| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Standard +0.3 This is a standard S2 CDF question requiring routine techniques: finding k using F(1.5)=1, differentiating to get pdf, computing E(T) by integration, and applying conditional probability. All steps are methodical with no novel insight 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 integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(1.5) = 1 \Rightarrow k(2 \times (1.5)^3 - (1.5)^4) = 1\); \(k\left(\frac{108-81}{16}\right) = 1 \therefore k = \frac{16}{27}\) | M1, A1 cso | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T > 1) = 1 - F(1) = 1 - \frac{16}{27}(2-1) = \frac{11}{27}\) | M1, A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(t) = F'(t) = \frac{16}{27}(6t^2 - 4t^3)\) | M1, A1 | |
| \(f(t) = \begin{cases} \frac{32}{27}(3t^2 - 2t^3) & 0 \leq t \leq 1.5 \\ 0 & \text{otherwise} \end{cases}\) | B1 | Full definition (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(T) = \int_0^{1.5} t\,f(t)\,dt = \frac{32}{27}\int_0^{1.5}(3t^3 - 2t^4)\,dt\) | M1 | \(\int t\,f(t)\) |
| \(= \frac{32}{27}\left[\frac{3t^4}{4} - \frac{2t^5}{5}\right]_0^{1.5}\) | A1 | |
| \(= \frac{32}{27}\left[\left(\frac{243}{64} - \frac{2}{5}\times\frac{243}{32}\right) - (0)\right]\) | ||
| \(= \frac{9}{2} - \frac{18}{5} = 0.9\) | A1 cso | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(E(T)) = \frac{16}{27}(2 \times 0.9^3 - 0.9^4) = 0.4752\) | B1 | evidence seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(T > 1 \mid T > 0.9) = \frac{P(T>1)}{P(T>0.9)} = \frac{\text{part }(b)}{1 - \text{part }(e)} = 0.7763\ldots\) | M1, M1, A1 | accept awrt 0.776 (3) |
## Question 5:
### Part (a):
| $F(1.5) = 1 \Rightarrow k(2 \times (1.5)^3 - (1.5)^4) = 1$; $k\left(\frac{108-81}{16}\right) = 1 \therefore k = \frac{16}{27}$ | M1, A1 cso | (2) |
|---|---|---|
### Part (b):
| $P(T > 1) = 1 - F(1) = 1 - \frac{16}{27}(2-1) = \frac{11}{27}$ | M1, A1 | (2) |
|---|---|---|
### Part (c):
| $f(t) = F'(t) = \frac{16}{27}(6t^2 - 4t^3)$ | M1, A1 | |
|---|---|---|
| $f(t) = \begin{cases} \frac{32}{27}(3t^2 - 2t^3) & 0 \leq t \leq 1.5 \\ 0 & \text{otherwise} \end{cases}$ | B1 | Full definition (3) |
### Part (d):
| $E(T) = \int_0^{1.5} t\,f(t)\,dt = \frac{32}{27}\int_0^{1.5}(3t^3 - 2t^4)\,dt$ | M1 | $\int t\,f(t)$ |
|---|---|---|
| $= \frac{32}{27}\left[\frac{3t^4}{4} - \frac{2t^5}{5}\right]_0^{1.5}$ | A1 | |
| $= \frac{32}{27}\left[\left(\frac{243}{64} - \frac{2}{5}\times\frac{243}{32}\right) - (0)\right]$ | | |
| $= \frac{9}{2} - \frac{18}{5} = 0.9$ | A1 cso | (3) |
### Part (e):
| $F(E(T)) = \frac{16}{27}(2 \times 0.9^3 - 0.9^4) = 0.4752$ | B1 | evidence seen |
|---|---|---|
### Part (f):
| $P(T > 1 \mid T > 0.9) = \frac{P(T>1)}{P(T>0.9)} = \frac{\text{part }(b)}{1 - \text{part }(e)} = 0.7763\ldots$ | M1, M1, A1 | accept awrt 0.776 (3) |
|---|---|---|
---5. The continuous random variable $T$ represents the time in hours that students spend on homework. The cumulative distribution function of $T$ is
$$\mathrm { F } ( t ) = \begin{cases} 0 , & t < 0 \\ k \left( 2 t ^ { 3 } - t ^ { 4 } \right) & 0 \leq t \leq 1.5 \\ 1 , & t > 1.5 \end{cases}$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 16 } { 27 }$.
\item Find the proportion of students who spend more than 1 hour on homework.
\item Find the probability density function $\mathrm { f } ( t )$ of $T$.
\item Show that $\mathrm { E } ( T ) = 0.9$.
\item Show that $\mathrm { F } ( \mathrm { E } ( T ) ) = 0.4752$.
A student is selected at random. Given that the student spent more than the mean amount of time on homework,
\item find the probability that this student spent more than 1 hour on homework.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q5 [14]}}