| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Standard +0.3 This is a straightforward application of standard variance formulas for continuous distributions. Part (i) requires computing E(T²) via integration and using Var(T) = E(T²) - [E(T)]², with E(T) already given. Part (ii) involves solving ∫ₙ¹⁰ f(t)dt = 0.1, requiring polynomial integration and equation solving. Both parts are routine S2 techniques with no conceptual challenges beyond careful algebraic manipulation. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int_0^{10} \frac{1}{2500}(100t^3 - t^5)\,dt\) | M1 | Attempt \(\int t^2 f(t)\) |
| \(= \frac{1}{2500}\left[25t^4 - \frac{t^6}{6}\right]_0^{10} = \frac{100}{3}\) | ||
| \(\frac{100}{3} - \left(\frac{16}{3}\right)^2\) | M1 | For \(E(T^2) - (E(T))^2\) |
| \(= \frac{44}{9}\) or \(4.89\) (3 sf) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int_n^{10} \frac{1}{2500}(100t - t^3)\,dt\) | M1 | Attempt integ \(f(t)\), ignore limits |
| \(\frac{1}{2500}\left[50t^2 - \frac{t^4}{4}\right] = 0.1\) | M1 | Attempt integ \(f(t)\), limits \(n\) to 10 or 0 to \(n\); equated to \(0.1\) or \(0.9\) |
| \(\frac{1}{2500}\left[2500 - \left(50n^2 - \frac{n^4}{4}\right)\right] = 0.1\) | M1 | \(0.1/0.9\) matched to correct limits and used |
| \((n^4 - 200n^2 + 9000 = 0)\) | M1 | Correct method of solution of a QE in \(n^2\) |
| \((n^2 = 68.3772,\ n = 8.27)\); \(n = 8\) | A1 | [5] Must be single answer only |
## Question 3(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_0^{10} \frac{1}{2500}(100t^3 - t^5)\,dt$ | M1 | Attempt $\int t^2 f(t)$ |
| $= \frac{1}{2500}\left[25t^4 - \frac{t^6}{6}\right]_0^{10} = \frac{100}{3}$ | | |
| $\frac{100}{3} - \left(\frac{16}{3}\right)^2$ | M1 | For $E(T^2) - (E(T))^2$ |
| $= \frac{44}{9}$ or $4.89$ (3 sf) | A1 | **[3]** |
## Question 3(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_n^{10} \frac{1}{2500}(100t - t^3)\,dt$ | M1 | Attempt integ $f(t)$, ignore limits |
| $\frac{1}{2500}\left[50t^2 - \frac{t^4}{4}\right] = 0.1$ | M1 | Attempt integ $f(t)$, limits $n$ to 10 or 0 to $n$; equated to $0.1$ or $0.9$ |
| $\frac{1}{2500}\left[2500 - \left(50n^2 - \frac{n^4}{4}\right)\right] = 0.1$ | M1 | $0.1/0.9$ matched to correct limits and used |
| $(n^4 - 200n^2 + 9000 = 0)$ | M1 | Correct method of solution of a QE in $n^2$ |
| $(n^2 = 68.3772,\ n = 8.27)$; $n = 8$ | A1 | **[5]** Must be single answer only |
---3 The waiting time, $T$ weeks, for a particular operation at a hospital has probability density function given by
$$f ( t ) = \begin{cases} \frac { 1 } { 2500 } \left( 100 t - t ^ { 3 } \right) & 0 \leqslant t \leqslant 10 \\ 0 & \text { otherwise } \end{cases}$$
(i) Given that $\mathrm { E } ( T ) = \frac { 16 } { 3 }$, find $\operatorname { Var } ( T )$.\\
(ii) $10 \%$ of patients have to wait more than $n$ weeks for their operation. Find the value of $n$, giving your answer correct to the nearest integer.
\hfill \mbox{\textit{CAIE S2 2013 Q3 [8]}}