| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard procedures: part (a) requires standardizing two values and reading from tables, while part (b) involves inverse normal lookup. Both are routine S1 exercises with no conceptual challenges or multi-step reasoning beyond basic z-score calculations. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(56 < X < 66) = P\left(\frac{56-62}{5} < z < \frac{66-62}{5}\right) = P(-1.2 < z < 0.8)\) | M1 | Using \(\pm\) standardisation formula at least once, no \(\sqrt{\sigma}\) or \(\sigma^2\), allow continuity correction |
| \(\Phi(0.8) + \Phi(1.2) - 1 = 0.7881 + 0.8849 - 1\) | M1 | Appropriate area \(\Phi\), from standardisation formula in final solution |
| \(0.673\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 1.127\) | B1 | \(\pm(1.126 - 1.127)\) seen, 4 sf or more |
| \(\frac{60t - 62}{5} = 1.127\), \(\quad 60t = 5.635 + 62 = 67.635\) | M1 | z-value \(= \pm\frac{(60t-62)}{5}\), condone z-value \(= \pm\frac{(t-62)}{5}\); no continuity correction, condone \(\sqrt{\sigma}\) or \(\sigma^2\) |
| \(t = 1.13\) | A1 | CAO |
| 3 |
**Question 1:**
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(56 < X < 66) = P\left(\frac{56-62}{5} < z < \frac{66-62}{5}\right) = P(-1.2 < z < 0.8)$ | M1 | Using $\pm$ standardisation formula at least once, no $\sqrt{\sigma}$ or $\sigma^2$, allow continuity correction |
| $\Phi(0.8) + \Phi(1.2) - 1 = 0.7881 + 0.8849 - 1$ | M1 | Appropriate area $\Phi$, from standardisation formula in final solution |
| $0.673$ | A1 | |
| | **3** | |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 1.127$ | B1 | $\pm(1.126 - 1.127)$ seen, 4 sf or more |
| $\frac{60t - 62}{5} = 1.127$, $\quad 60t = 5.635 + 62 = 67.635$ | M1 | z-value $= \pm\frac{(60t-62)}{5}$, condone z-value $= \pm\frac{(t-62)}{5}$; no continuity correction, condone $\sqrt{\sigma}$ or $\sigma^2$ |
| $t = 1.13$ | A1 | CAO |
| | **3** | |1 The times taken to swim 100 metres by members of a large swimming club have a normal distribution with mean 62 seconds and standard deviation 5 seconds.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen member of the club takes between 56 and 66 seconds to swim 100 metres.
\item $13 \%$ of the members of the club take more than $t$ minutes to swim 100 metres. Find the value of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q1 [6]}}