| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 expected frequency (z-score, table lookup, multiply by sample size). Part (b) requires setting up two equations from percentiles and solving simultaneously, which is slightly more demanding than standard one-parameter problems but still a well-practiced technique in S1. Overall slightly easier than average A-level difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\!\left(Z > \frac{20-14.6}{5.2}\right) = P(Z > 1.03846)\) | M1 | Use of \(\pm\) standardisation formula with 20, 14.6 and 5.2 not \(\sigma^2\), not \(\sqrt{\sigma}\), no continuity correction |
| \(1 - 0.8504\) | M1 | Calculating the appropriate probability area (leading to final answer) |
| \(0.150\) | A1 | \(0.1496, 0.149 < p \leq 0.15[0]\). Only dependent on 2nd M mark so M0M1A1 possible. SC B1 for \(0.149 < p \leq 0.15[0]\) if M0M0A0 awarded |
| \([250 \times \textit{their}\ 0.1496 =]\ 37, 38\) | B1 FT | Strict FT *their* at least 4-figure probability seen anywhere (give BOD if they go on to use 0.150). Final answer must be positive integer, no approximation or rounding stated |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z_1 = \frac{14.5 - \mu}{\sigma} = -0.842\) | B1 | \(-0.843 < z_1 < -0.841\) or \(0.841 < z_1 < 0.843\) |
| \(z_2 = \frac{18.5 - \mu}{\sigma} = -0.44\) | B1 | \(-0.441 < z_2 < -0.439\) or \(0.439 < z_2 < 0.441\) |
| Solve, obtaining values for \(\mu\) and \(\sigma\) | M1 | Use of \(\pm\) standardisation formula once with \(\mu\), \(\sigma\) and a \(z\)-value (not 0.20, 0.80, 0.67, 0.23, 0.5793, 0.7881, 0.7486, 0.591 or \(1\text{-}z\) i.e. 0.158 etc.). Condone continuity correction \(\pm 0.05\), not \(\sigma^2\), \(\sqrt{\sigma}\) |
| M1 | Solve using elimination/substitution or other appropriate approach to obtain values for both \(\mu\) and \(\sigma\) | |
| \(\mu = 22.9,\quad \sigma = 9.95\) | A1 | AWRT 22.9, 9.95 |
| 5 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(Z > \frac{20-14.6}{5.2}\right) = P(Z > 1.03846)$ | M1 | Use of $\pm$ standardisation formula with 20, 14.6 and 5.2 not $\sigma^2$, not $\sqrt{\sigma}$, no continuity correction |
| $1 - 0.8504$ | M1 | Calculating the appropriate probability area (leading to final answer) |
| $0.150$ | A1 | $0.1496, 0.149 < p \leq 0.15[0]$. Only dependent on 2nd M mark so M0M1A1 possible. **SC B1** for $0.149 < p \leq 0.15[0]$ if M0M0A0 awarded |
| $[250 \times \textit{their}\ 0.1496 =]\ 37, 38$ | B1 FT | Strict FT *their* at least 4-figure probability seen anywhere (give BOD if they go on to use 0.150). Final answer must be positive integer, no approximation or rounding stated |
| | **4** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z_1 = \frac{14.5 - \mu}{\sigma} = -0.842$ | B1 | $-0.843 < z_1 < -0.841$ or $0.841 < z_1 < 0.843$ |
| $z_2 = \frac{18.5 - \mu}{\sigma} = -0.44$ | B1 | $-0.441 < z_2 < -0.439$ or $0.439 < z_2 < 0.441$ |
| Solve, obtaining values for $\mu$ and $\sigma$ | M1 | Use of $\pm$ standardisation formula once with $\mu$, $\sigma$ and a $z$-value (not 0.20, 0.80, 0.67, 0.23, 0.5793, 0.7881, 0.7486, 0.591 or $1\text{-}z$ i.e. 0.158 etc.). Condone continuity correction $\pm 0.05$, not $\sigma^2$, $\sqrt{\sigma}$ |
| | M1 | Solve using elimination/substitution or other appropriate approach to obtain values for both $\mu$ and $\sigma$ |
| $\mu = 22.9,\quad \sigma = 9.95$ | A1 | AWRT 22.9, 9.95 |
| | **5** | |
---4 A mathematical puzzle is given to a large number of students. The times taken to complete the puzzle are normally distributed with mean 14.6 minutes and standard deviation 5.2 minutes.
\begin{enumerate}[label=(\alph*)]
\item In a random sample of 250 of the students, how many would you expect to have taken more than 20 minutes to complete the puzzle?\\
All the students are given a second puzzle to complete. Their times, in minutes, are normally distributed with mean $\mu$ and standard deviation $\sigma$. It is found that $20 \%$ of the students have times less than 14.5 minutes and $67 \%$ of the students have times greater than 18.5 minutes.
\item Find the value of $\mu$ and the value of $\sigma$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2023 Q4 [9]}}