| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find mode of distribution |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics 2 question requiring standard techniques: integrating a pdf to find the median (algebraic manipulation provided), differentiating to find the mode (with explanation), and computing variance using E(X²) - [E(X)]². While it involves multiple parts, each step follows routine procedures with no novel insight required. The difficulty is slightly above average (positive score) because it's Further Maths content, but the techniques are mechanical and well-practiced. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int_2^m (0.8-6.4x^{-3})\,dx = 0.5\) | M1 | Integral \(= 0.5\) (ignore limits) |
| \(\left[0.8x+3.2x^{-2}\right]_2^m = 0.5\) | M1 | Integration with limits |
| \(0.8m+\frac{3.2}{m^2}-\left(0.8(2)+\frac{3.2}{2^2}\right)=0.5 \Rightarrow m^3-3.625m^2+4=0\) | A1*cso | Given answer with at least one line of intermediate working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(f'(x)=19.2x^{-4}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Since \(f'(x)>0\), \(f(x)\) is increasing (the pdf has its maximum value at the upper end of the interval), the mode is 4 | B1 | Correct reasoning and conclusion; allow equivalent correct reasoning e.g. no turning points with a sketch of \(f(x)\); do not allow unsupported comments e.g. '\(x=4\) is the highest point on \(f(x)\)' |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X)=\int_2^4 x(0.8-6.4x^{-3})\,dx\) | M1 | Multiplying out \(xf(x)\) and attempt to integrate |
| \(E(X)=\left[0.4x^2+6.4x^{-1}\right]_2^4 = 0.4(4^2)+6.4(4^{-1})-(0.4(2^2)+6.4(2^{-1}))[=3.2]\) | M1 | Correct use of limits (implied by 3.2 oe) |
| \(\text{Var}(X)=10.5 - {'}3.2{'}^2\) | M1 | Use of \(E(X^2)-[E(X)]^2\) |
| \(\text{Var}(X)=0.26\) | A1 |
## Question 4:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int_2^m (0.8-6.4x^{-3})\,dx = 0.5$ | M1 | Integral $= 0.5$ (ignore limits) |
| $\left[0.8x+3.2x^{-2}\right]_2^m = 0.5$ | M1 | Integration with limits |
| $0.8m+\frac{3.2}{m^2}-\left(0.8(2)+\frac{3.2}{2^2}\right)=0.5 \Rightarrow m^3-3.625m^2+4=0$ | A1*cso | Given answer with at least one line of intermediate working |
**(3 marks)**
### Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $f'(x)=19.2x^{-4}$ | B1 | |
### Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Since $f'(x)>0$, $f(x)$ is increasing (the pdf has its maximum value at the upper end of the interval), the mode is 4 | B1 | Correct reasoning and conclusion; allow equivalent correct reasoning e.g. no turning points with a sketch of $f(x)$; do not allow unsupported comments e.g. '$x=4$ is the highest point on $f(x)$' |
**(2 marks)**
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X)=\int_2^4 x(0.8-6.4x^{-3})\,dx$ | M1 | Multiplying out $xf(x)$ and attempt to integrate |
| $E(X)=\left[0.4x^2+6.4x^{-1}\right]_2^4 = 0.4(4^2)+6.4(4^{-1})-(0.4(2^2)+6.4(2^{-1}))[=3.2]$ | M1 | Correct use of limits (implied by 3.2 oe) |
| $\text{Var}(X)=10.5 - {'}3.2{'}^2$ | M1 | Use of $E(X^2)-[E(X)]^2$ |
| $\text{Var}(X)=0.26$ | A1 | |
**(4 marks)**
---\begin{enumerate}
\item A random variable $X$ has probability density function given by
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c c }
0.8 - 6.4 x ^ { - 3 } & 2 \leqslant x \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$
The median of $X$ is $m$\\
(a) Show that $m ^ { 3 } - 3.625 m ^ { 2 } + 4 = 0$\\
(b) (i) Find $\mathrm { f } ^ { \prime } ( x )$\\
(ii) Explain why the mode of $X$ is 4
Given that $\mathrm { E } \left( X ^ { 2 } \right) = 10.5$ to 3 significant figures,\\
(c) find $\operatorname { Var } ( X )$, showing your working clearly.
\hfill \mbox{\textit{Edexcel FS2 AS 2022 Q4 [9]}}