| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Multi-part piecewise CDF |
| Difficulty | Standard +0.3 This is a standard S2 CDF question requiring routine techniques: finding probabilities using F(b)-F(a), differentiating piecewise functions to get the PDF, and using continuity conditions to find constants. While multi-part with several steps, it follows predictable patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F(6)=1 \Rightarrow 4k(12-7)=1\) | M1 | Using \(F(6)=1\) to get linear equation in \(k\) |
| \(k=\frac{1}{20}\) | A1 | A1 \(\frac{1}{20}\) or 0.05 |
| \(\alpha^2-2\alpha-3=4(2\alpha-7)\) | M1 | Using \(F(\alpha)\): \(\alpha^2-2\alpha-3=4(2\alpha-7)\) |
| \(\alpha^2-10\alpha+25=0\), \((\alpha-5)^2=0\), \(\alpha=5\) | A1cao | cao 5 |
\(P(4.5| M1 |
Writing or using \(F(5.5)-F(4.5)\) |
|
| \(=4\times\frac{1}{20}\times(11-7)-\frac{1}{20}\times(4.5^2-9-3)\) | dM1 | Dep on previous M1; substituting 4.5 and 5.5 into appropriate lines |
| \(=\frac{31}{80}\) or 0.3875 or awrt 0.388 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
\(f(y)=\begin{cases}\frac{1}{20}(2y-2) & 3\leq y\leq 5\\ \frac{2}{5} & 5| M1 A1ft A1 |
M1 attempt to differentiate \(x^n\to x^{n-1}\); A1ft either 1st or 2nd line correct (ft their \(k\) and \(\alpha\)); A1 fully correct including 0 otherwise |
|
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $F(6)=1 \Rightarrow 4k(12-7)=1$ | M1 | Using $F(6)=1$ to get linear equation in $k$ |
| $k=\frac{1}{20}$ | A1 | A1 $\frac{1}{20}$ or 0.05 |
| $\alpha^2-2\alpha-3=4(2\alpha-7)$ | M1 | Using $F(\alpha)$: $\alpha^2-2\alpha-3=4(2\alpha-7)$ |
| $\alpha^2-10\alpha+25=0$, $(\alpha-5)^2=0$, $\alpha=5$ | A1cao | cao 5 |
| $P(4.5<X\leq 5.5)=F(5.5)-F(4.5)$ | M1 | Writing or using $F(5.5)-F(4.5)$ |
| $=4\times\frac{1}{20}\times(11-7)-\frac{1}{20}\times(4.5^2-9-3)$ | dM1 | Dep on previous M1; substituting 4.5 and 5.5 into appropriate lines |
| $=\frac{31}{80}$ or 0.3875 or awrt 0.388 | A1 | |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(y)=\begin{cases}\frac{1}{20}(2y-2) & 3\leq y\leq 5\\ \frac{2}{5} & 5<y\leq 6\\ 0 & \text{otherwise}\end{cases}$ | M1 A1ft A1 | M1 attempt to differentiate $x^n\to x^{n-1}$; A1ft either 1st or 2nd line correct (ft their $k$ and $\alpha$); A1 fully correct including 0 otherwise |5. The continuous random variable $Y$ has cumulative distribution function $\mathrm { F } ( y )$ given by
$$\mathrm { F } ( y ) = \left\{ \begin{array} { l r }
0 & y < 3 \\
k \left( y ^ { 2 } - 2 y - 3 \right) & 3 \leqslant y \leqslant \alpha \\
4 k ( 2 y - 7 ) & \alpha < y \leqslant 6 \\
1 & y > 6
\end{array} \right.$$
where $k$ and $\alpha$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 4.5 < Y \leqslant 5.5 )$
\item Find the probability density function $\mathrm { f } ( \mathrm { y } )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q5 [10]}}