| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | March |
| Marks | 7 |
| 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 problem requiring inverse normal calculation for part (a) and a standard probability calculation for part (b). Both parts use routine techniques taught in S1 with no conceptual challenges—slightly easier than average due to the direct application of standard methods without multi-step reasoning or novel insight. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X > 87) = P\left(Z > \frac{87-82}{\sigma}\right) = 0.22\) | M1 | Using \(\pm\) standardisation formula, not \(\sigma^2\), not \(\sqrt{\sigma}\), no continuity correction |
| \(P\left(Z < \frac{5}{\sigma}\right) = 0.78\); \(\left(\frac{5}{\sigma} =\right) 0.772\) | B1 | AWRT \(\pm 0.772\) seen; B0 for \(\pm 0.228\) |
| \(\sigma = 6.48\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\left(-\frac{4}{\sigma} < Z < \frac{4}{\sigma}\right) = P(-0.6176 < Z < 0.6176)\) | M1 | Using \(\pm 4\) within a standardisation formula (SOI), allow \(\sigma^2\), \(\sqrt{\sigma}\) and continuity correction |
| M1 | Standardisation formula applied to both *their* \(\pm 4\) | |
| \(\Phi = 0.7317\); \(\text{Prob} = 2\Phi - 1 = 2(0.7317) - 1\) | M1 | Correct area \(2\Phi - 1\) oe linked to final solution |
| \(= 0.463\) | A1 | |
| Total: 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 87) = P\left(Z > \frac{87-82}{\sigma}\right) = 0.22$ | M1 | Using $\pm$ standardisation formula, not $\sigma^2$, not $\sqrt{\sigma}$, no continuity correction |
| $P\left(Z < \frac{5}{\sigma}\right) = 0.78$; $\left(\frac{5}{\sigma} =\right) 0.772$ | B1 | AWRT $\pm 0.772$ seen; B0 for $\pm 0.228$ |
| $\sigma = 6.48$ | A1 | |
| **Total: 3** | | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\left(-\frac{4}{\sigma} < Z < \frac{4}{\sigma}\right) = P(-0.6176 < Z < 0.6176)$ | M1 | Using $\pm 4$ within a standardisation formula (SOI), allow $\sigma^2$, $\sqrt{\sigma}$ and continuity correction |
| | M1 | Standardisation formula applied to **both** *their* $\pm 4$ |
| $\Phi = 0.7317$; $\text{Prob} = 2\Phi - 1 = 2(0.7317) - 1$ | M1 | Correct area $2\Phi - 1$ oe linked to final solution |
| $= 0.463$ | A1 | |
| **Total: 4** | | |3 The weights of apples of a certain variety are normally distributed with mean 82 grams. $22 \%$ of these apples have a weight greater than 87 grams.
\begin{enumerate}[label=(\alph*)]
\item Find the standard deviation of the weights of these apples.
\item Find the probability that the weight of a randomly chosen apple of this variety differs from the mean weight by less than 4 grams.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q3 [7]}}