| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This question involves standard normal distribution calculations: part (a) requires converting to z-scores and using tables for a straightforward probability between two values, while part (b) involves working backwards from a percentage to find a standard deviation using the inverse normal. Both are routine S1 techniques with no conceptual challenges, making this slightly easier than average but not trivial due to the two-part 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 |
| \(P(23 < X < 35) = P\!\left(\frac{23-28}{3.3} < Z < \frac{35-28}{3.3}\right)\) \([= P(-1.515 < Z < 2.121)]\) | M1 | Using \(\pm\) standardisation formula once with 23 or 35, 28 and 3.3; allow \(\sigma^2\), allow \(\sqrt{\sigma}\), no continuity correction |
| A1 | One fully correct \(\pm\) standardisation formula | |
| \(= \Phi(2.121) + \Phi(1.5151) - 1\) \(= 0.9830 + 0.9351 - 1\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability |
| \(= 0.918\) | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X>7.6) = P\!\left(Z > \frac{7.6-8.5}{\sigma}\right) = 0.75\) | B1 | \(0.674\) or \(-0.674\) seen; CAO as critical value |
| \(\frac{7.6-8.5}{\sigma} = -0.674\) | M1 | Use of \(\pm\) standardisation formula with 7.6, 8.5, \(\sigma\) and a \(z\)-value (not 0.75, 0.25, 0.7734, 0.2266, 0.5987 nor \(1-z\)-value: 0.326, 0.5987); condone use of \(\frac{\pm 0.9}{\sigma}\) |
| \(\sigma = 1.34\) | A1 | \(1.33 \leqslant \sigma \leqslant 1.34\) |
## Question 2:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(23 < X < 35) = P\!\left(\frac{23-28}{3.3} < Z < \frac{35-28}{3.3}\right)$ $[= P(-1.515 < Z < 2.121)]$ | M1 | Using $\pm$ standardisation formula once with 23 or 35, 28 and 3.3; allow $\sigma^2$, allow $\sqrt{\sigma}$, no continuity correction |
| | A1 | One fully correct $\pm$ standardisation formula |
| $= \Phi(2.121) + \Phi(1.5151) - 1$ $= 0.9830 + 0.9351 - 1$ | M1 | Appropriate area $\Phi$, from final process, must be a probability |
| $= 0.918$ | A1 | AWRT |
**Total: 4 marks**
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X>7.6) = P\!\left(Z > \frac{7.6-8.5}{\sigma}\right) = 0.75$ | B1 | $0.674$ or $-0.674$ seen; CAO as critical value |
| $\frac{7.6-8.5}{\sigma} = -0.674$ | M1 | Use of $\pm$ standardisation formula with 7.6, 8.5, $\sigma$ and a $z$-value (not 0.75, 0.25, 0.7734, 0.2266, 0.5987 nor $1-z$-value: 0.326, 0.5987); condone use of $\frac{\pm 0.9}{\sigma}$ |
| $\sigma = 1.34$ | A1 | $1.33 \leqslant \sigma \leqslant 1.34$ |
**Total: 3 marks**2 The lengths of the tails of adult raccoons of a certain species are normally distributed with mean 28 cm and standard deviation 3.3 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen adult raccoon of this species has a tail length between 23 cm and 35 cm .\\
The masses of adult raccoons of this species are normally distributed with mean 8.5 kg and standard deviation $\sigma \mathrm { kg } .75 \%$ of adult raccoons of this species have mass greater than 7.6 kg .
\item Find the value of $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q2 [7]}}