| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This question involves straightforward normal distribution calculations using standardization and tables/inverse tables. Part (a) is direct z-score calculation, part (b) is routine inverse normal (finding a value given a percentage), and part (c) requires recognizing a symmetric interval about the mean. All are standard S1 techniques with no problem-solving insight required, making this easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(X > 28.6) =]\ P\!\left(Z > \dfrac{28.6 - 32.2}{9.6}\right)\) \([= P(Z > -0.375)]\) | M1 | 28.6, 32.2 and 9.6 substituted appropriately in \(\pm\) standardisation formula once, allow continuity correction of \(\pm 0.05\), no \(\sigma^2\), \(\sqrt{\sigma}\) |
| \([\Phi(\textit{their}\ 0.375) =]\ \textit{their}\ 0.6462\) | M1 | Appropriate numerical area, from final process, must be probability, expect \(> 0.5\) |
| 0.646 | A1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = \pm 0.842\) | B1 | \(0.841 < z \leqslant 0.842\) or \(-0.842 \leqslant z < -0.841\) seen |
| \(\dfrac{t - 32.2}{9.6} = 0.842\) | M1 | Substituting 32.2 and 9.6 into \(\pm\) standardisation formula, no continuity correction, allow \(\sigma^2\), \(\sqrt{\sigma}\), must be equated to a \(z\)-value |
| \(t = 40.3\) | A1 | \(40.28 \leqslant t \leqslant 40.3\) WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P\!\left(-\dfrac{15}{9.6} < Z < \dfrac{15}{9.6}\right)\), \(P(-1.5625 < Z < 1.5625)\) | M1 | Identifying at least one of \(\dfrac{15}{9.6}\) and \(-\dfrac{15}{9.6}\) as the appropriate \(z\)-values, or substituting *their* \((32.2 \pm 15)\) into \(\pm\) standardisation formula once, no continuity correction, \(\sigma^2\) nor \(\sqrt{\sigma}\). Condone \(\pm 1.563\) for M1 |
| \([2\,\Phi\!\left(\dfrac{15}{9.6}\right) - 1]\) | A1 | \(p = 0.941\) AWRT SOI |
| \(= 2 \times 0.9409 - 1\) | M1 | Appropriate area \(2\Phi - 1\), e.g. \(1 - 2 \times 0.0591\), \(2\times(0.9409 - 0.5)\) or \(0.9409 - 0.0591\), from final process, must be probability \(> 0.5\) |
| 0.882 | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(X > 28.6) =]\ P\!\left(Z > \dfrac{28.6 - 32.2}{9.6}\right)$ $[= P(Z > -0.375)]$ | M1 | 28.6, 32.2 and 9.6 substituted appropriately in $\pm$ standardisation formula once, allow continuity correction of $\pm 0.05$, no $\sigma^2$, $\sqrt{\sigma}$ |
| $[\Phi(\textit{their}\ 0.375) =]\ \textit{their}\ 0.6462$ | M1 | Appropriate numerical area, from final process, must be probability, expect $> 0.5$ |
| 0.646 | A1 | AWRT |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \pm 0.842$ | B1 | $0.841 < z \leqslant 0.842$ or $-0.842 \leqslant z < -0.841$ seen |
| $\dfrac{t - 32.2}{9.6} = 0.842$ | M1 | Substituting 32.2 and 9.6 into $\pm$ standardisation formula, no continuity correction, allow $\sigma^2$, $\sqrt{\sigma}$, must be equated to a $z$-value |
| $t = 40.3$ | A1 | $40.28 \leqslant t \leqslant 40.3$ WWW |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P\!\left(-\dfrac{15}{9.6} < Z < \dfrac{15}{9.6}\right)$, $P(-1.5625 < Z < 1.5625)$ | M1 | Identifying at least one of $\dfrac{15}{9.6}$ and $-\dfrac{15}{9.6}$ as the appropriate $z$-values, or substituting *their* $(32.2 \pm 15)$ into $\pm$ standardisation formula once, no continuity correction, $\sigma^2$ nor $\sqrt{\sigma}$. Condone $\pm 1.563$ for M1 |
| $[2\,\Phi\!\left(\dfrac{15}{9.6}\right) - 1]$ | A1 | $p = 0.941$ AWRT SOI |
| $= 2 \times 0.9409 - 1$ | M1 | Appropriate area $2\Phi - 1$, e.g. $1 - 2 \times 0.0591$, $2\times(0.9409 - 0.5)$ or $0.9409 - 0.0591$, from final process, must be probability $> 0.5$ |
| 0.882 | A1 | |
---6 The times taken, in minutes, to complete a particular task by employees at a large company are normally distributed with mean 32.2 and standard deviation 9.6.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen employee takes more than 28.6 minutes to complete the task.
\item $20 \%$ of employees take longer than $t$ minutes to complete the task.
Find the value of $t$.
\item Find the probability that the time taken to complete the task by a randomly chosen employee differs from the mean by less than 15.0 minutes.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q6 [10]}}