| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | November |
| Marks | 7 |
| 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 two-part normal distribution problem requiring standard table lookups and inverse normal calculations. Part (i) involves finding σ from a given probability using z-tables (z ≈ 1.69), then solving (5.5-4.5)/σ = 1.69. Part (ii) requires standardizing two values and finding the difference between two probabilities. Both parts are routine S1 techniques with no conceptual challenges beyond basic normal distribution manipulation. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = \pm 1.68\) | B1 | Number rounding to 1.68 seen |
| \(z = \frac{5.5 - 4.5}{\sigma}\) | M1 | Standardising and attempting to solve with their \(z\); must be \(z\) value, no cc, no \(\sigma^2\), no \(\sqrt{\sigma}\) |
| \(\sigma = 0.595\), accept \(25/42\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z_1 = \frac{3.8 - 4.5}{0.5952} = -1.176\) | M1 | For standardising 3.8 or 4.8, mean 4.5 not 5.5, their \(\sigma\) or \(\sqrt{\sigma}\) or \(\sigma^2\) in denom |
| \(z_2 = \frac{4.8 - 4.5}{0.5952} = 0.504\) | A1ft | One correct \(z\)-value, ft on their \(\sigma\) |
| \(\text{prob} = \Phi(0.504) - (1 - \Phi(1.176))\) | M1 | Correct area ie \(\Phi_1 + \Phi_2 - 1\) or \(\Phi_1 - \Phi_2\) if \(\mu\) taken to be 5.5 |
| \(= 0.6929 - (1 - 0.8802) = 0.573\) | A1 [4] | Correct answer only |
## Question 4:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \pm 1.68$ | B1 | Number rounding to 1.68 seen |
| $z = \frac{5.5 - 4.5}{\sigma}$ | M1 | Standardising and attempting to solve with their $z$; must be $z$ value, no cc, no $\sigma^2$, no $\sqrt{\sigma}$ |
| $\sigma = 0.595$, accept $25/42$ | A1 **[3]** | Correct answer |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z_1 = \frac{3.8 - 4.5}{0.5952} = -1.176$ | M1 | For standardising 3.8 or 4.8, mean 4.5 not 5.5, their $\sigma$ or $\sqrt{\sigma}$ or $\sigma^2$ in denom |
| $z_2 = \frac{4.8 - 4.5}{0.5952} = 0.504$ | A1ft | One correct $z$-value, ft on their $\sigma$ |
| $\text{prob} = \Phi(0.504) - (1 - \Phi(1.176))$ | M1 | Correct area ie $\Phi_1 + \Phi_2 - 1$ or $\Phi_1 - \Phi_2$ if $\mu$ taken to be 5.5 |
| $= 0.6929 - (1 - 0.8802) = 0.573$ | A1 **[4]** | Correct answer only |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{1a10471c-5810-44ca-9353-c2c76e190a2b-2_542_876_1425_632}
The random variable $X$ has a normal distribution with mean 4.5. It is given that $\mathrm { P } ( X > 5.5 ) = 0.0465$ (see diagram).\\
(i) Find the standard deviation of $X$.\\
(ii) Find the probability that a random observation of $X$ lies between 3.8 and 4.8.
\hfill \mbox{\textit{CAIE S1 2007 Q4 [7]}}