| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Skewness from moments |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine CDF/PDF manipulation. Parts (a)-(d) involve direct application of formulas (reading F(x), differentiating for f(x), integrating for E(X)). Part (e) requires finding maximum of f(x) via differentiation. Part (f) compares mean/median/mode for skewness - a textbook exercise. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03d E(g(X)): general expectation formula5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - F(0.3) = 1 - (2 \times 0.3^3 - 0.3^3)\) | 'one minus' required | M1 |
| \(= 0.847\) | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(0.60) = 0.5040\) \(F(0.59) = 0.4908\) | both required awrt 0.5, 0.49 | M1A1 |
| 0.5 lies between therefore median value lies between 0.59 and 0.60. | B1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \begin{cases} -3x^2 + 4x, & 0 \leq x \leq 1, \\ 0, & \text{otherwise.} \end{cases}\) | attempt to differentiate, all correct | M1A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 xf(x)dx = \int_0^1 -3x^3 + 4x^2 dx\) | attempt to integrate xf(x) | M1 |
| \(= \left[\frac{-3x^4}{4} + \frac{4x^3}{3}\right]_0^1\) | sub in limits | M1 |
| \(= \frac{7}{12}\) or 0.583 or 0.583 or equivalent fraction | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{df(x)}{dx} = -6x + 4 = 0\) | attempt to differentiate f(x) and equate to 0 | M1 |
| \(x = \frac{2}{3}\) or 0.6 or 0.667 | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| mean < median < mode, therefore negative skew. | Any pair, cao | B1, B1 (2 marks) |
**Part (a)**
$1 - F(0.3) = 1 - (2 \times 0.3^3 - 0.3^3)$ | 'one minus' required | M1 |
$= 0.847$ | | A1 (2 marks) |
**Part (b)**
$F(0.60) = 0.5040$ $F(0.59) = 0.4908$ | both required awrt 0.5, 0.49 | M1A1 |
0.5 lies between therefore median value lies between 0.59 and 0.60. | | B1 (3 marks) |
**Part (c)**
$f(x) = \begin{cases} -3x^2 + 4x, & 0 \leq x \leq 1, \\ 0, & \text{otherwise.} \end{cases}$ | attempt to differentiate, all correct | M1A1 (2 marks) |
**Part (d)**
$\int_0^1 xf(x)dx = \int_0^1 -3x^3 + 4x^2 dx$ | attempt to integrate xf(x) | M1 |
$= \left[\frac{-3x^4}{4} + \frac{4x^3}{3}\right]_0^1$ | sub in limits | M1 |
$= \frac{7}{12}$ or 0.583 or 0.583 or equivalent fraction | | A1 (3 marks) |
**Part (e)**
$\frac{df(x)}{dx} = -6x + 4 = 0$ | attempt to differentiate f(x) and equate to 0 | M1 |
$x = \frac{2}{3}$ or 0.6 or 0.667 | | A1 (2 marks) |
**Part (f)**
mean < median < mode, therefore negative skew. | Any pair, cao | B1, B1 (2 marks) |
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# Total marks: 4 + 5 + 15 + 12 + 12 + 13 + 14 = **75 marks**7. The continuous random variable $X$ has cumulative distribution function
$$\mathrm { F } ( x ) = \begin{cases} 0 , & x < 0 \\ 2 x ^ { 2 } - x ^ { 3 } , & 0 \leqslant x \leqslant 1 \\ 1 , & x > 1 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 0.3 )$.
\item Verify that the median value of $X$ lies between $x = 0.59$ and $x = 0.60$.
\item Find the probability density function $\mathrm { f } ( x )$.
\item Evaluate $\mathrm { E } ( X )$.
\item Find the mode of $X$.
\item Comment on the skewness of $X$. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2007 Q7 [14]}}