| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 19 |
| 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 S2 probability density function question requiring routine integration techniques and application of standard formulas. Part (a) uses the integral-equals-1 property, parts (d) and (e) apply E(T) and Var(T) formulas mechanically. The only slight challenge is the algebraic manipulation in integration, but the question structure is highly formulaic 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 |
|---|---|
| \(\int_0^3 k(t^2 + 2) \, dt = 1\) | M1 |
| \(\therefore k\left[\frac{1}{3}t^3 + 2t\right] = 1\) | A1 |
| \(\therefore k[(9 + 6) - (0)] = 1\); \(15k = 1\); \(k = \frac{1}{15}\) | M1 A1 |
| Answer | Marks |
|---|---|
| Graph showing \(f(t)\) curve passing through approximately \((0, \frac{2}{15})\) and \((3, \frac{11}{15})\) | B3 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks |
|---|---|
| \(E(T) = \int_0^3 t \times \frac{1}{15}(t^2 + 2) \, dt = \frac{1}{15}\int_0^3 t^3 + 2t \, dt\) | M1 |
| \(= \frac{1}{15}\left[\frac{1}{4}t^4 + t^2\right]_0^3\) | M1 A1 |
| \(= \frac{1}{15}\left[\left(\frac{81}{4} + 9\right) - (0)\right] = \frac{39}{30}\) or 1.95 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(T^2) = \int_0^3 t^2 \times \frac{1}{15}(t^2 + 2) \, dt = \frac{1}{15}\int_0^3 t^4 + 2t^2 \, dt\) | M1 | |
| \(= \frac{1}{15}\left[\frac{1}{5}t^5 + \frac{2}{3}t^3\right]_0^3\) | A1 | |
| \(= \frac{1}{15}\left[\left(\frac{243}{5} + 18\right) - (0)\right] = \frac{111}{50}\) | M1 A1 | |
| \(\text{Var}(T) = \frac{111}{50} - \left(\frac{39}{30}\right)^2 = \frac{255}{400} = \frac{51}{80} = 0.6375\) | M1 | |
| \(\therefore \text{std. dev} = \sqrt{0.6375} = 0.798\) (3sf) | A1 | (19) |
**Part (a)**
$\int_0^3 k(t^2 + 2) \, dt = 1$ | M1 |
$\therefore k\left[\frac{1}{3}t^3 + 2t\right] = 1$ | A1 |
$\therefore k[(9 + 6) - (0)] = 1$; $15k = 1$; $k = \frac{1}{15}$ | M1 A1 |
**Part (b)**
Graph showing $f(t)$ curve passing through approximately $(0, \frac{2}{15})$ and $(3, \frac{11}{15})$ | B3 |
**Part (c)**
3 | A1 |
**Part (d)**
$E(T) = \int_0^3 t \times \frac{1}{15}(t^2 + 2) \, dt = \frac{1}{15}\int_0^3 t^3 + 2t \, dt$ | M1 |
$= \frac{1}{15}\left[\frac{1}{4}t^4 + t^2\right]_0^3$ | M1 A1 |
$= \frac{1}{15}\left[\left(\frac{81}{4} + 9\right) - (0)\right] = \frac{39}{30}$ or 1.95 | M1 A1 |
**Part (e)**
$E(T^2) = \int_0^3 t^2 \times \frac{1}{15}(t^2 + 2) \, dt = \frac{1}{15}\int_0^3 t^4 + 2t^2 \, dt$ | M1 |
$= \frac{1}{15}\left[\frac{1}{5}t^5 + \frac{2}{3}t^3\right]_0^3$ | A1 |
$= \frac{1}{15}\left[\left(\frac{243}{5} + 18\right) - (0)\right] = \frac{111}{50}$ | M1 A1 |
$\text{Var}(T) = \frac{111}{50} - \left(\frac{39}{30}\right)^2 = \frac{255}{400} = \frac{51}{80} = 0.6375$ | M1 |
$\therefore \text{std. dev} = \sqrt{0.6375} = 0.798$ (3sf) | A1 | (19)
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**Total: 75 marks**The continuous random variable $T$ has the following probability density function:
$$f(t) = \begin{cases}
k(t^2 + 2), & 0 \leq t \leq 3, \\
0, & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{1}{15}$. [4 marks]
\item Sketch $f(t)$ for all values of $t$. [3 marks]
\item State the mode of $T$. [1 mark]
\item Find $E(T)$. [5 marks]
\item Show that the standard deviation of $T$ is 0.798 correct to 3 significant figures. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [19]}}