| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Conditional probability with PDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard techniques: sketching a piecewise PDF, calculating conditional probability using P(A∩B)/P(B) with integration, and applying linearity of expectation with E(X²)=Var(X)+[E(X)]². While multi-part, each component is routine application of formulas without requiring problem-solving insight or novel approaches. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape with \(g = \frac{3}{20}\) below \(\frac{1}{3}\), lines not joining at \(x=2\), none below/touching \(x\)-axis | M1 | For correct shape (\(g=\frac{3}{20}\) must be below \(\frac{1}{3}\)) with lines not joining at \(x=2\) and none below/touch the \(x\)-axis; ignore broken/dotted lines |
| Fully correct graph with labels on \(x\)-axis | A1 | For fully correct graph with labels on \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[P(G \leq 2)=\right] 1 - 2 \times \frac{3}{20} [=0.7]\) or \(\frac{1}{2}\times 3 \times \left(\frac{2}{15}+\frac{1}{3}\right)\) or \(\frac{1}{15}\int_{-1}^{2}(g+3)\,dg[=0.7]\) or \(\frac{1}{30}\times 2^2 + \frac{1}{5}\times 2 + \frac{1}{6}[=0.7]\) | M1 | For correct method to find \(P(G \leq 2)\) or \(P\!\left(G \leq \frac{1}{2}\right)\) or \(P\!\left(\frac{1}{2} \leq G \leq 2\right)\); may be implied by \(0.7/\frac{7}{10}\) or \(0.425=\frac{17}{40}\) or \(0.275/\frac{11}{40}\) |
| \(\left[P(1 \leq 2G \leq 6 \mid G \leq 2)\right] = \frac{P\!\left(\frac{1}{2} \leq G \leq 2\right)}{P(G \leq 2)} = \frac{0.425}{0.7}\) or \(1-\frac{0.275}{0.7}\) | M1M1 | Second M1 for \(\frac{p}{0.7}\) where \(0 |
| \(= \frac{17}{28}\) or \(0.607...\) | A1 awrt 0.607 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[E(H^2)=\right] 2.4 + 12^2 [=146.4]\) | M1 | For correct method to find \(E(H^2)\) |
| \(\left[E(G)=\right] \int_{-1}^{2} \frac{1}{15}(g^2+3g)\,dg + \int_{2}^{4} \frac{3}{20}g\,dg\) | M1 | For realising \(\int xf(x)\,dx\) on both functions and adding together; ignore limits |
| \(\left[E(G)=\right] \left[\frac{1}{15}\left(\frac{1}{3}g^3+\frac{3}{2}g^2\right)\right]_{-1}^{2} + \left(\frac{3}{40}g^2\right)_{2}^{4}\) | M1 | For attempting to integrate (\(x^n \to x^{n+1}\)) at least one part of \(xf(x)\) |
| \(= \frac{1}{15}\left(\frac{8}{3}+\frac{12}{2}+\frac{1}{3}-\frac{3}{2}\right)+\left(\frac{48}{40}-\frac{12}{40}\right)[=1.4]\) | dM1 | Dep on previous M1; for use of correct limits in one part of \(xf(x)\); if working not shown may be implied by 0.5 or 0.9 or 1.4 |
| \(\left[E(2H^2+3G+3)=\right] 2\times"146.4"+3\times"1.4"+3\) | M1 | For using \(2\times E(H^2)+3\times E(G)+3\), provided \(E(H^2)\) and \(E(G)\) have been shown |
| \(= 300\) | A1 | Cao |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape with $g = \frac{3}{20}$ below $\frac{1}{3}$, lines not joining at $x=2$, none below/touching $x$-axis | M1 | For correct shape ($g=\frac{3}{20}$ must be below $\frac{1}{3}$) with lines not joining at $x=2$ and none below/touch the $x$-axis; ignore broken/dotted lines |
| Fully correct graph with labels on $x$-axis | A1 | For fully correct graph with labels on $x$-axis |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[P(G \leq 2)=\right] 1 - 2 \times \frac{3}{20} [=0.7]$ or $\frac{1}{2}\times 3 \times \left(\frac{2}{15}+\frac{1}{3}\right)$ or $\frac{1}{15}\int_{-1}^{2}(g+3)\,dg[=0.7]$ or $\frac{1}{30}\times 2^2 + \frac{1}{5}\times 2 + \frac{1}{6}[=0.7]$ | M1 | For correct method to find $P(G \leq 2)$ or $P\!\left(G \leq \frac{1}{2}\right)$ or $P\!\left(\frac{1}{2} \leq G \leq 2\right)$; may be implied by $0.7/\frac{7}{10}$ or $0.425=\frac{17}{40}$ or $0.275/\frac{11}{40}$ |
| $\left[P(1 \leq 2G \leq 6 \mid G \leq 2)\right] = \frac{P\!\left(\frac{1}{2} \leq G \leq 2\right)}{P(G \leq 2)} = \frac{0.425}{0.7}$ or $1-\frac{0.275}{0.7}$ | M1M1 | Second M1 for $\frac{p}{0.7}$ where $0<p<0.7$; third M1 for correct ratio of probabilities |
| $= \frac{17}{28}$ or $0.607...$ | A1 awrt 0.607 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[E(H^2)=\right] 2.4 + 12^2 [=146.4]$ | M1 | For correct method to find $E(H^2)$ |
| $\left[E(G)=\right] \int_{-1}^{2} \frac{1}{15}(g^2+3g)\,dg + \int_{2}^{4} \frac{3}{20}g\,dg$ | M1 | For realising $\int xf(x)\,dx$ on both functions and adding together; ignore limits |
| $\left[E(G)=\right] \left[\frac{1}{15}\left(\frac{1}{3}g^3+\frac{3}{2}g^2\right)\right]_{-1}^{2} + \left(\frac{3}{40}g^2\right)_{2}^{4}$ | M1 | For attempting to integrate ($x^n \to x^{n+1}$) at least one part of $xf(x)$ |
| $= \frac{1}{15}\left(\frac{8}{3}+\frac{12}{2}+\frac{1}{3}-\frac{3}{2}\right)+\left(\frac{48}{40}-\frac{12}{40}\right)[=1.4]$ | dM1 | Dep on previous M1; for use of correct limits in one part of $xf(x)$; if working not shown may be implied by 0.5 or 0.9 or 1.4 |
| $\left[E(2H^2+3G+3)=\right] 2\times"146.4"+3\times"1.4"+3$ | M1 | For using $2\times E(H^2)+3\times E(G)+3$, provided $E(H^2)$ and $E(G)$ have been shown |
| $= 300$ | A1 | Cao |
---\begin{enumerate}
\item The continuous random variable $G$ has probability density function $\mathrm { f } ( \mathrm { g } )$ given by
\end{enumerate}
$$f ( g ) = \begin{cases} \frac { 1 } { 15 } ( g + 3 ) & - 1 < g \leqslant 2 \\ \frac { 3 } { 20 } & 2 < g \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(a) Sketch the graph of $\mathrm { f } ( \mathrm { g } )$\\
(b) Find $\mathrm { P } ( ( 1 \leqslant 2 G \leqslant 6 ) \mid G \leqslant 2 )$
The continuous random variable $H$ is such that $\mathrm { E } ( H ) = 12$ and $\operatorname { Var } ( H ) = 2.4$\\
(c) Find $\mathrm { E } \left( 2 H ^ { 2 } + 3 G + 3 \right)$
Show your working clearly.\\
(Solutions relying on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel S2 2024 Q4 [12]}}