| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find standard deviation from probability |
| Difficulty | Standard +0.3 This is a slightly above-average S2 question requiring standardization and inverse normal lookup for part (i)(a), then symmetry/z-score manipulation for (i)(b), plus a conceptual comment about model validity in (ii). The techniques are standard for S2 with no novel insight required, but the multi-part structure and the less routine part (i)(b) elevate it slightly above typical drill exercises. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) \(\frac{15.0 - 20.0}{\sigma} = -0.253\) | M1 | For standardising and equating to \(\Phi^{-1}(p)\) |
| B1 | For correct value 0.253 (or 0.254) seen | |
| Hence \(\sigma = \frac{5}{0.253} = 19.8\) | M1 | For solving equation for \(\sigma\) |
| A1 | For correct value 19.8 | |
| (i)(b) \(g = 25.0\), using symmetry | B1 | For stating (or finding) the value of \(g\) |
| Hence \(\text{P}(G > 2g) = 1 - \Phi\left(\frac{50.0 - 20.0}{19.8}\right)\) | M1 | For correct process for upper tail prob |
| \(= 1 - 0.935 = 0.065\) | A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Hence the assumption seems unjustified | M1 | For considering relevant normal probability |
| A1 | For stating the appropriate conclusion |
(i)(a) $\frac{15.0 - 20.0}{\sigma} = -0.253$ | M1 | For standardising and equating to $\Phi^{-1}(p)$
| B1 | For correct value 0.253 (or 0.254) seen
Hence $\sigma = \frac{5}{0.253} = 19.8$ | M1 | For solving equation for $\sigma$
| A1 | For correct value 19.8
(i)(b) $g = 25.0$, using symmetry | B1 | For stating (or finding) the value of $g$
Hence $\text{P}(G > 2g) = 1 - \Phi\left(\frac{50.0 - 20.0}{19.8}\right)$ | M1 | For correct process for upper tail prob
$= 1 - 0.935 = 0.065$ | A1 | For correct answer
(ii) If normal, $\text{P}(G < 0)$ is substantial
Hence the assumption seems unjustified | M1 | For considering relevant normal probability
| A1 | For stating the appropriate conclusion
**Total: 9 marks**
---4 The random variable $G$ has mean 20.0 and standard deviation $\sigma$. It is given that $\mathrm { P } ( G > 15.0 ) = 0.6$. Assume that $G$ is normally distributed.
\begin{enumerate}[label=(\roman*)]
\item (a) Find the value of $\sigma$.\\
(b) Given that $\mathrm { P } ( G > g ) = 0.4$, find the value of $\mathrm { P } ( G > 2 g )$.
\item It is known that no values of $G$ are ever negative. State with a reason what this tells you about the assumption that $G$ is normally distributed.
\end{enumerate}
\hfill \mbox{\textit{OCR S2 Q4 [9]}}