| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Standard +0.3 Part (a) is a routine normal distribution calculation requiring standardization and table lookup. Part (b) requires working backwards from a probability to find σ using inverse normal methods, which is slightly less routine but still a standard S1 technique with clear structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X < 18.2) = P(Z <] \frac{18.2 - 19.8}{2.4})\) | M1 | Use of \(\pm\) standardisation formula with 18.2, 19.8 and 2.4 substituted appropriately, no continuity correction. Condone \(\sigma^2(2.4^2)\) or \(\sqrt{\sigma}(\sqrt{2.4})\) |
| \(= [\Phi(-0.6667) = 1 - \Phi(0.6667)]\) \(= 1 - 0.7477\) | M1 | Calculating appropriate probability areas leading to final answer, expect \(< 0.5\). Note: 0.432 is z-value of 0.667 so not an appropriate probability area (M0) |
| \(= 0.252(3)\) | A1 | AWRT 0.252 SOI, accept 0.2525. If one or both M marks not awarded, SC B1 for AWRT 0.252 SOI |
| Expected number \(= 0.2523 \times 450 = 113.5\), \(= 113\) or \(114\) | B1FT | Strict FT *their* at least 4 figure probability \(\times 450\). One integer answer only. No indication of approximation e.g. \(\approx, \cong,\) *about* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X > 25.5) = 0.26\) so \(P(Z > \frac{25.5 - 23.4}{\sigma}) = 0.26]\) | B1 | \(0.643 \leqslant z \leqslant 0.6435\) or \(-0.6435 \leqslant z \leqslant -0.643\) seen |
| \(\frac{25.5 - 23.4}{\sigma} = 0.643\) | M1 | \(\pm\) standardisation formula with 25.5, 23.4, \(\sigma\) equating to a z-value (not \(1 -\) *their* z-value). Condone continuity correction \(\pm 0.05\), not \(\sigma^2\), not \(\sqrt{\sigma}\) |
| \(\sigma = 3.27\) | A1 | \(3.26 \leqslant \sigma \leqslant 3.27\). Do not award for improper fractions |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X < 18.2) = P(Z <] \frac{18.2 - 19.8}{2.4})$ | M1 | Use of $\pm$ standardisation formula with 18.2, 19.8 and 2.4 substituted appropriately, no continuity correction. Condone $\sigma^2(2.4^2)$ or $\sqrt{\sigma}(\sqrt{2.4})$ |
| $= [\Phi(-0.6667) = 1 - \Phi(0.6667)]$ $= 1 - 0.7477$ | M1 | Calculating appropriate probability areas leading to final answer, expect $< 0.5$. Note: 0.432 is z-value of 0.667 so not an appropriate probability area (M0) |
| $= 0.252(3)$ | A1 | AWRT 0.252 SOI, accept 0.2525. If one or both M marks not awarded, **SC B1** for AWRT 0.252 SOI |
| Expected number $= 0.2523 \times 450 = 113.5$, $= 113$ or $114$ | B1FT | Strict FT *their* at least 4 figure probability $\times 450$. One integer answer only. No indication of approximation e.g. $\approx, \cong,$ *about* |
**Total: 4 marks**
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X > 25.5) = 0.26$ so $P(Z > \frac{25.5 - 23.4}{\sigma}) = 0.26]$ | B1 | $0.643 \leqslant z \leqslant 0.6435$ or $-0.6435 \leqslant z \leqslant -0.643$ seen |
| $\frac{25.5 - 23.4}{\sigma} = 0.643$ | M1 | $\pm$ standardisation formula with 25.5, 23.4, $\sigma$ equating to a z-value (not $1 -$ *their* z-value). Condone continuity correction $\pm 0.05$, not $\sigma^2$, not $\sqrt{\sigma}$ |
| $\sigma = 3.27$ | A1 | $3.26 \leqslant \sigma \leqslant 3.27$. Do not award for improper fractions |
**Total: 3 marks**
---4 The heights, in metres, of white pine trees are normally distributed with mean 19.8 and standard deviation 2.4 .
In a certain forest there are 450 white pine trees.
\begin{enumerate}[label=(\alph*)]
\item How many of these trees would you expect to have height less than 18.2 metres?\\
The heights, in metres, of red pine trees are normally distributed with mean 23.4 and standard deviation $\sigma$. It is known that $26 \%$ of red pine trees have height greater than 25.5 metres.
\item Find the value of $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q4 [7]}}