| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard techniques: evaluating F(x) at given points, differentiating to find f(x), finding the mode by setting f'(x)=0, and comparing median/mode for skewness. All steps are routine applications of learned procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(1.23) = \) awrt 0.495 | M1 | |
| \(F(1.24) = \) awrt 0.501 | A1 | |
| 0.5 lies between therefore median value lies between 1.23 and 1.24 | A1 | (3) |
| \(f(x) = \begin{cases}\frac{9x}{10} - \frac{3x^2}{10} & 0 \leq x \leq 2 \\ 0 & \text{otherwise}\end{cases}\) | M1A1, B1 | (3) |
| \(\frac{18}{20} - \frac{12x}{20} = 0\) or completing square so: \(\frac{3}{10}\left[\frac{9}{4} - \left(x - \frac{3}{2}\right)^2\right]\) | M1 | |
| \(x = 1.5\) | A1 | (2) |
| Median < mode, negative skew | M1, A1 | (2) |
| [10] |
| Answer | Marks |
|---|---|
| Item | Guidance |
| (a) | M1: attempt at both \(F(1.23)\) and \(F(1.24)\) and at least one correct or \(\frac{x^2}{20}(9 - 2x) = 0.5\) |
| 1st A1: both awrt 0.495 and awrt 0.501 or 1.238 | |
| 2nd A1: correct comment about the value of the median (not just \(0.495 < F(m) < 0.501\)) | |
| (b) | M1: attempting to differentiate. Multiply out and at least one term \(y^n \to y^{n-1}\) |
| A1: correct differentiation. Allow \(\frac{18x}{20} - \frac{6x^2}{10}\) or \(\frac{3}{10}x(3-x)\) or any exact equivalent. | |
| B1: correct pdf, including 0 otherwise and \(0 \leq x \leq 2\) | |
| (c) | M1: for an attempt to differentiate pdf and put = 0 or complete the square or a sketch. Sketch should have the correct shape and show some positive values on \(x\) – axis. |
| Answer only scores M1A1 | |
| A1: at completing the square should get to \(p \pm q(x - 1.5)^2\) | |
| (d) | M1: reason must match their values/sketch (NB mean = 1.2). Their values must be in [0, 2]. No mode or median will score M0 unless their reason is based on their sketch |
| A1: no ft correct answer only | |
| e.g. If their mode = 1 and they say "mode < median" score M1 for a correct reason but A0 even if they say "positive skew" since there is no ft and "negative skew" would follow incorrect working. |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(1.23) = $ awrt 0.495 | M1 | |
| $F(1.24) = $ awrt 0.501 | A1 | |
| 0.5 lies between therefore median value lies between 1.23 and 1.24 | A1 | (3) |
| $f(x) = \begin{cases}\frac{9x}{10} - \frac{3x^2}{10} & 0 \leq x \leq 2 \\ 0 & \text{otherwise}\end{cases}$ | M1A1, B1 | (3) |
| $\frac{18}{20} - \frac{12x}{20} = 0$ or completing square so: $\frac{3}{10}\left[\frac{9}{4} - \left(x - \frac{3}{2}\right)^2\right]$ | M1 | |
| $x = 1.5$ | A1 | (2) |
| Median < mode, negative skew | M1, A1 | (2) |
| | [10] | |
**Notes:**
| Item | Guidance |
|---|---|
| (a) | M1: attempt at both $F(1.23)$ and $F(1.24)$ and at least one correct or $\frac{x^2}{20}(9 - 2x) = 0.5$ |
| | 1st A1: both awrt 0.495 and awrt 0.501 or 1.238 |
| | 2nd A1: correct comment about the value of the **median** (not just $0.495 < F(m) < 0.501$) |
| (b) | M1: attempting to differentiate. Multiply out and at least one term $y^n \to y^{n-1}$ |
| | A1: correct differentiation. Allow $\frac{18x}{20} - \frac{6x^2}{10}$ or $\frac{3}{10}x(3-x)$ or any exact equivalent. |
| | B1: correct pdf, including 0 otherwise and $0 \leq x \leq 2$ |
| (c) | M1: for an attempt to differentiate pdf and put = 0 or complete the square or a sketch. Sketch should have the correct shape and show some positive values on $x$ – axis. |
| | | Answer only scores M1A1 |
| | A1: at completing the square should get to $p \pm q(x - 1.5)^2$ |
| (d) | M1: reason must match their values/sketch (NB mean = 1.2). Their values must be in [0, 2]. No mode or median will score M0 unless their reason is based on their sketch |
| | A1: no ft correct answer only |
| | e.g. If their mode = 1 and they say "mode < median" score M1 for a correct reason but A0 even if they say "positive skew" since there is no ft and "negative skew" would follow incorrect working. |
---6. A continuous random variable $X$ has cumulative distribution function $\mathrm { F } ( x )$ given by
$$F ( x ) = \left\{ \begin{array} { l c }
0 & x < 0 \\
\frac { x ^ { 2 } } { 20 } ( 9 - 2 x ) & 0 \leqslant x \leqslant 2 \\
1 & x > 2
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Verify that the median of $X$ lies between 1.23 and 1.24
\item Specify fully the probability density function $\mathrm { f } ( x )$.
\item Find the mode of $X$.
\item Describe the skewness of this distribution. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q6 [10]}}