| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard table lookups. Part (i) requires inverse normal calculation (finding h given P(X<h)=0.67), which is routine. Part (ii) involves calculating P(μ-0.5σ < X < μ+0.5σ) and multiplying by 120—both are textbook exercises requiring no problem-solving insight, just mechanical application of z-scores and tables. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(h < 148) = 0.67\) | B1 | \(z = \pm 0.44\) seen |
| \(\frac{h-148}{8} = 0.44\) | M1 | \(z\text{-value} = \pm\frac{(h-148)}{8}\) |
| \(151.52 \approx 152\) | A1 | CAO |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(144 < X < 152) = P\left(\frac{144-148}{8} < Z < \frac{152-148}{8}\right)\) | M1 | Using \(\pm\) standardisation formula for either 144 or 152, \(\mu = 148\), \(\sigma = 8\) and no continuity correction, allow \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(= P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = 0.6915 - (1-0.6915) = 2 \times 0.6915 - 1\) | M1 | Correct final area legitimately obtained from \(\phi(\textit{their } z_2) - \phi(\textit{their } z_1)\) |
| \(= 0.383\) | A1 | Final probability answer |
| \(0.383 \times 120 = 45.96\); Accept 45 or 46 only | B1FT | *Their* prob (to 3 or 4 sf) \(\times 120\), rounded to a whole number or truncated |
| 4 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(h < 148) = 0.67$ | B1 | $z = \pm 0.44$ seen |
| $\frac{h-148}{8} = 0.44$ | M1 | $z\text{-value} = \pm\frac{(h-148)}{8}$ |
| $151.52 \approx 152$ | A1 | CAO |
| | **3** | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(144 < X < 152) = P\left(\frac{144-148}{8} < Z < \frac{152-148}{8}\right)$ | M1 | Using $\pm$ standardisation formula for either 144 or 152, $\mu = 148$, $\sigma = 8$ and no continuity correction, allow $\sigma^2$ or $\sqrt{\sigma}$ |
| $= P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = 0.6915 - (1-0.6915) = 2 \times 0.6915 - 1$ | M1 | Correct final area legitimately obtained from $\phi(\textit{their } z_2) - \phi(\textit{their } z_1)$ |
| $= 0.383$ | A1 | Final probability answer |
| $0.383 \times 120 = 45.96$; Accept 45 or 46 only | B1FT | *Their* prob (to 3 or 4 sf) $\times 120$, rounded to a whole number or truncated |
| | **4** | |
---4 The heights of students at the Mainland college are normally distributed with mean 148 cm and standard deviation 8 cm .\\
(i) The probability that a Mainland student chosen at random has a height less than $h \mathrm {~cm}$ is 0.67 . Find the value of $h$.\\
120 Mainland students are chosen at random.\\
(ii) Find the number of these students that would be expected to have a height within half a standard deviation of the mean.\\
\hfill \mbox{\textit{CAIE S1 2019 Q4 [7]}}