| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF to PDF derivation |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard CDF-to-PDF conversion (differentiation), basic probability calculation, and expectation/variance formulas. The linear CDF makes all calculations routine with no conceptual challenges, placing it slightly below average difficulty for A-level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = \begin{cases} \frac{1}{9} & -4 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases}\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Horizontal line from \(-4\) to \(5\) | B1 | horizontal line from \(-4\) to \(5\) |
| Line drawn at \(\frac{1}{9}\) | B1 | for drawn at \(\frac{1}{9}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 2) = \frac{1}{9} \times 3\) | M1 | \(F(5) - F(2) = 1 - \frac{2}{3} = \frac{1}{3}\) |
| \(= \frac{1}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= \frac{1}{2}\) | B1 | |
| Variance \(= \frac{1}{12} \times 81 = 6.75\) | B1 |
# Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = \begin{cases} \frac{1}{9} & -4 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases}$ | M1, A1 | |
**Total: 2 marks**
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# Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontal line from $-4$ to $5$ | B1 | horizontal line from $-4$ to $5$ |
| Line drawn at $\frac{1}{9}$ | B1 | for drawn at $\frac{1}{9}$ |
**Total: 2 marks**
---
# Question 8(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 2) = \frac{1}{9} \times 3$ | M1 | $F(5) - F(2) = 1 - \frac{2}{3} = \frac{1}{3}$ |
| $= \frac{1}{3}$ | A1 | |
**Total: 2 marks**
---
# Question 8(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= \frac{1}{2}$ | B1 | |
| Variance $= \frac{1}{12} \times 81 = 6.75$ | B1 | |
**Total: 2 marks**8 The continuous random variable $X$ has the cumulative distribution function
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c c }
0 & x \leqslant - 4 \\
\frac { x + 4 } { 9 } & - 4 \leqslant x \leqslant 5 \\
1 & x \geqslant 5
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Determine the probability density function, $\mathrm { f } ( x )$, of $X$.
\item Sketch the graph of f .
\item Determine $\mathrm { P } ( X > 2 )$.
\item Evaluate the mean and variance of $X$.
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2007 Q8 [8]}}