| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard integration to find E(T), integration to find the CDF, and then using the CDF to find the median. All steps are routine applications of formulas with no conceptual challenges beyond basic continuous probability distribution theory. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \[E(T) = \int_0^1 t f(t)\,dt\] | ||
| \[= \int_0^1 4t^3(1-t^2)\,dt\] | M1 | |
| \[= \left[\frac{4t^3}{3} - \frac{4t^5}{5}\right]_0^1\] | A1 | |
| \[= \frac{4}{3} - \frac{4}{5}\] | A1 | |
| \[= \frac{8}{15}\] | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \[F(t) = P(T \leq t) = \int_0^t f(t)\,dt\] | ||
| \[= \int_0^t 4t(1-t^2)\,dt\] | M1 | |
| \[= [2t^2 - t^4]_0^t\] | A1 | 2 marks |
| \[= 2t^2 - t^4\] |
| Answer | Marks | Guidance |
|---|---|---|
| \[P(\mu < T < m) = F(m) - F(\mu)\] | M1 | |
| \[F(m) = 0.5\] | B1 | |
| \[F(\mu) = F\left(\frac{8}{15}\right) = 0.4880\] | B1 | |
| \[\therefore P(\mu < T < m) = 0.5 - 0.4880 = 0.012\] | M1√, A1 | 5 marks |
**7(a)**
$$E(T) = \int_0^1 t f(t)\,dt$$ | |
$$= \int_0^1 4t^3(1-t^2)\,dt$$ | M1 | |
$$= \left[\frac{4t^3}{3} - \frac{4t^5}{5}\right]_0^1$$ | A1 | |
$$= \frac{4}{3} - \frac{4}{5}$$ | A1 | |
$$= \frac{8}{15}$$ | | 3 marks | AG
**7(b)(i)**
$$F(t) = P(T \leq t) = \int_0^t f(t)\,dt$$ | |
$$= \int_0^t 4t(1-t^2)\,dt$$ | M1 | |
$$= [2t^2 - t^4]_0^t$$ | A1 | 2 marks
$$= 2t^2 - t^4$$ | |
**7(b)(ii)**
$$P(\mu < T < m) = F(m) - F(\mu)$$ | M1 | |
$$F(m) = 0.5$$ | B1 | |
$$F(\mu) = F\left(\frac{8}{15}\right) = 0.4880$$ | B1 | |
$$\therefore P(\mu < T < m) = 0.5 - 0.4880 = 0.012$$ | M1√, A1 | 5 marks | 0.5 − their F(μ)
**Question 7 Total: 10 marks**
---7 Engineering work on the railway network causes an increase in the journey time of commuters travelling into work each morning.
The increase in journey time, $T$ hours, is modelled by a continuous random variable with probability density function
$$\mathrm { f } ( t ) = \begin{cases} 4 t \left( 1 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { E } ( T ) = \frac { 8 } { 15 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the cumulative distribution function, $\mathrm { F } ( t )$, for $0 \leqslant t \leqslant 1$.
\item Hence, or otherwise, for a commuter selected at random, find
$$\mathrm { P } ( \text { mean } < T < \text { median } )$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2006 Q7 [10]}}