| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine integration of a piecewise PDF to find E(X) and F(x), then solving F(x)=0.5 for the median. While it has multiple parts and requires careful handling of the piecewise function, all techniques are straightforward applications of standard formulas with no novel insight required. Slightly easier than average due to the simple polynomial forms. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(X) = \int_0^1 \frac{2x^2}{9}dx + \int_1^4 \frac{2x}{9}dx + \int_4^6 \frac{2x}{3} - \frac{x^2}{9}dx\) | M1 | Using \(\int xf(x)dx\), ignore limits. Must have at least one \(x^n \to x^{n+1}\) |
| \(= \left[\frac{2x^3}{27}\right]_0^1 + \left[\frac{2x^2}{18}\right]_1^4 + \left[\frac{x^2}{3} - \frac{x^3}{27}\right]_4^6\) | A1 | All integration correct; correct limits |
| \(= \left[\frac{2}{27}\right] + \left[\frac{32}{18} - \frac{2}{18}\right] + \left[4 - \frac{80}{27}\right]\) | M1d | Dependent on previous M. Substitute in correct limits — no need to see zero substituted |
| \(= 2\frac{7}{9}\) or awrt \(2.78\) | A1 | oe or awrt 2.78 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F(x) = 0\), \(x < 0\) | B1 | For first and last line — allow \(\leq\) instead of \(<\) and \(\geq\) instead of \(>\); allow "otherwise" for one of \(x<0\) and \(x>6\) |
| \(F(x) = \frac{x^2}{9}\), \(0 \leq x \leq 1\) | B1 | Allow use of \(<\) instead of \(\leq\) |
| \(F(x) = \frac{2x}{9} - \frac{1}{9}\), \(1 < x < 4\) | M1A1 | For \(1 < x < 4\): \(F(x) = \int_1^x \frac{2}{9}dx + \frac{1}{9}\); limits are needed |
| \(F(x) = \frac{2x}{3} - \frac{x^2}{18} - 1\), \(4 \leq x \leq 6\) | M1A1 | For \(4 \leq x \leq 6\): \(F(x) = \int_4^x \frac{2}{3} - \frac{x}{9}dx + \frac{7}{9}\); limits needed |
| \(F(x) = 1\), \(x > 6\) | B1 | Allow use of \(<\) instead of \(\leq\) for 4th line |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F(x) = 0.5\) | M1 | Setting any one of their lines \(= 0.5\) |
| \(\frac{2m}{9} - \frac{1}{9} = 0.5\) | A1ft | Their 3rd line \(= 0.5\) |
| \(m = 2.75\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Median \(<\) mean therefore positive skew Or Mean \(\approx\) median therefore no skewness | M1A1cao | M1: reason must match values / a correctly shaped and labelled sketch; must compare median and mean, ignore references to mode. A1: no ft — correct answer only from correct values of mean and median, or a correct and fully labelled sketch |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(X) = \int_0^1 \frac{2x^2}{9}dx + \int_1^4 \frac{2x}{9}dx + \int_4^6 \frac{2x}{3} - \frac{x^2}{9}dx$ | M1 | Using $\int xf(x)dx$, ignore limits. Must have at least one $x^n \to x^{n+1}$ |
| $= \left[\frac{2x^3}{27}\right]_0^1 + \left[\frac{2x^2}{18}\right]_1^4 + \left[\frac{x^2}{3} - \frac{x^3}{27}\right]_4^6$ | A1 | All integration correct; correct limits |
| $= \left[\frac{2}{27}\right] + \left[\frac{32}{18} - \frac{2}{18}\right] + \left[4 - \frac{80}{27}\right]$ | M1d | Dependent on previous M. Substitute in correct limits — no need to see zero substituted |
| $= 2\frac{7}{9}$ or awrt $2.78$ | A1 | oe or awrt 2.78 |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F(x) = 0$, $x < 0$ | B1 | For first and last line — allow $\leq$ instead of $<$ and $\geq$ instead of $>$; allow "otherwise" for one of $x<0$ and $x>6$ |
| $F(x) = \frac{x^2}{9}$, $0 \leq x \leq 1$ | B1 | Allow use of $<$ instead of $\leq$ |
| $F(x) = \frac{2x}{9} - \frac{1}{9}$, $1 < x < 4$ | M1A1 | For $1 < x < 4$: $F(x) = \int_1^x \frac{2}{9}dx + \frac{1}{9}$; limits are needed |
| $F(x) = \frac{2x}{3} - \frac{x^2}{18} - 1$, $4 \leq x \leq 6$ | M1A1 | For $4 \leq x \leq 6$: $F(x) = \int_4^x \frac{2}{3} - \frac{x}{9}dx + \frac{7}{9}$; limits needed |
| $F(x) = 1$, $x > 6$ | B1 | Allow use of $<$ instead of $\leq$ for 4th line |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F(x) = 0.5$ | M1 | Setting any one of their lines $= 0.5$ |
| $\frac{2m}{9} - \frac{1}{9} = 0.5$ | A1ft | Their 3rd line $= 0.5$ |
| $m = 2.75$ | A1 | |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Median $<$ mean therefore positive skew **Or** Mean $\approx$ median therefore no skewness | M1A1cao | M1: reason must match values / a correctly shaped and labelled sketch; must compare median and mean, ignore references to mode. A1: no ft — correct answer only from correct values of mean and median, or a correct and fully labelled sketch |6. The continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ given by
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 2 x } { 9 } & 0 \leqslant x \leqslant 1 \\
\frac { 2 } { 9 } & 1 < x < 4 \\
\frac { 2 } { 3 } - \frac { x } { 9 } & 4 \leqslant x \leqslant 6 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( X )$.
\item Find the cumulative distribution function $\mathrm { F } ( x )$ for all values of $x$.
\item Find the median of $X$.
\item Describe the skewness. Give a reason for your answer.\\
\includegraphics[max width=\textwidth, alt={}, center]{caaa0133-5b13-4ca7-8e65-8543327c33fd-12_104_61_2412_1884}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q6 [15]}}