| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| 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 straightforward S2 question requiring standard techniques: evaluating F(x), differentiating to find f(x), computing E(X) and Var(X) by integration, finding the mode from f'(x)=0, then applying a given formula. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 0.7) = 1 - F(0.7) = 0.4267\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \frac{d}{dx}F(x) = \frac{4}{3} \times 2x - \frac{4x^2}{3}\) | M1 | |
| \(= \frac{4x}{3}(2 - x^2)\) for \(0 \leq x \leq 1\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int_0^1 \frac{4}{3}(2x^2 - x^4)\, dx = \left[\frac{4}{3}\left(\frac{2x^3}{3} - \frac{x^5}{5}\right)\right]_0^1\) | M1 A1 | |
| \(= \frac{28}{45} = 0.622\) | A1 | |
| \(\text{Var}(X) = \int_0^1 \frac{4}{3}(2x^3 - x^5)\, dx - \left(\frac{28}{45}\right)^2\) | M1 | |
| \(= \left[\frac{4}{3}\left(\frac{2x^4}{4} - \frac{x^6}{6}\right)\right]_0^1 - \left(\frac{28}{45}\right)^2\) | A1 | |
| \(= \frac{116}{2025} = 0.05728\) | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = \frac{4}{3}(2 - 3x^2) = 0\) | M1 | |
| \(\Rightarrow \text{mode} = \sqrt{\frac{2}{3}} = 0.816496\) | A1 | |
| \(\text{skewness} = \dfrac{\frac{28}{45} - \sqrt{\frac{2}{3}}}{\sqrt{\frac{116}{2025}}} = -0.81170\) | M1 A1 | (4 marks) |
# Question 4:
## Part (a)
| $P(X > 0.7) = 1 - F(0.7) = 0.4267$ | M1 A1 | (2 marks) |
## Part (b)
| $f(x) = \frac{d}{dx}F(x) = \frac{4}{3} \times 2x - \frac{4x^2}{3}$ | M1 | |
| $= \frac{4x}{3}(2 - x^2)$ for $0 \leq x \leq 1$ | A1 | (2 marks) |
## Part (c)
| $E(X) = \int_0^1 \frac{4}{3}(2x^2 - x^4)\, dx = \left[\frac{4}{3}\left(\frac{2x^3}{3} - \frac{x^5}{5}\right)\right]_0^1$ | M1 A1 | |
| $= \frac{28}{45} = 0.622$ | A1 | |
| $\text{Var}(X) = \int_0^1 \frac{4}{3}(2x^3 - x^5)\, dx - \left(\frac{28}{45}\right)^2$ | M1 | |
| $= \left[\frac{4}{3}\left(\frac{2x^4}{4} - \frac{x^6}{6}\right)\right]_0^1 - \left(\frac{28}{45}\right)^2$ | A1 | |
| $= \frac{116}{2025} = 0.05728$ | A1 | (6 marks) |
## Part (d)
| $f'(x) = \frac{4}{3}(2 - 3x^2) = 0$ | M1 | |
| $\Rightarrow \text{mode} = \sqrt{\frac{2}{3}} = 0.816496$ | A1 | |
| $\text{skewness} = \dfrac{\frac{28}{45} - \sqrt{\frac{2}{3}}}{\sqrt{\frac{116}{2025}}} = -0.81170$ | M1 A1 | (4 marks) |
---4. The continuous random variable $X$ has cumulative distribution function
$$\mathrm { F } ( x ) = \begin{cases} 0 , & x < 0 \\ \frac { 1 } { 3 } x ^ { 2 } \left( 4 - x ^ { 2 } \right) , & 0 \leq x \leq 1 \\ 1 & x > 1 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 0.7 )$.
\item Find the probability density function $\mathrm { f } ( x )$ of $X$.
\item Calculate $\mathrm { E } ( X )$ and show that, to 3 decimal places, $\operatorname { Var } ( X ) = 0.057$.
One measure of skewness is
$$\frac { \text { Mean - Mode } } { \text { Standard deviation } } .$$
\item Evaluate the skewness of the distribution of $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2003 Q4 [14]}}