| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Standard applied PDF calculations |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic pdf properties: finding k by integration (standard technique), calculating probabilities and moments using simple polynomial integration, and identifying the mode. Part (e) requires minimal real-world reasoning about telephone call distributions. All parts are routine textbook exercises with no problem-solving insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{10} krdt = 1\) or Area of triangle = 1 | M1 | |
| \(\left[\frac{kt^2}{2}\right]_0^{10} = 1\) or \(10 \times 0.5 \times 10k = 1\) or linear equation in \(k\) | M1 | |
| \(50k = 1\), \(k = \frac{1}{50}\) | cso | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{10} ktdt = \left[\frac{kt^2}{2}\right]_0^{10} = \frac{16}{25}\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(T) = \int_0^{10} kt^2dt = \left[\frac{kt^3}{3}\right]_0^{10} = 6\frac{2}{3}\) | M1, A1 | |
| \(\text{Var}(T) = \int_0^{10} kt^4dt - (6\frac{2}{3})^2 = \left[\frac{kt^4}{4}\right]_0^{10} - (6\frac{2}{3})^2 = 50 - (6\frac{2}{3})^2 = 5\frac{5}{9}\) | M1; M1dep | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of pdf: bell-shaped curve starting at origin, peaking before \(t=10\), returning to zero at \(t=10\) | B1 | (1 mark) |
## Part (a)
$\int_0^{10} krdt = 1$ or Area of triangle = 1 | M1 |
$\left[\frac{kt^2}{2}\right]_0^{10} = 1$ or $10 \times 0.5 \times 10k = 1$ or linear equation in $k$ | M1 |
$50k = 1$, $k = \frac{1}{50}$ | cso | A1 | (3 marks)
## Part (b)
$\int_0^{10} ktdt = \left[\frac{kt^2}{2}\right]_0^{10} = \frac{16}{25}$ | M1, A1 | (2 marks)
## Part (c)
$E(T) = \int_0^{10} kt^2dt = \left[\frac{kt^3}{3}\right]_0^{10} = 6\frac{2}{3}$ | M1, A1 |
$\text{Var}(T) = \int_0^{10} kt^4dt - (6\frac{2}{3})^2 = \left[\frac{kt^4}{4}\right]_0^{10} - (6\frac{2}{3})^2 = 50 - (6\frac{2}{3})^2 = 5\frac{5}{9}$ | M1; M1dep | A1 | (5 marks)
## Part (d)
10 | B1 | (1 mark)
## Part (e)
Sketch of pdf: bell-shaped curve starting at origin, peaking before $t=10$, returning to zero at $t=10$ | B1 | (1 mark)
---The length of a telephone call made to a company is denoted by the continuous random variable $T$. It is modelled by the probability density function
$$\text{f}(t) = \begin{cases} kt & 0 \leqslant t \leqslant 10 \\ 0 & \text{otherwise} \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $k$ is $\frac{1}{50}$. [3]
\item Find P($T > 6$). [2]
\item Calculate an exact value for E($T$) and for Var($T$). [5]
\item Write down the mode of the distribution of $T$. [1]
\end{enumerate}
It is suggested that the probability density function, f($t$), is not a good model for $T$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Sketch the graph of a more suitable probability density function for $T$. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2009 Q4 [12]}}