| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing standard procedures: differentiating a piecewise CDF to get PDF, using symmetry to find E(X), applying variance formula with routine integration, and using E(X²) from variance. All parts follow textbook methods with no problem-solving or novel insight required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = \begin{cases} \frac{1}{4k} & -k \leqslant x \leqslant 3k \\ 0 & \text{otherwise} \end{cases}\) | M1 A1 | M1: For 1st line of pdf including inequality, allow \(<\) instead of \(\leqslant\). A1: Fully correct, allow \(<\) instead of \(\leqslant\), allow equivalent for 0 otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([E(X)] = k\) | B1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{Var}(X)] = \frac{(3k-(-k))^2}{12} = \frac{16k^2}{12}\) or \(\left[\frac{x^3}{3}\text{"f}(x)\text{"}\right]_{-k}^{3k} - (\text{"k"})^2\) | M1 | Use of \(\text{Var}(X) = \frac{(\beta-\alpha)^2}{12}\) or integration method |
| \(= \frac{4k^2}{3}\)* | A1* cso | Answer given. Correct solution only with no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X^2) = \text{Var}(X) + E(X)^2 = \frac{4k^2}{3} + (\text{"k"})^2\) | M1 | Use of \(E(X^2) = \text{Var}(X) + E(X)^2\) ft their \(E(X)\); or \(\left[\frac{x^3}{3}\text{"f}(x)\text{"}\right]_{-k}^{3k}\) |
| \(= \frac{7k^2}{3}\) | A1 | \(\frac{7k^2}{3}\) must be seen in part (d), may be implied by \(7k^2\) |
| \(E(3X^2) = 3E(X^2) = 3 \times \frac{7k^2}{3} = 7k^2\) | A1 | Cao |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = \begin{cases} \frac{1}{4k} & -k \leqslant x \leqslant 3k \\ 0 & \text{otherwise} \end{cases}$ | M1 A1 | M1: For 1st line of pdf including inequality, allow $<$ instead of $\leqslant$. A1: Fully correct, allow $<$ instead of $\leqslant$, allow equivalent for 0 otherwise |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[E(X)] = k$ | B1 | Cao |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Var}(X)] = \frac{(3k-(-k))^2}{12} = \frac{16k^2}{12}$ or $\left[\frac{x^3}{3}\text{"f}(x)\text{"}\right]_{-k}^{3k} - (\text{"k"})^2$ | M1 | Use of $\text{Var}(X) = \frac{(\beta-\alpha)^2}{12}$ or integration method |
| $= \frac{4k^2}{3}$* | A1* cso | Answer given. Correct solution only with no incorrect working |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X^2) = \text{Var}(X) + E(X)^2 = \frac{4k^2}{3} + (\text{"k"})^2$ | M1 | Use of $E(X^2) = \text{Var}(X) + E(X)^2$ ft their $E(X)$; or $\left[\frac{x^3}{3}\text{"f}(x)\text{"}\right]_{-k}^{3k}$ |
| $= \frac{7k^2}{3}$ | A1 | $\frac{7k^2}{3}$ must be seen in part (d), may be implied by $7k^2$ |
| $E(3X^2) = 3E(X^2) = 3 \times \frac{7k^2}{3} = 7k^2$ | A1 | Cao |
---2 The continuous random variable $X$ has cumulative distribution function given by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 & x < - k \\
\frac { x + k } { 4 k } & - k \leqslant x \leqslant 3 k \\
1 & x > 3 k
\end{array} \right.$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Specify fully, in terms of $k$, the probability density function of $X$
\item Write down, in terms of $k$, the value of $\mathrm { E } ( X )$
\item Show that $\operatorname { Var } ( X ) = \frac { 4 } { 3 } k ^ { 2 }$
\item Find, in terms of $k$, the value of $\mathrm { E } \left( 3 X ^ { 2 } \right)$\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2022 Q2 [8]}}