| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | PDF with multiple constants |
| Difficulty | Standard +0.8 This S2 question requires understanding PDF properties, using continuity conditions, equating mode and median (requiring integration and solving a cubic equation), and finding multiple related constants. While systematic, it demands careful algebraic manipulation across multiple parts with interdependent steps, making it moderately challenging but still within standard S2 scope. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{Mode} =] 4\) | B1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a\int_0^4 x^3\,dx = \frac{1}{2} \Rightarrow a\left[\frac{x^4}{4}\right]_0^4 = \frac{1}{2}\) | M1 | For integrating the 1st line of the pdf and setting \(= 0.5\). Ignore limits |
| \(64a = \frac{1}{2} \Rightarrow a = \frac{1}{128}\) | A1* | Answer given so correct solution must be seen with no errors. At least one line of correct working from M mark to final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}\times\frac{1}{2}\times(d-4) = \frac{1}{2}\) or \(\frac{1}{2}\times\frac{1}{2}\times(d-4)+\int_0^4 ax^3\,dx = 1\) | M1 | For setting the area of the triangle \(= 0.5\) |
| \(d = 6\) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(b = \frac{-\frac{1}{2}}{6-4} = -\frac{1}{4}\) or \(4b+c = 0.5\) oe | M1 | A correct method for finding \(b\) ft their \(d\) value; or \(4b+c=0.5\) oe. Allow \(4b+c=64a\) |
| \(0 = -\frac{1}{4}\times 6 + c\) or \(\frac{1}{2} = -\frac{1}{4}\times 4+c\); \(10b+2c=0.5\) oe or \(6b+c=0\) oe | M1 | A correct method for finding \(c\) ft their \(b\) and \(d\) value; or \(10b+2c=0.5\) oe or \(d\times b+c=0\) oe. Allow \(db+c=0\) |
| \(b = -\frac{1}{4}\) and \(c = \frac{3}{2}\) | A1 | For both \(b\) and \(c\) correct. NB \(b=-0.25\) oe and \(c=1.5\) oe scores 3/3 |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Mode} =] 4$ | B1 | Cao |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a\int_0^4 x^3\,dx = \frac{1}{2} \Rightarrow a\left[\frac{x^4}{4}\right]_0^4 = \frac{1}{2}$ | M1 | For integrating the 1st line of the pdf and setting $= 0.5$. Ignore limits |
| $64a = \frac{1}{2} \Rightarrow a = \frac{1}{128}$ | A1* | Answer given so correct solution must be seen with no errors. At least one line of correct working from M mark to final answer |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}\times\frac{1}{2}\times(d-4) = \frac{1}{2}$ or $\frac{1}{2}\times\frac{1}{2}\times(d-4)+\int_0^4 ax^3\,dx = 1$ | M1 | For setting the area of the triangle $= 0.5$ |
| $d = 6$ | A1 | Cao |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $b = \frac{-\frac{1}{2}}{6-4} = -\frac{1}{4}$ or $4b+c = 0.5$ oe | M1 | A correct method for finding $b$ ft their $d$ value; or $4b+c=0.5$ oe. Allow $4b+c=64a$ |
| $0 = -\frac{1}{4}\times 6 + c$ or $\frac{1}{2} = -\frac{1}{4}\times 4+c$; $10b+2c=0.5$ oe or $6b+c=0$ oe | M1 | A correct method for finding $c$ ft their $b$ and $d$ value; or $10b+2c=0.5$ oe or $d\times b+c=0$ oe. Allow $db+c=0$ |
| $b = -\frac{1}{4}$ and $c = \frac{3}{2}$ | A1 | For both $b$ and $c$ correct. NB $b=-0.25$ oe and $c=1.5$ oe scores 3/3 |
---\begin{enumerate}
\item The continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ given by
\end{enumerate}
$$\mathrm { f } ( x ) = \begin{cases} a x ^ { 3 } & 0 \leqslant x \leqslant 4 \\ b x + c & 4 < x \leqslant d \\ 0 & \text { otherwise } \end{cases}$$
where $a$, $b$, $c$ and $d$ are constants such that
\begin{itemize}
\item $b x + c = a x ^ { 3 }$ at $x = 4$
\item $b x + c$ is a straight line segment with end coordinates ( $4,64 a$ ) and ( $d , 0$ )\\
(a) State the mode of $X$
\end{itemize}
Given that the mode of $X$ is equal to the median of $X$\\
(b) use algebraic integration to show that $a = \frac { 1 } { 128 }$\\
(c) Find the value of $d$\\
(d) Hence find the value of $b$ and the value of $c$
\hfill \mbox{\textit{Edexcel S2 2023 Q2 [8]}}