| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Find quantiles from CDF |
| Difficulty | Standard +0.3 This is a straightforward multi-part S2 question testing standard CDF operations: finding median (solving F(t)=0.5), conditional probability, differentiating to find PDF, and finding mode. Part (a) is given as 'show that', making it easier. The polynomial is manageable and all techniques are routine for this specification. Slightly above average due to the algebraic manipulation required, but no novel insight needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) When \(F(t) = 0.5\), \(t^3(3t^2 - 16t + 24) = 8\) | M1 | |
| When \(t = 0.77\), L.H.S. \(= 7.98 \approx 8\) | M1 A1 | |
| (b) \(F(1.2) = 0.8208\); \(F(1.5) = 0.9492\) | B1 B1 | |
| \(P(\text{Wait } 1.5 | \text{wait } 1.2) = (1 - 0.9492) + (1 - 0.8208) = 0.283\) | M1 A1 A1 |
| (c) \(f(t) = \frac{1}{15}(12t^3 - 48t^2 + 48t)\) \((0 \le t \le 2)\), \(f(t) = 0\) otherwise | M1 A1 A1 | |
| Graph sketched | B1 | |
| (d) Mode is at max. point where \(f'(t) = 0\); \(36t^2 - 96t + 48 = 0\) | M1 A1 | |
| \(12(3t - 2)(t - 2) = 0\); Mode is \(t = \frac{2}{3}\) \((6\frac{2}{3} \text{ minutes})\) | M1 A1 A1 | |
| (e) Unlikely to be no delays above 20 minutes | B1 |
(a) When $F(t) = 0.5$, $t^3(3t^2 - 16t + 24) = 8$ | M1 |
When $t = 0.77$, L.H.S. $= 7.98 \approx 8$ | M1 A1 |
(b) $F(1.2) = 0.8208$; $F(1.5) = 0.9492$ | B1 B1 |
$P(\text{Wait } 1.5 | \text{wait } 1.2) = (1 - 0.9492) + (1 - 0.8208) = 0.283$ | M1 A1 A1 |
(c) $f(t) = \frac{1}{15}(12t^3 - 48t^2 + 48t)$ $(0 \le t \le 2)$, $f(t) = 0$ otherwise | M1 A1 A1 |
Graph sketched | B1 |
(d) Mode is at max. point where $f'(t) = 0$; $36t^2 - 96t + 48 = 0$ | M1 A1 |
$12(3t - 2)(t - 2) = 0$; Mode is $t = \frac{2}{3}$ $(6\frac{2}{3} \text{ minutes})$ | M1 A1 A1 |
(e) Unlikely to be no delays above 20 minutes | B1 |
**Total: 18 marks**7. Each day on the way to work, a commuter encounters a similar traffic jam. The length of time, in 10-minute units, spent waiting in the traffic jam is modelled by the random variable $T$ with the cumulative distribution function:
$$\begin{array} { l l }
\mathrm { F } ( t ) = 0 & t < 0 , \\
\mathrm {~F} ( t ) = \frac { t ^ { 2 } \left( 3 t ^ { 2 } - 16 t + 24 \right) } { 16 } & 0 \leq t \leq 2 , \\
\mathrm {~F} ( t ) = 1 & t > 2 .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Show that 0.77 is approximately the median value of $T$.
\item Given that he has already waited for 12 minutes, find the probability that he will have to wait another 3 minutes.
\item Find, and sketch, the probability density function of $T$.
\item Hence find the modal value of $T$.
\item Comment on the validity of this model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [18]}}