| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This question tests understanding of PDF symmetry and basic probability calculations. Part (b) requires recognizing symmetry to deduce E(T)=10 without calculation—a conceptual insight but straightforward once the parabola is sketched. Parts (c) and (e) involve standard integration techniques. Part (d) tests understanding of symmetry properties. The multi-part structure and symmetry recognition elevate it slightly above routine drill exercises, but all techniques are standard S2 material with no novel problem-solving 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 'Tails down' parabola only from \(x = 0\) to \(20\) shown | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Symmetrical | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{3}{4000}\displaystyle\int_0^{20}(20t^3 - t^4)\,dx = \dfrac{3}{4000}\left[20\dfrac{t^4}{4} - \dfrac{t^5}{5}\right]_0^{20}\) | M1 | |
| \(\text{Var}(T) = \dfrac{3}{4000} \times 160000 - 10^2\) | M1 | |
| \(20\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((p - 0.5) \times 2\) or \(1 - 2(1-p)\) | M1 | |
| \(2p - 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{3}{4000}\displaystyle\int_{8}^{12}(20t - t^2)\,dx\) | M1 | |
| \(\dfrac{3}{4000}\left[20\dfrac{t^2}{2} - \dfrac{t^3}{3}\right]_{8}^{12} = \dfrac{3}{4000}\left(1440 - 576 - 640 + \dfrac{512}{3}\right)\) | A1 | |
| \(\dfrac{37}{125}\) or \(0.296\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Does not allow times greater than 20 minutes | B1 |
## Question 6:
**Part 6(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| 'Tails down' parabola only from $x = 0$ to $20$ shown | B1 | |
**Part 6(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Symmetrical | B1 | |
**Part 6(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{3}{4000}\displaystyle\int_0^{20}(20t^3 - t^4)\,dx = \dfrac{3}{4000}\left[20\dfrac{t^4}{4} - \dfrac{t^5}{5}\right]_0^{20}$ | M1 | |
| $\text{Var}(T) = \dfrac{3}{4000} \times 160000 - 10^2$ | M1 | |
| $20$ | A1 | |
**Part 6(d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(p - 0.5) \times 2$ or $1 - 2(1-p)$ | M1 | |
| $2p - 1$ | A1 | |
## Question 6(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{3}{4000}\displaystyle\int_{8}^{12}(20t - t^2)\,dx$ | M1 | |
| $\dfrac{3}{4000}\left[20\dfrac{t^2}{2} - \dfrac{t^3}{3}\right]_{8}^{12} = \dfrac{3}{4000}\left(1440 - 576 - 640 + \dfrac{512}{3}\right)$ | A1 | |
| $\dfrac{37}{125}$ or $0.296$ | A1 | |
**Total: 3 marks**
---
## Question 6(f):
| Answer | Mark | Guidance |
|--------|------|----------|
| Does not allow times greater than 20 minutes | B1 | |
**Total: 1 mark**6 The length of time, $T$ minutes, that a passenger has to wait for a bus at a certain bus stop is modelled by the probability density function given by
$$\mathrm { f } ( t ) = \begin{cases} \frac { 3 } { 4000 } \left( 20 t - t ^ { 2 } \right) & 0 \leqslant t \leqslant 20 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = \mathrm { f } ( t )$.
\item Hence explain, without calculation, why $\mathrm { E } ( T ) = 10$.
\item Find $\operatorname { Var } ( T )$.
\item It is given that $\mathrm { P } ( T < 10 + a ) = p$, where $0 < a < 10$.
Find $\mathrm { P } ( 10 - a < T < 10 + a )$ in terms of $p$.
\item Find $\mathrm { P } ( 8 < T < 12 )$.
\item Give one reason why this model may be unrealistic.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q6 [11]}}