| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Cumulative distribution function |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic understanding of continuous uniform distributions. Part (a) requires simple differentiation of the CDF (1 mark), and part (b) involves routine calculations: using the CDF directly, recognizing P(X≠7)=1 for continuous variables, and applying standard formulas for expectation. All techniques are direct applications with no problem-solving insight required, making it 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 |
|---|---|---|
| for \(-5 \leq x \leq 15\): \(f(x) = \frac{d}{dx}F(x) = \frac{d}{dx}\left(\frac{x+5}{20}\right) = \frac{1}{20}\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(X \geq 7) = 1 - F(7) = 1 - \frac{12}{20} = \frac{2}{5}\) or \(\left[\frac{8}{20}, \frac{4}{10}, 0.4\right]\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbb{P}(X \neq 7) = 1\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{1}{2}(-5+15) = 5\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(3X^2) = \int_{-5}^{15} \frac{3x^2}{20}dx\) (ignore limits) | M1 | |
| \(\left[\frac{x^3}{20}\right]_{-5}^{15}\) | ||
| \(\frac{1}{20}(3375+125)\) | A1 | correct limits seen / used |
| \(168\frac{3}{4} + 6\frac{1}{4}\) | ||
| \(= 175\) | A1 | 3 |
| Alternative: \(\text{Var}(X) = \frac{1}{12}(15--5)^2 = \frac{400}{12}\) (oe) | (B1) | |
| \(E(3X^2) = 3 \times \left[\frac{400}{12} + 5^2\right]\) | (M1) | \(E(3X^2) = 3E(X^2) = 3 \times \{\text{their Var}(X) > 0\} + \{\text{their } E(X)\}^2\}\) *used* |
| \(= 175\) | (A1) | |
| 7 |
### 3(a)
| for $-5 \leq x \leq 15$: $f(x) = \frac{d}{dx}F(x) = \frac{d}{dx}\left(\frac{x+5}{20}\right) = \frac{1}{20}$ | B1 | 1 | AG |
### 3(b)(i)
| $\mathbb{P}(X \geq 7) = 1 - F(7) = 1 - \frac{12}{20} = \frac{2}{5}$ or $\left[\frac{8}{20}, \frac{4}{10}, 0.4\right]$ | B1 | 1 | **Alternative:** Use of $f(x) = \frac{1}{20}$ or graph $\Rightarrow \mathbb{P}(X \geq 7) = \frac{1}{20} \times (15-7) = \frac{2}{5}$ (oe) |
### 3(b)(ii)
| $\mathbb{P}(X \neq 7) = 1$ | B1 | 1 | cao |
### 3(b)(iii)
| $E(X) = \frac{1}{2}(-5+15) = 5$ | B1 | 1 | **Alternative:** $E(X) = \int_{-5}^{15} \frac{x}{20}dx = \left[\frac{x^2}{40}\right]_{-5}^{15} = \frac{1}{40}(225-25) = \frac{1}{40} \times 200 = 5$ **B1 (cao)** |
### 3(b)(iv)
| $E(3X^2) = \int_{-5}^{15} \frac{3x^2}{20}dx$ (ignore limits) | M1 | |
| --- | --- | --- |
| $\left[\frac{x^3}{20}\right]_{-5}^{15}$ | | |
| $\frac{1}{20}(3375+125)$ | A1 | correct limits seen / used |
| $168\frac{3}{4} + 6\frac{1}{4}$ | | |
| $= 175$ | A1 | 3 | (cao) (allow 174.9) |
| **Alternative:** $\text{Var}(X) = \frac{1}{12}(15--5)^2 = \frac{400}{12}$ (oe) | (B1) | |
| $E(3X^2) = 3 \times \left[\frac{400}{12} + 5^2\right]$ | (M1) | $E(3X^2) = 3E(X^2) = 3 \times \{\text{their Var}(X) > 0\} + \{\text{their } E(X)\}^2\}$ *used* |
| $= 175$ | (A1) | |
| | | **7** | |3 The continuous random variable $X$ has a cumulative distribution function defined by
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c l }
0 & x < - 5 \\
\frac { x + 5 } { 20 } & - 5 \leqslant x \leqslant 15 \\
1 & x > 15
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $- 5 \leqslant x \leqslant 15$, the probability density function, $\mathrm { f } ( x )$, of $X$ is given by $\mathrm { f } ( x ) = \frac { 1 } { 20 }$.\\
(1 mark)
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X \geqslant 7 )$;
\item $\mathrm { P } ( X \neq 7 )$;
\item $\mathrm { E } ( X )$;
\item $\mathrm { E } \left( 3 X ^ { 2 } \right)$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S2 2012 Q3 [7]}}