| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Challenging +1.2 This is a Further Statistics 2 question requiring integration of a power function PDF, solving simultaneous equations from normalization and a given probability, then finding the median. While it involves multiple steps and algebraic manipulation, the techniques are standard for FS2: integrating x^(-2) and x^(-3) is routine, the normalization condition is a standard constraint, and finding the median requires solving F(m)=0.5. The algebra is somewhat involved but follows predictable patterns. This is moderately above average difficulty due to the multi-step nature and being from Further Maths, but doesn't require novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int ax^{-2} - bx^{-3}\,dx = -\frac{a}{x} + \frac{b}{2x^2}\) | M1 | Attempt to integrate one term correct; power increasing by 1 |
| \(\left[-\frac{a}{x}+\frac{b}{2x^2}\right]_2^{\infty} = 0 - \left(-\frac{a}{2}+\frac{b}{8}\right) = \frac{a}{2}-\frac{b}{8}\) | M1 | Integrating both terms and substituting limits correctly; any one of \((2,\infty)\), \((4,\infty)\), or \((2,4)\) |
| \(\left[-\frac{a}{x}+\frac{b}{2x^2}\right]_2^{4} = \left(-\frac{a}{4}+\frac{b}{32}\right)-\left(-\frac{a}{2}+\frac{b}{8}\right) = \frac{a}{4}-\frac{3b}{32}\) | M1 | Integrating and substituting limits correctly for second expression |
| \(\frac{a}{2}-\frac{b}{8}=1\) or \(4a-b=8\) | dM1 | Dependent on 2nd M; one expression equal to correct value from \(1, \frac{3}{8}\) or \(\frac{5}{8}\) |
| \(\frac{a}{4}-\frac{3b}{32}=\frac{3}{8}\) or \(8a-3b=12\) | ||
| \(\therefore a = 3\) | A1*cso | Fully correct solution achieving \(a=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b = 4\) | B1 | Writing or using \(b=4\); may be seen in (a) |
| \(\left[-\frac{3}{x}+\frac{4}{2x^2}\right]_2^{m} = 0.5\) or e.g. \(-3x^{-1}+2x^{-2}+1=0.5\) | M1 | Equating integral with \(b\), limits 2 and \(m\), equated to 0.5; must be of form \(\alpha x^{-1}+\beta x^{-2}+\gamma = 0.5\) |
| \(m^2 - 6m + 4 = 0\) | A1 | Correct 3-term quadratic \(= 0\); terms do not need to be on same side |
| \((m=)\ 3+\sqrt{5}\) | A1 | \(3+\sqrt{5}\); any other solutions should be eliminated |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int ax^{-2} - bx^{-3}\,dx = -\frac{a}{x} + \frac{b}{2x^2}$ | M1 | Attempt to integrate one term correct; power increasing by 1 |
| $\left[-\frac{a}{x}+\frac{b}{2x^2}\right]_2^{\infty} = 0 - \left(-\frac{a}{2}+\frac{b}{8}\right) = \frac{a}{2}-\frac{b}{8}$ | M1 | Integrating both terms and substituting limits correctly; any one of $(2,\infty)$, $(4,\infty)$, or $(2,4)$ |
| $\left[-\frac{a}{x}+\frac{b}{2x^2}\right]_2^{4} = \left(-\frac{a}{4}+\frac{b}{32}\right)-\left(-\frac{a}{2}+\frac{b}{8}\right) = \frac{a}{4}-\frac{3b}{32}$ | M1 | Integrating and substituting limits correctly for second expression |
| $\frac{a}{2}-\frac{b}{8}=1$ or $4a-b=8$ | dM1 | Dependent on 2nd M; one expression equal to correct value from $1, \frac{3}{8}$ or $\frac{5}{8}$ |
| $\frac{a}{4}-\frac{3b}{32}=\frac{3}{8}$ or $8a-3b=12$ | | |
| $\therefore a = 3$ | A1*cso | Fully correct solution achieving $a=3$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = 4$ | B1 | Writing or using $b=4$; may be seen in (a) |
| $\left[-\frac{3}{x}+\frac{4}{2x^2}\right]_2^{m} = 0.5$ or e.g. $-3x^{-1}+2x^{-2}+1=0.5$ | M1 | Equating integral with $b$, limits 2 and $m$, equated to 0.5; must be of form $\alpha x^{-1}+\beta x^{-2}+\gamma = 0.5$ |
| $m^2 - 6m + 4 = 0$ | A1 | Correct 3-term quadratic $= 0$; terms do not need to be on same side |
| $(m=)\ 3+\sqrt{5}$ | A1 | $3+\sqrt{5}$; any other solutions should be eliminated |\begin{enumerate}
\item A continuous random variable $X$ has probability density function
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c l }
a x ^ { - 2 } - b x ^ { - 3 } & 2 \leqslant x < \infty \\
0 & \text { otherwise }
\end{array} \right.$$
where $a$ and $b$ are constants.
Given that $\mathrm { P } ( X \leqslant 4 ) = \frac { 3 } { 8 }$\\
(a) use algebraic integration to show that $a = 3$
Show your working clearly.\\
(b) Find the exact value of the median of $X$
\hfill \mbox{\textit{Edexcel FS2 2024 Q5 [10]}}