| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| 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 piecewise CDF question requiring routine techniques: continuity at boundaries for part (a), conditional probability formula for part (b), and differentiation of each piece for part (c). While multi-step, it involves straightforward application of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{3}{10}d - \frac{1}{75}d^2 - \frac{2}{3} = 1\) | M1 | For this equation |
| \(45d - 2d^2 - 100 = 150\) or \(\frac{3}{10}d - \frac{1}{75}d^2 - \frac{5}{3} = 0 \rightarrow 2d^2 - 45d + 250 = 0\) | A1* | cso at least one step seen before given answer e.g. removing denominator or correct 3 term quadratic = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(1 < H < 4.5) = \left(\frac{4.5}{6} - \frac{1}{3}\right) - \left(\frac{1}{48}\right) = \frac{19}{48}\) or 0.39583... | M1 | Correct method to find \(P(1 < H < 4.5)\), implied by \(\frac{19}{48}\) or awrt 0.396 |
| \(P(1 < H < 1.5) = \left(\frac{1.5^2}{48}\right) - \left(\frac{1}{48}\right) = \frac{5}{192}\) or 0.02604... | M1 | For writing or finding \(P(1 < H < 1.5)\), implied by \(\frac{5}{192}\) or awrt 0.026 |
| \(P(H < 1.5 \mid 1 < H < 4.5) = \frac{\text{"0.02604..."}}{\text{"0.3958..."}}\) | M1 | For \(\frac{p}{\text{"0.3958"}}\) where \(0 < p <\) "0.3958" |
| \(= \frac{5}{76}\) or 0.06578... awrt 0.0658 | A1 | awrt 0.0658 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([f(h)=]\begin{cases} \frac{h}{24} & 0 < h \leq 4 \\ \frac{1}{6} & 4 < h \leq 5 \\ \frac{3}{10} - \frac{2}{75}h & 5 < h \leq d \\ 0 & \text{otherwise} \end{cases}\) | M1 | For one of rows 1, 2 or 3 correct. Allow any letters. Condone missing/incorrect range |
| M1 | For any two rows correct with ranges. Allow any letters and \(\leq\) for \(<\) signs | |
| A1 | Fully correct, all the same letter in rows 1 to 3. Allow \(\leq\) for \(<\) signs. Condone \(d=10\) but not \(d=12.5\) |
# Question 2:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{3}{10}d - \frac{1}{75}d^2 - \frac{2}{3} = 1$ | M1 | For this equation |
| $45d - 2d^2 - 100 = 150$ or $\frac{3}{10}d - \frac{1}{75}d^2 - \frac{5}{3} = 0 \rightarrow 2d^2 - 45d + 250 = 0$ | A1* | cso at least one step seen before given answer e.g. removing denominator or correct 3 term quadratic = 0 |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(1 < H < 4.5) = \left(\frac{4.5}{6} - \frac{1}{3}\right) - \left(\frac{1}{48}\right) = \frac{19}{48}$ or 0.39583... | M1 | Correct method to find $P(1 < H < 4.5)$, implied by $\frac{19}{48}$ or awrt 0.396 |
| $P(1 < H < 1.5) = \left(\frac{1.5^2}{48}\right) - \left(\frac{1}{48}\right) = \frac{5}{192}$ or 0.02604... | M1 | For writing or finding $P(1 < H < 1.5)$, implied by $\frac{5}{192}$ or awrt 0.026 |
| $P(H < 1.5 \mid 1 < H < 4.5) = \frac{\text{"0.02604..."}}{\text{"0.3958..."}}$ | M1 | For $\frac{p}{\text{"0.3958"}}$ where $0 < p <$ "0.3958" |
| $= \frac{5}{76}$ or 0.06578... awrt **0.0658** | A1 | awrt 0.0658 |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[f(h)=]\begin{cases} \frac{h}{24} & 0 < h \leq 4 \\ \frac{1}{6} & 4 < h \leq 5 \\ \frac{3}{10} - \frac{2}{75}h & 5 < h \leq d \\ 0 & \text{otherwise} \end{cases}$ | M1 | For one of rows 1, 2 or 3 correct. Allow any letters. Condone missing/incorrect range |
| | M1 | For any two rows correct with ranges. Allow any letters and $\leq$ for $<$ signs |
| | A1 | Fully correct, all the same letter in rows 1 to 3. Allow $\leq$ for $<$ signs. Condone $d=10$ but not $d=12.5$ |
---2 The continuous random variable $H$ has cumulative distribution function given by
$$\mathrm { F } ( h ) = \left\{ \begin{array} { l r }
0 & h \leqslant 0 \\
\frac { h ^ { 2 } } { 48 } & 0 < h \leqslant 4 \\
\frac { h } { 6 } - \frac { 1 } { 3 } & 4 < h \leqslant 5 \\
\frac { 3 } { 10 } h - \frac { h ^ { 2 } } { 75 } - \frac { 2 } { 3 } & 5 < h \leqslant d \\
1 & h > d
\end{array} \right.$$
where $d$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $2 d ^ { 2 } - 45 d + 250 = 0$
\item Find $\mathrm { P } ( H < 1.5 \mid 1 < H < 4.5 )$
\item Find the probability density function $\mathrm { f } ( h )$
You may leave the limits of $h$ in terms of $d$ where necessary.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2024 Q2 [9]}}