| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Standard applied PDF calculations |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard integration techniques. Parts involve computing E(T) by integration, finding the CDF by integrating the pdf, and solving F(t)=0.5 for the median. While it has multiple parts (13 marks total) and requires careful algebraic manipulation of polynomials, all techniques are routine for Further Maths students with no novel problem-solving or conceptual insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(E(T) = \frac{1}{2500}\int_0^{10} t^2(100-t^2)dt = \frac{1}{2500}\left[\frac{100t^3}{3} - \frac{t^5}{5}\right]_0^{10} = 5.33(333...)\) | M1, A1, A1 | AO3, AO1, AO1 |
| (b)(i) \(F(t) = \frac{1}{2500}\int_0^t u(100-u^2)du = \frac{1}{2500}\left[50u^2 - \frac{u^4}{4}\right]_0^t = \frac{1}{2500}\left(50t^2 - \frac{t^4}{4}\right)\) (for \(0 \leq t \leq 10\)); \(= 1\) for \(t > 10\); \((F(t) = 0\) for \(t < 0)\) | M1, A1, A1, B1 | AO3, AO1, AO1, AO1 |
| (ii) \(P(T > 5) = 1 - F(5) = 0.563\) (0.5625) | M1, A1 | AO3, AO1 |
| (iii) The median \(m\) satisfies \(F(m) = 0.5\); \(m^4 - 200m^2 + 5000 = 0\); \(m^2 = \frac{200 + \sqrt{40000 - 20000}}{2}\) (= 29.289...); \(m = 5.41(1961...)\) | M1, A1, A1, A1, A1 | AO3, AO3, AO1, AO1, AO1 |
**(a)** $E(T) = \frac{1}{2500}\int_0^{10} t^2(100-t^2)dt = \frac{1}{2500}\left[\frac{100t^3}{3} - \frac{t^5}{5}\right]_0^{10} = 5.33(333...)$ | M1, A1, A1 | AO3, AO1, AO1
**(b)(i)** $F(t) = \frac{1}{2500}\int_0^t u(100-u^2)du = \frac{1}{2500}\left[50u^2 - \frac{u^4}{4}\right]_0^t = \frac{1}{2500}\left(50t^2 - \frac{t^4}{4}\right)$ (for $0 \leq t \leq 10$); $= 1$ for $t > 10$; $(F(t) = 0$ for $t < 0)$ | M1, A1, A1, B1 | AO3, AO1, AO1, AO1 | Allow omission of $t < 0$
**(ii)** $P(T > 5) = 1 - F(5) = 0.563$ (0.5625) | M1, A1 | AO3, AO1
**(iii)** The median $m$ satisfies $F(m) = 0.5$; $m^4 - 200m^2 + 5000 = 0$; $m^2 = \frac{200 + \sqrt{40000 - 20000}}{2}$ (= 29.289...); $m = 5.41(1961...)$ | M1, A1, A1, A1, A1 | AO3, AO3, AO1, AO1, AO1 | [13]
---The queueing times, $T$ minutes, of customers at a local Post Office are modelled by the probability density function
$$f(t) = \frac{1}{2500}t(100-t^2) \quad \text{for } 0 \leq t \leq 10,$$
$$f(t) = 0 \quad \text{otherwise.}$$
\begin{enumerate}[label=(\alph*)]
\item Determine the mean queueing time. [3]
\item
\begin{enumerate}[label=(\roman*)]
\item Find the cumulative distribution function, $F(t)$, of $T$.
\item Find the probability that a randomly chosen customer queues for more than 5 minutes.
\item Find the median queueing time. [10]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 Q2 [13]}}