| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 14 |
| 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 standard S2 question requiring routine integration to find E(T), E(T²), and Var(T), plus a straightforward probability calculation. The pdf is polynomial making integration mechanical, and all parts follow textbook procedures with no novel problem-solving required. Slightly easier than average due to the straightforward polynomial form. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Mean \(= \int_0^4 \frac{3}{64}t^3(4-t)\,dt = \frac{3}{64}[t^4 - t^5/5]_0^4 = 2.4\) | M1 A1 | |
| \(\text{Var}(T) = \int_0^4 \frac{3}{64}t^4(4-t)\,dt - 2 \cdot 4^2 = \frac{3}{64}[4t^5/5 - t^6/6]_0^4 - 5.76 = 0.64\) | M1 A1 A1 | |
| Standard deviation \(= \sqrt{0.64} = 0.8\) | M1 A1 | |
| (b) \(P(T \leq 3) = \int_0^3 \frac{3}{64}t^3(4-t)\,dt = 0.738\); \(P(T > 3) = 0.262\) | M1 A1 M1 A1 | |
| (c) \(0.738^2 = 0.545\) | M1 A1 | |
| (d) Unlikely that all recover within 4 days | M1; B1 | Total: 14 marks |
(a) Mean $= \int_0^4 \frac{3}{64}t^3(4-t)\,dt = \frac{3}{64}[t^4 - t^5/5]_0^4 = 2.4$ | M1 A1 |
$\text{Var}(T) = \int_0^4 \frac{3}{64}t^4(4-t)\,dt - 2 \cdot 4^2 = \frac{3}{64}[4t^5/5 - t^6/6]_0^4 - 5.76 = 0.64$ | M1 A1 A1 |
Standard deviation $= \sqrt{0.64} = 0.8$ | M1 A1 |
(b) $P(T \leq 3) = \int_0^3 \frac{3}{64}t^3(4-t)\,dt = 0.738$; $P(T > 3) = 0.262$ | M1 A1 M1 A1 |
(c) $0.738^2 = 0.545$ | M1 A1 |
(d) Unlikely that all recover within 4 days | M1; B1 | Total: 14 marks6. Patients suffering from 'flu are treated with a drug. The number of days, $t$, that it then takes for them to recover is modelled by the continuous random variable $T$ with the probability density function
$$\begin{array} { l l }
\mathrm { f } ( t ) = \frac { 3 t ^ { 2 } ( 4 - t ) } { 64 } & 0 \leq t \leq 4 \\
\mathrm { f } ( t ) = 0 & \text { otherwise. }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the mean and standard deviation of $T$.
\item Find the probability that a patient takes more than 3 days to recover.
\item Two patients are selected at random. Find the probability that they both recover within three days.
\item Comment on the suitability of the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [14]}}