| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 Part (i) is a straightforward normal distribution probability calculation requiring a single standardization and table lookup. Part (ii) involves working backwards from a probability to find a standard deviation using the inverse normal, which is slightly less routine but still a standard S1 technique. Both parts are typical textbook exercises with no novel problem-solving required, making this slightly 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 |
| \(P(X > 120) = 1 - \Phi\!\left(\dfrac{120-112}{17.2}\right)\) | M1 | For standardising with or without the \(\sqrt{\,}\), \(17.2^2\), but no cc. |
| \(= 1 - \Phi(0.4651)\) | M1 | For finding the correct area, \(1 - \Phi(z)\), NOT \(\Phi(1 - \text{their } z(0.4651))\) |
| \(= 1 - 0.6790 = 0.321\) | A1 (3) | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = -0.842\) | B1 | For \(z\), \(\pm 0.842\) or \(\pm 0.84\) |
| \(-0.842 = \dfrac{103 - 115}{\sigma}\) | M1 | For solving an equation involving their \(z\) or \(z = 0.7881\) or \(0.5793\) only, 103, 115 and \(\sigma\) or \(\sqrt{\sigma}\) or \(\sigma^2\); must have used tables |
| \(\sigma = 14.3\) | A1 (3) | For correct answer |
## Question 3:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 120) = 1 - \Phi\!\left(\dfrac{120-112}{17.2}\right)$ | M1 | For standardising with or without the $\sqrt{\,}$, $17.2^2$, but no cc. |
| $= 1 - \Phi(0.4651)$ | M1 | For finding the correct area, $1 - \Phi(z)$, NOT $\Phi(1 - \text{their } z(0.4651))$ |
| $= 1 - 0.6790 = 0.321$ | A1 (3) | For correct answer |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = -0.842$ | B1 | For $z$, $\pm 0.842$ or $\pm 0.84$ |
| $-0.842 = \dfrac{103 - 115}{\sigma}$ | M1 | For solving an equation involving their $z$ or $z = 0.7881$ or $0.5793$ only, 103, 115 and $\sigma$ or $\sqrt{\sigma}$ or $\sigma^2$; must have used tables |
| $\sigma = 14.3$ | A1 (3) | For correct answer |
---3 (i) The height of sunflowers follows a normal distribution with mean 112 cm and standard deviation 17.2 cm . Find the probability that the height of a randomly chosen sunflower is greater than 120 cm .\\
(ii) When a new fertiliser is used, the height of sunflowers follows a normal distribution with mean 115 cm . Given that $80 \%$ of the heights are now greater than 103 cm , find the standard deviation.
\hfill \mbox{\textit{CAIE S1 2003 Q3 [6]}}