| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a standard continuous probability distribution question requiring routine integration techniques: finding k using the normalization condition, calculating P(T>10) with definite integration, and computing E(T). All three parts follow textbook procedures with straightforward polynomial integration, making it slightly easier than average despite requiring multiple steps. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| (i) \(k \int_0^{15} (225 - t^2) dt = 1\) | M1 |
| \(k\left[225t - \frac{t^3}{3}\right]_0^{15} = 1\) | A1 |
| \(k \times [3375 - 1125] = 1\) or \(k \times 2250 = 1\) | A1 |
| \(k = \frac{1}{2250}\) (AG) |
| Answer | Marks |
|---|---|
| (ii) \(\frac{1}{2250} \int_{10}^{15} (225 - t^2) dt\) | M1 |
| \(= \frac{1}{2250} \left[225t - \frac{t^3}{3}\right]_{10}^{15}\) | A1 |
| \(= \frac{1}{2250}\left[2250 - \left(2250 - \frac{1000}{3}\right)\right] = \frac{4}{27}\) or 0.148 (3 sf) | A1 |
| Answer | Marks |
|---|---|
| (iii) \(\frac{1}{2250} \int_0^{15} (225t - t^3) dt\) | M1* |
| \(= \frac{1}{2250} \left[\frac{225t^2}{2} - \frac{t^4}{4}\right]_0^{15}\) | A1 |
| \(= \frac{1}{2250}\left[\frac{50625}{2} - \frac{50625}{4}\right] = \frac{45}{8}\) or 5.625 or 5.63 (3 sf) | M1*dep A1 |
(i) $k \int_0^{15} (225 - t^2) dt = 1$ | M1
$k\left[225t - \frac{t^3}{3}\right]_0^{15} = 1$ | A1
$k \times [3375 - 1125] = 1$ or $k \times 2250 = 1$ | A1
$k = \frac{1}{2250}$ (AG) |
**Total: 3**
(ii) $\frac{1}{2250} \int_{10}^{15} (225 - t^2) dt$ | M1
$= \frac{1}{2250} \left[225t - \frac{t^3}{3}\right]_{10}^{15}$ | A1
$= \frac{1}{2250}\left[2250 - \left(2250 - \frac{1000}{3}\right)\right] = \frac{4}{27}$ or 0.148 (3 sf) | A1
**Total: 3**
Alternatively: $1 - \int_0^{10}$
(iii) $\frac{1}{2250} \int_0^{15} (225t - t^3) dt$ | M1*
$= \frac{1}{2250} \left[\frac{225t^2}{2} - \frac{t^4}{4}\right]_0^{15}$ | A1
$= \frac{1}{2250}\left[\frac{50625}{2} - \frac{50625}{4}\right] = \frac{45}{8}$ or 5.625 or 5.63 (3 sf) | M1*dep A1
**Total: 4**
Accept 5 mins 37 or 38 secs
**Total: 10**
---6 The waiting time, $T$ minutes, for patients at a doctor's surgery has probability density function given by
$$\mathrm { f } ( t ) = \begin{cases} k \left( 225 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 15 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 2250 }$.\\
(ii) Find the probability that a patient has to wait for more than 10 minutes.\\
(iii) Find the mean waiting time.
\hfill \mbox{\textit{CAIE S2 2015 Q6 [10]}}