| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate and compare mean, median, mode |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question covering routine techniques: finding k by integration, calculating E(X), finding mode by differentiation, median by solving an integral equation, and interpreting skewness. All parts follow textbook methods with straightforward polynomial integration and no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^2 k(4x-x^3)dx = 1\) | M1, A1 | \(\int f(x)dx = 1\), all correct |
| \(k\left[2x^2 - \frac{1}{4}x^4\right]_0^2 = 1\) | A1 | [*] |
| \(k(8-4) = 1\) | A1 | cso |
| \(k = \frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int_0^2 x \cdot \frac{1}{4}(4x-x^3)dx\) | M1 | \(\int xf(x)dx\) |
| \(= \left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]_0^2\) | A1 | [*] |
| \(= \frac{16}{15}\) | A1 | \(1.07\) or \(1\frac{1}{15}\) or \(\frac{16}{15}\) or \(1.06\) |
| Answer | Marks | Guidance |
|---|---|---|
| At mode, \(f'(x) = 0\) | M1 | Implied |
| \(4 - 3x^2 = 0\) | M1 | Attempt to differentiate |
| \(x = \frac{2}{\sqrt{3}}\) | A1 | \(\sqrt{\frac{4}{3}}\) or \(1.15\) or \(\frac{2}{\sqrt{3}}\) or \(\frac{2\sqrt{3}}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| At median, \(\int_0^m \frac{1}{4}(4t-t^3)dt = \frac{1}{2}\) | M1 | \(F(x) = \frac{1}{2}\) or \(\int f(x)dx = \frac{1}{2}\) |
| \(\frac{1}{4}\left(2x^2 - \frac{1}{4}x^4\right) = \frac{1}{2}\) | M1 | Attempt to integrate |
| \(x^4 - 8x^2 + 8 = 0\) | M1 | Attempt to solve quadratic |
| \(x^2 = 4 \pm 2\sqrt{2}\); \(x = 1.08\) | A1 | awrt 1.08 |
| Answer | Marks | Guidance |
|---|---|---|
| mean \((1.07) <\) median \((1.08) <\) mode \((1.15)\) \(\Rightarrow\) negative skew | M1 | any pair cao |
| A1 |
| Answer | Marks |
|---|---|
| Lines \(x < 0\) and \(x > 2\), labels, 0 and 2 | B1 |
| negative skew between 0 and 2 | B1 |
**6(a)**
| $\int_0^2 k(4x-x^3)dx = 1$ | M1, A1 | $\int f(x)dx = 1$, all correct |
| $k\left[2x^2 - \frac{1}{4}x^4\right]_0^2 = 1$ | A1 | [*] |
| $k(8-4) = 1$ | A1 | cso |
| $k = \frac{1}{4}$ | | |
**6(b)**
| $E(X) = \int_0^2 x \cdot \frac{1}{4}(4x-x^3)dx$ | M1 | $\int xf(x)dx$ |
| $= \left[\frac{1}{3}x^3 - \frac{1}{20}x^5\right]_0^2$ | A1 | [*] |
| $= \frac{16}{15}$ | A1 | $1.07$ or $1\frac{1}{15}$ or $\frac{16}{15}$ or $1.06$ |
**6(c)**
| At mode, $f'(x) = 0$ | M1 | Implied |
| $4 - 3x^2 = 0$ | M1 | Attempt to differentiate |
| $x = \frac{2}{\sqrt{3}}$ | A1 | $\sqrt{\frac{4}{3}}$ or $1.15$ or $\frac{2}{\sqrt{3}}$ or $\frac{2\sqrt{3}}{3}$ |
**6(d)**
| At median, $\int_0^m \frac{1}{4}(4t-t^3)dt = \frac{1}{2}$ | M1 | $F(x) = \frac{1}{2}$ or $\int f(x)dx = \frac{1}{2}$ |
| $\frac{1}{4}\left(2x^2 - \frac{1}{4}x^4\right) = \frac{1}{2}$ | M1 | Attempt to integrate |
| $x^4 - 8x^2 + 8 = 0$ | M1 | Attempt to solve quadratic |
| $x^2 = 4 \pm 2\sqrt{2}$; $x = 1.08$ | A1 | awrt 1.08 |
**6(e)**
| mean $(1.07) <$ median $(1.08) <$ mode $(1.15)$ $\Rightarrow$ negative skew | M1 | any pair cao |
| | A1 | |
**6(f)**
| Lines $x < 0$ and $x > 2$, labels, 0 and 2 | B1 | |
| negative skew between 0 and 2 | B1 | |
---6. A continuous random variable $X$ has probability density function $\mathrm { f } ( x )$ where
$$f ( x ) = \begin{cases} k \left( 4 x - x ^ { 3 } \right) , & 0 \leqslant x \leqslant 2 \\ 0 , & \text { otherwise } \end{cases}$$
where $k$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 4 }$.
Find
\item $\mathrm { E } ( X )$,
\item the mode of $X$,
\item the median of $X$.
\item Comment on the skewness of the distribution.
\item Sketch f(x).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2005 Q6 [18]}}