| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics 2 question requiring standard techniques: integrating a polynomial pdf to find the CDF, using the CDF for probability calculations, applying the expectation formula with a simple transformation, and finding the mode by differentiation. While it involves multiple parts and some algebraic manipulation, each step follows routine procedures without requiring novel insight or particularly challenging integration. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{9x}{16} - \frac{x^3}{16}\,dx = \left[\frac{9x^2}{32} - \frac{x^4}{64}\right]\) or \(\frac{9x^2}{32} - \frac{x^4}{64} [+c]\) | M1 | 1.1b – correct method to integrate f(x), e.g. \(\frac{(9-x^2)^2}{-64}\) or \(1 - \frac{(9-x^2)^2}{64}\), allow one sign error |
| \(F(x) = \begin{cases} 0 & x < 1 \\ \frac{9x^2}{32} - \frac{x^4}{64} - \frac{17}{64} & 1 \leqslant x \leqslant 3 \\ 1 & x > 3 \end{cases}\) | B1 | 1.1b – 1st and 3rd lines correct, need \(x < 1\) and \(x > 3\) |
| A1 | 1.1b – correct middle line including \(1 \leqslant x \leqslant 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 1.8) = 1 - F(1.8)\) oe | M1 | 1.1b – correct method to find prob. Must have \(1 - \ldots\) and 1.8 substituted into their \(F(x)\) |
| \(= 0.5184\) | A1 | 1.1b – awrt 0.518 or \(\frac{324}{625}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^{-1}) = \int_1^3 x^{-1}\left(\frac{9}{16}x - \frac{1}{16}x^3\right)dx\) or \(\int_1^3 \frac{9}{16} - \frac{1}{16}x^2\,dx\) | M1 | 1.1b – correct expression to find \(E(X^{-1})\) or \(E(3X^{-1}+2)\) |
| \(E(3X^{-1}+2) = 3\times\left[\frac{9}{16}x - \frac{1}{48}x^3\right]_1^3 + 2\) or \(3\times\frac{7}{12} + 2\) | dM1 | 1.1b – attempt to integrate correct expression (at least 1 term correct) |
| \(= 3.75\) | A1 | 1.1b – [NB \(E(X) = 1.85\) leading to \(E(3X^{-1}+2) = \frac{134}{37} = 3.62\ldots\) scores M0M0A0] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(3X^{-1}+2) = \int_1^3 \left(\frac{9}{16}x - \frac{1}{16}x^3\right)(3x^{-1}+2)\,dx\) | M1 | |
| \(= \left[\frac{27}{16}x - \frac{1}{16}x^3 + \frac{9}{16}x^2 - \frac{1}{32}x^4\right]_1^3\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{df(x)}{dx} = \frac{9}{16} - \frac{3}{16}x^2\) | M1 | 1.1b – differentiating \(f(x)\) with at least one term correct |
| \(\frac{9}{16} - \frac{3}{16}x^2 = 0\) | dM1 | 1.1b – setting up an equation to find the mode, allow subst of \(\sqrt{3}\) |
| \(\frac{3}{16}x^2 = \frac{9}{16} \Rightarrow x = \sqrt{3}\) and either a sketch or full statement | A1cso* | 2.1 – \(x = \sqrt{3}\) and either a sketch through \((0,0)\), \((3,0)\) and showing max between them, or statement to show mode not on boundary: \(f(\sqrt{3}) = 0.6495\ldots > 0.5\), \(f(1) = 0.5\) and \(f(3) = 0\) |
## Question 2:
### Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{9x}{16} - \frac{x^3}{16}\,dx = \left[\frac{9x^2}{32} - \frac{x^4}{64}\right]$ or $\frac{9x^2}{32} - \frac{x^4}{64} [+c]$ | M1 | 1.1b – correct method to integrate f(x), e.g. $\frac{(9-x^2)^2}{-64}$ or $1 - \frac{(9-x^2)^2}{64}$, allow one sign error |
| $F(x) = \begin{cases} 0 & x < 1 \\ \frac{9x^2}{32} - \frac{x^4}{64} - \frac{17}{64} & 1 \leqslant x \leqslant 3 \\ 1 & x > 3 \end{cases}$ | B1 | 1.1b – 1st and 3rd lines correct, need $x < 1$ and $x > 3$ |
| | A1 | 1.1b – correct middle line including $1 \leqslant x \leqslant 3$ |
### Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 1.8) = 1 - F(1.8)$ oe | M1 | 1.1b – correct method to find prob. Must have $1 - \ldots$ and 1.8 substituted into their $F(x)$ |
| $= 0.5184$ | A1 | 1.1b – awrt 0.518 or $\frac{324}{625}$ |
### Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^{-1}) = \int_1^3 x^{-1}\left(\frac{9}{16}x - \frac{1}{16}x^3\right)dx$ or $\int_1^3 \frac{9}{16} - \frac{1}{16}x^2\,dx$ | M1 | 1.1b – correct expression to find $E(X^{-1})$ or $E(3X^{-1}+2)$ |
| $E(3X^{-1}+2) = 3\times\left[\frac{9}{16}x - \frac{1}{48}x^3\right]_1^3 + 2$ or $3\times\frac{7}{12} + 2$ | dM1 | 1.1b – attempt to integrate correct expression (at least 1 term correct) |
| $= 3.75$ | A1 | 1.1b – [NB $E(X) = 1.85$ leading to $E(3X^{-1}+2) = \frac{134}{37} = 3.62\ldots$ scores M0M0A0] |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(3X^{-1}+2) = \int_1^3 \left(\frac{9}{16}x - \frac{1}{16}x^3\right)(3x^{-1}+2)\,dx$ | M1 | |
| $= \left[\frac{27}{16}x - \frac{1}{16}x^3 + \frac{9}{16}x^2 - \frac{1}{32}x^4\right]_1^3$ | M1 | |
### Part 2(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{df(x)}{dx} = \frac{9}{16} - \frac{3}{16}x^2$ | M1 | 1.1b – differentiating $f(x)$ with at least one term correct |
| $\frac{9}{16} - \frac{3}{16}x^2 = 0$ | dM1 | 1.1b – setting up an equation to find the mode, allow subst of $\sqrt{3}$ |
| $\frac{3}{16}x^2 = \frac{9}{16} \Rightarrow x = \sqrt{3}$ **and** either a sketch or full statement | A1cso* | 2.1 – $x = \sqrt{3}$ and either a sketch through $(0,0)$, $(3,0)$ and showing max between them, or statement to show mode not on boundary: $f(\sqrt{3}) = 0.6495\ldots > 0.5$, $f(1) = 0.5$ and $f(3) = 0$ |
---\begin{enumerate}
\item A continuous random variable $X$ has probability density function
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { x } { 16 } \left( 9 - x ^ { 2 } \right) & 1 \leqslant x \leqslant 3 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Find the cumulative distribution function of $X$\\
(b) Calculate $\mathrm { P } ( X > 1.8 )$\\
(c) Use calculus to find $\mathrm { E } \left( \frac { 3 } { X } + 2 \right)$\\
(d) Show that the mode of $X$ is $\sqrt { 3 }$
\hfill \mbox{\textit{Edexcel FS2 AS 2023 Q2 [11]}}