| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Geometric applications |
| Difficulty | Moderate -0.8 This is a straightforward S2 uniform distribution question requiring only standard formulas and basic probability calculations. Parts (a)-(d) involve direct application of uniform distribution properties (PDF, mean, variance), while part (e) requires simple geometric probability with independence—all routine techniques with no novel problem-solving or insight needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{\alpha-6}{\alpha}=0.6\) | M1 | \(\frac{\alpha-6}{\alpha}=0.6\) (oe) or \(\frac{6}{\alpha}=0.4\) (oe) |
| \(\alpha=15\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
\(P(4| M1 A1 |
M1 \(\frac{10-4}{\text{their (a)}}\) |
|
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Mean \(=10\) | B1 | |
| Standard deviation \(=\frac{10\sqrt{3}}{3}\) or awrt 5.77 or \(\frac{20}{\sqrt{12}}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
\(P(\lvert Y-4\rvert<2)=P(2| M1 |
Writing or using \(P(2 | |
| \(=\frac{1}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P(X\text{ in middle 4cm})\times P(Y\text{ in middle 4cm})=\frac{4}{15}\times\frac{4}{20}\) | M1 | M1 \(\frac{4}{\text{their(a)}}\times\frac{4}{20}\) |
| \(=\frac{4}{75}\) | A1 | A1 \(\frac{4}{75}\) or awrt 0.0533 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P(X\text{ in middle 5cm})\times P(Y\text{ in middle 10cm})=\frac{5}{15}\times\frac{10}{20}=\frac{1}{6}\) | M1 A1 | M1 \(\frac{5}{\text{their(a)}}\times\frac{10}{20}\); A1 \(\frac{1}{6}\) or awrt 0.167 |
| \(P(\text{within 5cm of sides})=1-\frac{1}{6}=\frac{5}{6}\) | dM1 A1 | dM1 dep on previous M1 for \(1-\)"their 0.167"; A1 \(\frac{5}{6}\) or awrt 0.833 |
## Question 4:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{\alpha-6}{\alpha}=0.6$ | M1 | $\frac{\alpha-6}{\alpha}=0.6$ (oe) or $\frac{6}{\alpha}=0.4$ (oe) |
| $\alpha=15$ | A1 | |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(4<X<10)=\frac{10-4}{15}=\frac{2}{5}$ oe | M1 A1 | M1 $\frac{10-4}{\text{their (a)}}$ |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Mean $=10$ | B1 | |
| Standard deviation $=\frac{10\sqrt{3}}{3}$ or awrt 5.77 or $\frac{20}{\sqrt{12}}$ | B1 | |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(\lvert Y-4\rvert<2)=P(2<Y<6)$ | M1 | Writing or using $P(2<Y<6)$ |
| $=\frac{1}{5}$ | A1 | |
### Part (e)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(X\text{ in middle 4cm})\times P(Y\text{ in middle 4cm})=\frac{4}{15}\times\frac{4}{20}$ | M1 | M1 $\frac{4}{\text{their(a)}}\times\frac{4}{20}$ |
| $=\frac{4}{75}$ | A1 | A1 $\frac{4}{75}$ or awrt 0.0533 |
### Part (e)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(X\text{ in middle 5cm})\times P(Y\text{ in middle 10cm})=\frac{5}{15}\times\frac{10}{20}=\frac{1}{6}$ | M1 A1 | M1 $\frac{5}{\text{their(a)}}\times\frac{10}{20}$; A1 $\frac{1}{6}$ or awrt 0.167 |
| $P(\text{within 5cm of sides})=1-\frac{1}{6}=\frac{5}{6}$ | dM1 A1 | dM1 dep on previous M1 for $1-$"their 0.167"; A1 $\frac{5}{6}$ or awrt 0.833 |
---4. In a computer game, a ship appears randomly on a rectangular screen.
The continuous random variable $X \mathrm {~cm}$ is the distance of the centre of the ship from the bottom of the screen. The random variable $X$ is uniformly distributed over the interval $[ 0 , \alpha ]$ where $\alpha \mathrm { cm }$ is the height of the screen.
Given that $\mathrm { P } ( X > 6 ) = 0.6$
\begin{enumerate}[label=(\alph*)]
\item find the value of $\alpha$
\item find $\mathrm { P } ( 4 < X < 10 )$
The continuous random variable $Y$ cm is the distance of the centre of the ship from the left-hand side of the screen. The random variable $Y$ is uniformly distributed over the interval [ 0,20 ] where 20 cm is the width of the screen.
\item Find the mean and the standard deviation of $Y$.
\item Find $\mathrm { P } ( | Y - 4 | < 2 )$
\item Given that $X$ and $Y$ are independent, find the probability that the centre of the ship appears
\begin{enumerate}[label=(\roman*)]
\item in a square of side 4 cm which is at the centre of the screen,
\item within 5 cm of a side or the top or the bottom of the screen.
\begin{center}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q4 [14]}}