| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 7 |
| 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 normal distribution question requiring standard z-score calculations and inverse normal lookup. Part (a) is direct standardization and table reading, while part (b) requires combining two probability regions but uses routine techniques. Both parts are textbook exercises with no problem-solving insight needed, making this easier than average for A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(X > 170) = P\!\left(Z > \frac{170 - 176}{4.8}\right)]\) | M1 | Using \(\pm\) standardisation formula with 170, 176 and 4.8 substituted appropriately. Condone \(\sigma^2\) and \(\sqrt{\sigma}\). No continuity correction |
| \([\Phi(1.25)] = 0.894\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability. Expect final answer \(> 0.5\) |
| A1 | \(0.894\) or \(0.89435 \leqslant p \leqslant 0.8944\). If A0 scored, SC B1 for 0.894 or \(0.89435 \leqslant p \leqslant 0.8944\), WWW | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(h < 170) = 1 - 0.8944 = 0.1056\) | M1 | \(1 -\) *their* 3(a) seen or implied by 0.7056 or 0.2944 |
| \(\frac{k - 176}{4.8}\left[= \Phi^{-1}(0.1056 + 0.6)\right] = 0.541\) | B1 | \(0.540 < z \leqslant 0.541\) or \(-0.541 \leqslant z < 0.540\) seen |
| M1 | Use of \(\pm\)standardisation formula with \(k\), 176, 4.8 equated to a \(z\)-value (not 1.25, 0.7601, 0.2399, 0.7056, 0.7257, 0.8313, \(0.253 \pm 0.894\), 0.6, 0.4), not \(4.8^2\), not \(\sqrt{4.8}\), no continuity correction | |
| \(k = 178.6\) | A1 | CAO (answer required to 1 dp) |
| 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(X > 170) = P\!\left(Z > \frac{170 - 176}{4.8}\right)]$ | **M1** | Using $\pm$ standardisation formula with 170, 176 and 4.8 substituted appropriately. Condone $\sigma^2$ and $\sqrt{\sigma}$. No continuity correction |
| $[\Phi(1.25)] = 0.894$ | **M1** | Appropriate area $\Phi$, from final process, must be a probability. Expect final answer $> 0.5$ |
| | **A1** | $0.894$ or $0.89435 \leqslant p \leqslant 0.8944$. If A0 scored, **SC B1** for 0.894 or $0.89435 \leqslant p \leqslant 0.8944$, WWW |
| | **3** | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(h < 170) = 1 - 0.8944 = 0.1056$ | **M1** | $1 -$ *their* **3(a)** seen or implied by 0.7056 or 0.2944 |
| $\frac{k - 176}{4.8}\left[= \Phi^{-1}(0.1056 + 0.6)\right] = 0.541$ | **B1** | $0.540 < z \leqslant 0.541$ or $-0.541 \leqslant z < 0.540$ seen |
| | **M1** | Use of $\pm$standardisation formula with $k$, 176, 4.8 equated to a $z$-value (**not** 1.25, 0.7601, 0.2399, 0.7056, 0.7257, 0.8313, $0.253 \pm 0.894$, 0.6, 0.4), not $4.8^2$, not $\sqrt{4.8}$, no continuity correction |
| $k = 178.6$ | **A1** | CAO (answer required to 1 dp) |
| | **4** | |
---3 In Molimba, the heights, in cm , of adult males are normally distributed with mean 176 cm and standard deviation 4.8 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen adult male in Molimba has a height greater than 170 cm .\\
60\% of adult males in Molimba have a height between 170 cm and $k \mathrm {~cm}$, where $k$ is greater than 170 .
\item Find the value of $k$, giving your answer correct to 1 decimal place.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q3 [7]}}