| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a straightforward normal distribution question requiring standard z-score calculations and inverse normal lookup. Part (a) involves routine standardization and table reading, while part (b) requires working backwards from a probability to find a parameter—both are standard S1 techniques with no conceptual challenges or multi-step reasoning beyond basic manipulation. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(1.42 < X < 1.52) =] P\left(\frac{1.42-1.5}{0.05} < Z < \frac{1.52-1.5}{0.05}\right)\) | M1 | Use of \(\pm\) standardisation formula once with 1.5, 0.05 and either 1.42 or 1.52, allow \(\sigma^2\) or \(\sqrt{\sigma}\), no continuity correction |
| \([= P(-1.6 < Z < 0.4) = \Phi(0.4) + \Phi(1.6) - 1]\) \(= 0.6554 + 0.9452 - 1\) or \(0.6554 - 0.0548\) | M1 | Calculating the appropriate probability area (leading to their final answer, expect \(> 0.5\)) |
| \(= 0.601\) | A1 | \(0.6005 \leq p \leq 0.601\); SC B1 for 0.601 with no standardisation seen |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[P(X < 0.9) = P\left(Z < \frac{0.9-0.75}{\sigma}\right) = 0.68\right]\) | B1 | \(0.467 < z \leq 0.468\) or \(-0.468 \leq z < -0.467\) seen |
| \(\frac{0.9-0.75}{\sigma} = 0.468\) | M1 | \(\pm\) standardisation formula with 0.9, 0.75, \(\sigma\) equating to a \(z\)-value (not 0.32, 0.68, 0.532, 0.7517, 0.2483, 0.6255). Condone continuity correct \(\pm 0.05\), not \(\sigma^2, \sqrt{\sigma}\). Condone \(\pm\frac{0.15}{\sigma} = 0.468\) |
| \(\sigma = 0.321, \frac{25}{78}\) | A1 | \(0.3205 \leq \sigma < 0.3215\); SC B1 if M0 www |
| 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(1.42 < X < 1.52) =] P\left(\frac{1.42-1.5}{0.05} < Z < \frac{1.52-1.5}{0.05}\right)$ | M1 | Use of $\pm$ standardisation formula once with 1.5, 0.05 and either 1.42 or 1.52, allow $\sigma^2$ or $\sqrt{\sigma}$, no continuity correction |
| $[= P(-1.6 < Z < 0.4) = \Phi(0.4) + \Phi(1.6) - 1]$ $= 0.6554 + 0.9452 - 1$ or $0.6554 - 0.0548$ | M1 | Calculating the appropriate probability area (leading to their final answer, expect $> 0.5$) |
| $= 0.601$ | A1 | $0.6005 \leq p \leq 0.601$; **SC B1** for 0.601 with no standardisation seen |
| | **3** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[P(X < 0.9) = P\left(Z < \frac{0.9-0.75}{\sigma}\right) = 0.68\right]$ | B1 | $0.467 < z \leq 0.468$ or $-0.468 \leq z < -0.467$ seen |
| $\frac{0.9-0.75}{\sigma} = 0.468$ | M1 | $\pm$ standardisation formula with 0.9, 0.75, $\sigma$ equating to a $z$-value (not 0.32, 0.68, 0.532, 0.7517, 0.2483, 0.6255). Condone continuity correct $\pm 0.05$, not $\sigma^2, \sqrt{\sigma}$. Condone $\pm\frac{0.15}{\sigma} = 0.468$ |
| $\sigma = 0.321, \frac{25}{78}$ | A1 | $0.3205 \leq \sigma < 0.3215$; **SC B1** if M0 www |
| | **3** | |2 The weights of large bags of pasta produced by a company are normally distributed with mean 1.5 kg and standard deviation 0.05 kg .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen large bag of pasta weighs between 1.42 kg and 1.52 kg .\\
The weights of small bags of pasta produced by the company are normally distributed with mean 0.75 kg and standard deviation $\sigma \mathrm { kg }$. It is found that $68 \%$ of these small bags have weight less than 0.9 kg .
\item Find the value of $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q2 [6]}}