| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Estimate from percentile/frequency data |
| Difficulty | Standard +0.3 Part (a) requires setting up two equations using inverse normal distribution from given percentiles (42/500 and 100/500), then solving simultaneously—a standard S1 technique but with multiple steps. Part (b) is straightforward recall of the empirical rule (95% within 2 standard deviations). Overall slightly easier than average due to the routine nature of both parts. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z_1 = \frac{4-\mu}{\sigma} = -1.378\) | B1 | \(1.378 \leqslant z_1 \leqslant 1.379\) or \(-1.379 \leqslant z_1 \leqslant -1.378\) |
| \(z_2 = \frac{10-\mu}{\sigma} = 0.842\) | B1 | \(0.841 \leqslant z_2 \leqslant 0.842\) or \(-0.842 \leqslant z_2 \leqslant -0.841\) |
| Solve: \(\frac{4-\mu}{\sigma} = -1.378\); \(\frac{10-\mu}{\sigma} = 0.842\) | M1 | Use of \(\pm\)standardisation formula once with \(\mu\), \(\sigma\), a \(z\)-value and 4 or 10; allow continuity correction, not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| M1 | Use elimination or substitution to solve two equations in \(\mu\) and \(\sigma\) | |
| \(\sigma = 2.70\), \(\mu = 7.72\) | A1 | \(2.70 \leqslant \sigma \leqslant 2.71\), \(7.72 \leqslant \mu \leqslant 7.73\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\Phi(2) - \Phi(-2) = 2\Phi(2) - 1\) | M1 | Identifying 2 and \(-2\) as the appropriate \(z\)-values |
| \(2 \times their\ 0.9772 - 1\) | B1 | Calculating appropriate area from stated phis of \(z\)-values which must be \(\pm\) the same number |
| 0.9544 or 0.9545 | A1 | Accept AWRT 0.954 |
| \(0.9544 \times 800 = 763.52\); 763 or 764 | B1FT | FT *their* 4SF (or better) probability, final answer must be positive integer |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z_1 = \frac{4-\mu}{\sigma} = -1.378$ | B1 | $1.378 \leqslant z_1 \leqslant 1.379$ or $-1.379 \leqslant z_1 \leqslant -1.378$ |
| $z_2 = \frac{10-\mu}{\sigma} = 0.842$ | B1 | $0.841 \leqslant z_2 \leqslant 0.842$ or $-0.842 \leqslant z_2 \leqslant -0.841$ |
| Solve: $\frac{4-\mu}{\sigma} = -1.378$; $\frac{10-\mu}{\sigma} = 0.842$ | M1 | Use of $\pm$standardisation formula once with $\mu$, $\sigma$, a $z$-value and 4 or 10; allow continuity correction, not $\sigma^2$ or $\sqrt{\sigma}$ |
| | M1 | Use elimination or substitution to solve two equations in $\mu$ and $\sigma$ |
| $\sigma = 2.70$, $\mu = 7.72$ | A1 | $2.70 \leqslant \sigma \leqslant 2.71$, $7.72 \leqslant \mu \leqslant 7.73$ |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\Phi(2) - \Phi(-2) = 2\Phi(2) - 1$ | M1 | Identifying 2 and $-2$ as the appropriate $z$-values |
| $2 \times their\ 0.9772 - 1$ | B1 | Calculating appropriate area from stated phis of $z$-values which must be $\pm$ the same number |
| 0.9544 or 0.9545 | A1 | Accept AWRT 0.954 |
| $0.9544 \times 800 = 763.52$; 763 or 764 | B1FT | FT *their* 4SF (or better) probability, final answer must be positive integer |
---5 The lengths of the leaves of a particular type of tree are modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves from this type of tree and finds that 42 are less than 4 cm long and 100 are more than 10 cm long.
\begin{enumerate}[label=(\alph*)]
\item Find estimates for the mean and standard deviation of the lengths of leaves from this type of tree.\\
The lengths, in cm , of the leaves of a different type of tree have the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. The scientist takes a random sample of 800 leaves from this type of tree.
\item Find how many of these leaves the scientist would expect to have lengths, in cm , between $\mu - 2 \sigma$ and $\mu + 2 \sigma$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q5 [9]}}