| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | March |
| Marks | 6 |
| 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 question requiring standard table lookups and inverse normal calculations. Part (a) involves standardizing two values and finding P(85 < X < 100), while part (b) requires finding a percentile using inverse normal tables—both are routine S1 procedures with no conceptual challenges or multi-step reasoning. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\left(\left(\frac{85-96}{18}\right) < z < \left(\frac{100-96}{18}\right)\right)\) | M1 | Use of \(\pm\)standardisation formula once with appropriate values substituted, no continuity correction, not \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(P(-0.6111 < z < 0.2222) = \Phi(0.2222)+\Phi(0.6111)-1 = 0.5879+0.7294-1\) | M1 | Appropriate area \(\Phi\), from final process, must be probability. Use of \((1-z)\) implies M0 |
| \(0.317\) | A1 | Final answer which rounds to \(0.317\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = \pm1.175\) | B1 | \(1.17 \leqslant z \leqslant 1.18\) or \(-1.18 \leqslant z \leqslant -1.17\) |
| \(-1.175 = \frac{t-96}{18}\) | M1 | An equation using \(\pm\)standardisation formula with a \(z\)-value, condone \(\sigma^2\), \(\sqrt{\sigma}\) or continuity correction. E.g. equating to \(0.88, 0.12, 0.8106, 0.1894, 0.5478, 0.4522\), \(\pm0.175\) or \(\pm2.175\) implies M0 |
| \(74.85\) or \(74.9\) | A1 | \(74.85 \leqslant t \leqslant 74.9\) |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\left(\left(\frac{85-96}{18}\right) < z < \left(\frac{100-96}{18}\right)\right)$ | M1 | Use of $\pm$standardisation formula once with appropriate values substituted, no continuity correction, not $\sigma^2$ or $\sqrt{\sigma}$ |
| $P(-0.6111 < z < 0.2222) = \Phi(0.2222)+\Phi(0.6111)-1 = 0.5879+0.7294-1$ | M1 | Appropriate area $\Phi$, from final process, must be probability. Use of $(1-z)$ implies M0 |
| $0.317$ | A1 | Final answer which rounds to $0.317$ |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = \pm1.175$ | B1 | $1.17 \leqslant z \leqslant 1.18$ or $-1.18 \leqslant z \leqslant -1.17$ |
| $-1.175 = \frac{t-96}{18}$ | M1 | An equation using $\pm$standardisation formula with a $z$-value, condone $\sigma^2$, $\sqrt{\sigma}$ or continuity correction. E.g. equating to $0.88, 0.12, 0.8106, 0.1894, 0.5478, 0.4522$, $\pm0.175$ or $\pm2.175$ implies M0 |
| $74.85$ or $74.9$ | A1 | $74.85 \leqslant t \leqslant 74.9$ |3 The time spent by shoppers in a large shopping centre has a normal distribution with mean 96 minutes and standard deviation 18 minutes.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a shopper chosen at random spends between 85 and 100 minutes in the shopping centre.\\
$88 \%$ of shoppers spend more than $t$ minutes in the shopping centre.
\item Find the value of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q3 [6]}}