| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF with additional constraints |
| Difficulty | Standard +0.3 This is a standard Further Statistics 2 CDF question requiring routine techniques: finding k from F(2)=1, differentiating to get pdf for E(X) and mode, and comparing mode/mean for skewness. All steps are algorithmic with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k\left(2^3 - \frac{3}{8}2^4\right) = 1\), giving \(2k=1\), \(k=\frac{1}{2}\) | B1* | Substituting \(x=2\) into \(F(x)\) and equating to 1; minimum substitution seen is \(k(8-6)=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = k\left[3x^2 - \frac{3}{2}x^3\right]\) | M1 | Realising need to find pdf and differentiating; at least 1 correct term |
| \(\int_0^2 xf(x)\,dx = k\int_0^2\left(3x^3 - \frac{3}{2}x^4\right)dx\) | M1d | Dependent on 1st M1; at least one correct term ft their pdf |
| \(= \left[\frac{3x^4}{8} - \frac{3x^5}{20}\right]_0^2\) | ||
| \(= \frac{6}{5}\) or \(1.2\) | A1 | NB 1.2 with no working gains M0M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x - \frac{9x^2}{4} = 0\) | M1d | Dependent on 1st M1; at least one correct term ft their pdf |
| \(x\left(3 - \frac{9x}{4}\right) = 0\) | M1d | Dep on 3rd M1; correct method for solving; pdf must be of form \(ax^2+bx\) |
| \(x=0\) or \(\frac{4}{3}\), \(\therefore\) mode \(= \frac{4}{3}\) | A1 | \(\therefore\) mode \(= \frac{4}{3}\) only; must eliminate 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mode \(>\) mean implies negative skew | B1ft | ft their mode and mean or a correct sketch |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k\left(2^3 - \frac{3}{8}2^4\right) = 1$, giving $2k=1$, $k=\frac{1}{2}$ | B1* | Substituting $x=2$ into $F(x)$ and equating to 1; minimum substitution seen is $k(8-6)=1$ |
# Question 4(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = k\left[3x^2 - \frac{3}{2}x^3\right]$ | M1 | Realising need to find pdf and differentiating; at least 1 correct term |
| $\int_0^2 xf(x)\,dx = k\int_0^2\left(3x^3 - \frac{3}{2}x^4\right)dx$ | M1d | Dependent on 1st M1; at least one correct term ft their pdf |
| $= \left[\frac{3x^4}{8} - \frac{3x^5}{20}\right]_0^2$ | | |
| $= \frac{6}{5}$ or $1.2$ | A1 | NB 1.2 with no working gains M0M0A0 |
# Question 4(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x - \frac{9x^2}{4} = 0$ | M1d | Dependent on 1st M1; at least one correct term ft their pdf |
| $x\left(3 - \frac{9x}{4}\right) = 0$ | M1d | Dep on 3rd M1; correct method for solving; pdf must be of form $ax^2+bx$ |
| $x=0$ or $\frac{4}{3}$, $\therefore$ mode $= \frac{4}{3}$ | A1 | $\therefore$ mode $= \frac{4}{3}$ only; must eliminate 0 |
# Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mode $>$ mean implies negative skew | B1ft | ft their mode and mean or a correct sketch |
---4 The continuous random variable $X$ has cumulative distribution function given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x \leqslant 0 \\
k \left( x ^ { 3 } - \frac { 3 } { 8 } x ^ { 4 } \right) & 0 < x \leqslant 2 \\
1 & x > 2
\end{array} \right.$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 2 }$
\item Showing your working clearly, use calculus to find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( X )$
\item the mode of $X$
\end{enumerate}\item Describe, giving a reason, the skewness of the distribution of $X$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2019 Q4 [8]}}