| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate probability P(X in interval) |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: integration to find probabilities, median calculation, and expectation of a function of X. The piecewise PDF is given explicitly, and all parts follow routine procedures taught in Further Stats. Slightly above average difficulty only due to the piecewise nature and multiple parts, but no novel insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X>1) = 1 - \frac{1}{-1} \times 1 \times \frac{6}{5}^{-1/5}\) OR via \(\int_1^\infty \frac{6}{5}x^{-4}\,dx\) | 1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Median satisfies \(\frac{1}{2} \times m \times \frac{6}{5}m = \frac{1}{2}\), OR equivalent use of integration | 1 | M1 |
| \(m = \sqrt{\frac{5}{6}}\); accept 0.913 | 1 | A1 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X^2) = \int_0^1 \frac{6}{5}x^3\,dx + \int_1^\infty \frac{6}{5}x^{-2}\,dx\) | 1 | M1 |
| \(E(X^2) = 1.5\) | 1 | A1 |
| \(\text{Var}(X) = 1.5 - 1^2 = 0.5\) | 1 | A1FT |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(\sqrt{X}) = \int_0^1 \frac{6}{5}x^2\,dx + \int_1^\infty \frac{6}{5}x^{-2}\,dx\) | 1 | M1 |
| \(\frac{24}{25}\) or 0.96 | 1 | A1 |
| Total | 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X>1) = 1 - \frac{1}{-1} \times 1 \times \frac{6}{5}^{-1/5}$ **OR** via $\int_1^\infty \frac{6}{5}x^{-4}\,dx$ | 1 | B1 | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median satisfies $\frac{1}{2} \times m \times \frac{6}{5}m = \frac{1}{2}$, **OR** equivalent use of integration | 1 | M1 | |
| $m = \sqrt{\frac{5}{6}}$; accept 0.913 | 1 | A1 | Correct value; accept 0.913 |
| **Total** | **2** | | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X^2) = \int_0^1 \frac{6}{5}x^3\,dx + \int_1^\infty \frac{6}{5}x^{-2}\,dx$ | 1 | M1 | Attempt evaluation of the sum of these two integrals |
| $E(X^2) = 1.5$ | 1 | A1 | Correct value seen or implied |
| $\text{Var}(X) = 1.5 - 1^2 = 0.5$ | 1 | A1FT | Follow through their 1.5 providing variance is positive |
| **Total** | **3** | | |
---
## Question 5(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(\sqrt{X}) = \int_0^1 \frac{6}{5}x^2\,dx + \int_1^\infty \frac{6}{5}x^{-2}\,dx$ | 1 | M1 | Attempt evaluation of the sum of these two integrals |
| $\frac{24}{25}$ or 0.96 | 1 | A1 | |
| **Total** | **2** | | |5 The continuous random variable $X$ has probability density function f given by
$$f(x) = \begin{cases} 0 & x < 0 \\ \frac{6}{5} x & 0 \leqslant x \leqslant 1 \\ \frac{6}{5} x^{-4} & x > 1 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm{P}(X > 1)$.
\item Find the median value of $X$.
\item Given that $\mathrm{E}(X) = 1$, find the variance of $X$.
\item Find $\mathrm{E}(\sqrt{X})$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q5 [8]}}