| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward normal distribution problem requiring standard z-score lookups and inverse normal calculations. Part (i) involves basic probability calculation with given boundaries, while part (ii) requires finding percentiles (33rd and 67th), which are routine inverse normal operations covered in any S1 course. The problem is slightly easier than average due to clear structure and standard techniques, though not trivial as it requires proper understanding of percentiles and normal tables. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
\(P(85| B1 |
\(\pm\frac{85-100}{7}\) seen oe or \(\pm2.14\) |
|
| \(= 0.5 - P(z < -2.143) = 0.5-(1-\Phi(2.143))\) | M1 | \(\Phi - 0.5\) |
| \(= 0.9839 - 0.5 = 0.484\) | A1 [3] | Correct answer rounding to |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = \Phi^{-1}(0.67) = 0.44\) | B1 | \(\pm0.44\) seen |
| \(0.44 = \frac{a-100}{7}\) | M1 | Standardising, with or without sq rt, no cc, no \(7^2\); must be \(z\)-value e.g. could be \(0.412\) or \(0.413\) |
| \(103.1\) min \((103)\) = upper limit | A1 | Correct upper or lower boundary allow even if obtained from \(z = 0.412\) |
| \(96.9\) min = lower limit | A1 [4] | Correct other boundary |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(85<x<100) = 0.5 - P\left(z < \left(\frac{85-100}{7}\right)\right)$ | B1 | $\pm\frac{85-100}{7}$ seen oe or $\pm2.14$ |
| $= 0.5 - P(z < -2.143) = 0.5-(1-\Phi(2.143))$ | M1 | $\Phi - 0.5$ |
| $= 0.9839 - 0.5 = 0.484$ | A1 **[3]** | Correct answer rounding to |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = \Phi^{-1}(0.67) = 0.44$ | B1 | $\pm0.44$ seen |
| $0.44 = \frac{a-100}{7}$ | M1 | Standardising, with or without sq rt, no cc, no $7^2$; must be $z$-value e.g. could be $0.412$ or $0.413$ |
| $103.1$ min $(103)$ = upper limit | A1 | Correct upper or lower boundary allow even if obtained from $z = 0.412$ |
| $96.9$ min = lower limit | A1 **[4]** | Correct other boundary |
---3 The times for a certain car journey have a normal distribution with mean 100 minutes and standard deviation 7 minutes. Journey times are classified as follows:
\begin{displayquote}
'short' (the shortest $33 \%$ of times),\\
'long' (the longest $33 \%$ of times),\\
'standard' (the remaining 34\% of times).\\
(i) Find the probability that a randomly chosen car journey takes between 85 and 100 minutes.\\
(ii) Find the least and greatest times for 'standard' journeys.
\end{displayquote}
\hfill \mbox{\textit{CAIE S1 2009 Q3 [7]}}