| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring multiple techniques: finding constants using CDF boundary conditions, deriving PDF by differentiation, computing variance using integration (with non-trivial square root terms), and solving a quadratic equation involving surds. While systematic, the algebraic manipulation with √x terms and the multi-part nature elevate it above standard A-level, though it follows a predictable structure for FS2. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(4) = 0 \Rightarrow 4p - 2k = 0\) or \(F(9) = 1 \Rightarrow 9p - 3k = 1\) | M1, A1 | For selecting correct approach and getting 1 correct equation; A1 for 2 correct equations in \(p\) and \(k\) |
| Solving e.g. sub \(k = 2p \Rightarrow 9p - 6p = 1\) | M1 | For solving two simultaneous equations – reducing to linear equation in 1 variable |
| \(p = \dfrac{1}{3}, \quad k = \dfrac{2}{3}\) | A1 | For both correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Find \(f(x)\): \(f(x) = F'(x) = \dfrac{1}{3} - \dfrac{2}{3} \times \dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{3}\!\left(1 - x^{-\frac{1}{2}}\right)\) | M1 | For realising need to find \(f(x)\) first and attempt to differentiate \(F(x)\) |
| \(E(X^2) = \dfrac{1}{3}\displaystyle\int_4^9 x^2(1 - x^{-\frac{1}{2}})\,dx\) or \(\dfrac{1}{3}\displaystyle\int_4^9 (x^2 - x^{\frac{3}{2}})\,dx\) | M1 | For attempting \(\int x^2 f(x)\,dx\) with \(f(x)\) different from \(F(x)\) |
| \(= \dfrac{1}{3}\!\left[\dfrac{x^3}{3} - \dfrac{2x^{\frac{5}{2}}}{5}\right]_4^9 = \dfrac{1}{3}\!\left[\left(\dfrac{9^3}{3} - \dfrac{2\times3^5}{5}\right) - \left(\dfrac{64}{3} - \dfrac{2\times2^5}{5}\right)\right] = \dfrac{2059}{45}\) | M1, A1 | For correct integration; A1 for correct value exact fraction or at least 45.755… |
| \(\text{Var}(X) = \dfrac{2059}{45} - \left(\dfrac{119}{18}\right)^2\) | M1 | For correct method for \(\text{Var}(X)\) using their \(E(X^2)\) |
| \(= 2.048765\ldots = 2.05\) (3sf) | A1* | For fully correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[f(x) = \dfrac{1}{3}\!\left(1 - \dfrac{1}{\sqrt{x}}\right)\right]\) so max is when \(x\) is greatest, so mode \(= \mathbf{9}\) | B1 | For 9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(a) = \dfrac{7}{27} \Rightarrow a - 2\sqrt{a} - \dfrac{7}{9} = 0\) | M1 | For realising need to use \(F(x)\) and forming correct equation in \(a\) |
| \(y^2 - 2y - \dfrac{7}{9} = 0 \Rightarrow (y-1)^2 = \dfrac{16}{9}\) | M1 | For recognising equation as quadratic and trying to solve |
| \(y = \dfrac{7}{3}\) or \(\left(-\dfrac{1}{3}\text{ not valid}\right)\) so \(a = \dfrac{49}{9}\) | A1 | For exact working and selection of appropriate value |
# Question 3(a):
$F(4) = 0 \Rightarrow 4p - 2k = 0$ or $F(9) = 1 \Rightarrow 9p - 3k = 1$ | M1, A1 | For selecting correct approach and getting 1 correct equation; A1 for 2 correct equations in $p$ and $k$
Solving e.g. sub $k = 2p \Rightarrow 9p - 6p = 1$ | M1 | For solving two simultaneous equations – reducing to linear equation in 1 variable
$p = \dfrac{1}{3}, \quad k = \dfrac{2}{3}$ | A1 | For both correct values
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# Question 3(b):
Find $f(x)$: $f(x) = F'(x) = \dfrac{1}{3} - \dfrac{2}{3} \times \dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{1}{3}\!\left(1 - x^{-\frac{1}{2}}\right)$ | M1 | For realising need to find $f(x)$ first and attempt to differentiate $F(x)$
$E(X^2) = \dfrac{1}{3}\displaystyle\int_4^9 x^2(1 - x^{-\frac{1}{2}})\,dx$ or $\dfrac{1}{3}\displaystyle\int_4^9 (x^2 - x^{\frac{3}{2}})\,dx$ | M1 | For attempting $\int x^2 f(x)\,dx$ with $f(x)$ different from $F(x)$
$= \dfrac{1}{3}\!\left[\dfrac{x^3}{3} - \dfrac{2x^{\frac{5}{2}}}{5}\right]_4^9 = \dfrac{1}{3}\!\left[\left(\dfrac{9^3}{3} - \dfrac{2\times3^5}{5}\right) - \left(\dfrac{64}{3} - \dfrac{2\times2^5}{5}\right)\right] = \dfrac{2059}{45}$ | M1, A1 | For correct integration; A1 for correct value exact fraction or at least 45.755…
$\text{Var}(X) = \dfrac{2059}{45} - \left(\dfrac{119}{18}\right)^2$ | M1 | For correct method for $\text{Var}(X)$ using their $E(X^2)$
$= 2.048765\ldots = 2.05$ (3sf) | A1* | For fully correct solution
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# Question 3(c):
$\left[f(x) = \dfrac{1}{3}\!\left(1 - \dfrac{1}{\sqrt{x}}\right)\right]$ so max is when $x$ is greatest, so mode $= \mathbf{9}$ | B1 | For 9
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# Question 3(d):
$F(a) = \dfrac{7}{27} \Rightarrow a - 2\sqrt{a} - \dfrac{7}{9} = 0$ | M1 | For realising need to use $F(x)$ and forming correct equation in $a$
$y^2 - 2y - \dfrac{7}{9} = 0 \Rightarrow (y-1)^2 = \dfrac{16}{9}$ | M1 | For recognising equation as quadratic and trying to solve
$y = \dfrac{7}{3}$ or $\left(-\dfrac{1}{3}\text{ not valid}\right)$ so $a = \dfrac{49}{9}$ | A1 | For exact working and selection of appropriate value
---\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x < 4 \\
p x - k \sqrt { x } & 4 \leqslant x \leqslant 9 \\
1 & x > 9
\end{array} \right.$$
where $p$ and $k$ are constants.\\
(a) Find the value of $p$ and the value of $k$.
Given that $\mathrm { E } ( X ) = \frac { 119 } { 18 }$\\
(b) show that $\operatorname { Var } ( X ) = 2.05$ to 3 significant figures.\\
(c) Write down the mode of $X$.\\
(d) Find the exact value of the constant $a$ such that $\mathrm { P } ( X \leqslant a ) = \frac { 7 } { 27 }$
\hfill \mbox{\textit{Edexcel FS2 AS 2020 Q3 [14]}}