| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.8 This is a straightforward application of normal distribution with standard procedures: (i) requires finding P(79 < X < 91) using standardization and tables, and (ii) involves finding an inverse normal value from a given probability. Both parts are routine S1 calculations with no conceptual challenges or multi-step reasoning beyond basic z-score manipulation. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(79 < X < 91) = P\left(\frac{79-85}{6.8} < Z < \frac{91-85}{6.8}\right)\) | M1 | Using \(\pm\) standardisation formula for either 79 or 91, no continuity correction |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \Phi(0.8824) - \Phi(-0.8824)\) | M1 | Correct area \((\Phi - \Phi)\) with one +ve and one -ve z-value or \(2\Phi - 1\) or \(2(\Phi - 0.5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.622\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = -1.751\) | B1 | \(\pm 1.751\) seen |
| \(-1.751 = \frac{t - 85}{6.8}\) | M1 | An equation using \(\pm\) standardisation formula with a z-value, condone \(\sigma^2\) or \(\sqrt{\sigma}\) |
| \(t = 73.1\) | A1 | Correct answer |
**Question 1(i):**
$P(79 < X < 91) = P\left(\frac{79-85}{6.8} < Z < \frac{91-85}{6.8}\right)$ | M1 | Using $\pm$ standardisation formula for either 79 or 91, no continuity correction
$= P(-0.8824 < Z < 0.8824)$
$= \Phi(0.8824) - \Phi(-0.8824)$ | M1 | Correct area $(\Phi - \Phi)$ with one +ve and one -ve z-value or $2\Phi - 1$ or $2(\Phi - 0.5)$
$= 0.8111 - (1 - 0.8111)$
$= 0.622$ | A1 | Correct answer
**Total: 3 marks**
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**Question 1(ii):**
$z = -1.751$ | B1 | $\pm 1.751$ seen
$-1.751 = \frac{t - 85}{6.8}$ | M1 | An equation using $\pm$ standardisation formula with a z-value, condone $\sigma^2$ or $\sqrt{\sigma}$
$t = 73.1$ | A1 | Correct answer
**Total: 3 marks**1 The time taken, in minutes, by a ferry to cross a lake has a normal distribution with mean 85 and standard deviation 6.8.\\
(i) Find the probability that, on a randomly chosen occasion, the time taken by the ferry to cross the lake is between 79 and 91 minutes.\\
(ii) Over a long period it is found that $96 \%$ of ferry crossings take longer than a certain time $t$ minutes. Find the value of $t$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q1 [6]}}