| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Moderate -0.3 This question tests basic PDF properties (symmetry, integration to find constants) and interpretation of models. Part (a) uses simple symmetry reasoning, part (b)(i) is routine integration to find k, part (b)(ii) is standard expectation calculation, and parts (iii)-(iv) require only conceptual understanding of model constraints. All techniques are standard S2 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.3\) or \(1-0.6\) or \(0.4\) or \(0.2\) seen | M1 | |
| \(0.8\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k\int_0^{1.5}(2.25-x^2)dx = 1\) | M1 | attempt integ \(f(x)\) and \(= 1\). Ignore limits |
| \(k\left[2.25x - \frac{x^3}{3}\right]_0^{1.5} = 1\) | A1 | correct integration and limits |
| \(k \times [3.375 - 1.125] = 1\) or \(k \times \frac{9}{4} = 1\) oe | ||
| \(k = \frac{4}{9}\) AG | A1 [3] | No errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4}{9}\int_0^{1.5}(2.25x - x^3)dx\) | M1 | attempt integ \(xf(x)\), ignore limits, condone missing \(k\) |
| \(= \frac{4}{9}\left[2.25\frac{x^2}{2} - \frac{x^4}{4}\right]_0^{1.5}\) | A1 | correct integration and limits, condone missing \(k\) |
| \(= 0.5625\) or \(0.563\) | A1 | |
| Mean no. of hours \(= 56.25\) or \(56.3\) 56 hrs 15 mins | A1\(\checkmark\) [4] | ft their \(0.5625\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max \(x\) is \(1.5\), less than \(2.9\) or \(150 < 290\) | B1 [1] | Needs numerical justification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| any \(a\) such that \(2.9 \leq a \leq 5\) | B1 [1] |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.3$ or $1-0.6$ or $0.4$ or $0.2$ seen | M1 | |
| $0.8$ | A1 [2] | |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k\int_0^{1.5}(2.25-x^2)dx = 1$ | M1 | attempt integ $f(x)$ and $= 1$. Ignore limits |
| $k\left[2.25x - \frac{x^3}{3}\right]_0^{1.5} = 1$ | A1 | correct integration and limits |
| $k \times [3.375 - 1.125] = 1$ or $k \times \frac{9}{4} = 1$ oe | | |
| $k = \frac{4}{9}$ **AG** | A1 [3] | No errors seen |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4}{9}\int_0^{1.5}(2.25x - x^3)dx$ | M1 | attempt integ $xf(x)$, ignore limits, condone missing $k$ |
| $= \frac{4}{9}\left[2.25\frac{x^2}{2} - \frac{x^4}{4}\right]_0^{1.5}$ | A1 | correct integration and limits, condone missing $k$ |
| $= 0.5625$ or $0.563$ | A1 | |
| Mean no. of hours $= 56.25$ or $56.3$ 56 hrs 15 mins | A1$\checkmark$ [4] | ft their $0.5625$ |
### Part (b)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max $x$ is $1.5$, less than $2.9$ or $150 < 290$ | B1 [1] | Needs numerical justification |
### Part (b)(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| any $a$ such that $2.9 \leq a \leq 5$ | B1 [1] | |7
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{1060d9f5-cf40-419e-b212-7266885c6617-3_465_1127_954_550}
The diagram shows the graph of the probability density function of a variable $X$. Given that the graph is symmetrical about the line $x = 1$ and that $\mathrm { P } ( 0 < X < 2 ) = 0.6$, find $\mathrm { P } ( X > 0 )$.
\item A flower seller wishes to model the length of time that tulips last when placed in a jug of water. She proposes a model using the random variable $X$ (in hundreds of hours) with probability density function given by
$$f ( x ) = \begin{cases} k \left( 2.25 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 1.5 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac { 4 } { 9 }$.
\item Use this model to find the mean number of hours that a tulip lasts in a jug of water.
The flower seller wishes to create a similar model for daffodils. She places a large number of daffodils in jugs of water and the longest time that any daffodil lasts is found to be 290 hours.
\item Give a reason why $\mathrm { f } ( x )$ would not be a suitable model for daffodils.
\item The flower seller considers a model for daffodils of the form
$$g ( x ) = \begin{cases} c \left( a ^ { 2 } - x ^ { 2 } \right) & 0 \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $c$ are constants. State a suitable value for $a$. (There is no need to evaluate $c$.)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2016 Q7 [11]}}