ so $2m^2 - 20m + 35 = 0$ | M1 |
$m = 5 - \frac{1}{2}\sqrt{30} = 2.26$ | A1 |
**Part 3(c):**
$E(X^2) = \int_0^2 x^2 \cdot \frac{x}{5}\,dx + \int_2^5 \frac{2}{15}(5-x) \cdot x^2\,dx = \left[\frac{x^4}{20}\right] + \left[\frac{2}{15}\left(\frac{5x^3}{3} - \frac{x^4}{4}\right)\right]$ | M1 |
$= 6.5$ | A1 |
**Part 3(d):**
$F(3) - F(1)$ | M1 |
$= \frac{11}{15} - \frac{1}{10} = \frac{19}{30}$ | A1 |
*Alternative method:*
$\int_1^2 \frac{x}{5}\,dx + \int_2^3 \frac{2}{15}(5-x)\,dx = \left[\frac{x^2}{10}\right] + \left[\frac{2}{15}\left(5x - \frac{x^2}{2}\right)\right]$ | M1 |
$= \frac{3}{10} + \frac{1}{3} = \frac{19}{30}$ | A1 |
---3 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} \frac { 1 } { 5 } x & 0 \leqslant x < 2 \\ \frac { 2 } { 15 } ( 5 - x ) & 2 \leqslant x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the cumulative distribution function of $X$.
\item Find the median value of $X$.
\item Find $\mathrm { E } \left( X ^ { 2 } \right)$.
\item Find $\mathrm { P } ( 1 \leqslant x \leqslant 3 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q3 [9]}}