| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Find p then binomial probability |
| Difficulty | Standard +0.8 Part (i) requires inverse normal calculation using the 0.25 probability condition, which is standard. Part (ii) requires recognizing independence, calculating a single-day probability P(40 < X < 46), then raising it to the fourth power—this multi-step reasoning with compound probability across multiple days elevates it above routine questions but remains within typical A-level scope. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 0.674\) or \(0.675\), allow \(0.67\) to \(0.675\) | B1 | For correct \(z\), can be \(+\) or \(-\) |
| \(\frac{52 - \mu}{5} = 0.674\) | M1 | For an equation relating 52, 5, \(\mu\) and any \(z \neq 0.5987\) or \(0.7734\) ish |
| \(\mu = 48.6\) | A1 3 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z_1 = \frac{40 - 48.63}{5} = -1.726\) | M1 | For standardising 40 or 46, 5 or \(\sqrt{5}\) in denom or \(5^2\) with their mean, no cc |
| \(z_2 = \frac{46 - 48.63}{5} = 0.526\) | ||
| \(\text{prob} = 0.9578 - 0.7005 = 0.2573\) | M1 | For subtracting two probs consistent with their mean ie usually \(\Phi_1 - \Phi_2\) or \((1-\Phi_1)-(1-\Phi_2)\) but could be of type \(\Phi_1 - (1-\Phi_2)\) if their mean is in between 40 and 46 |
| \((0.2573)^4\) | M1 | For raising their answer above to a power 4 |
| \(= 0.00438\) or \(4.38 \times 10^{-3}\); accept \(0.00449 \times 10^{-3}\); NB \(0.0045\) gets A0 and RE #1 | A1ft 4 | For correct answer |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 0.674$ or $0.675$, allow $0.67$ to $0.675$ | B1 | For correct $z$, can be $+$ or $-$ |
| $\frac{52 - \mu}{5} = 0.674$ | M1 | For an equation relating 52, 5, $\mu$ and any $z \neq 0.5987$ or $0.7734$ ish |
| $\mu = 48.6$ | A1 **3** | For correct answer |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z_1 = \frac{40 - 48.63}{5} = -1.726$ | M1 | For standardising 40 or 46, 5 or $\sqrt{5}$ in denom or $5^2$ with their mean, no cc |
| $z_2 = \frac{46 - 48.63}{5} = 0.526$ | | |
| $\text{prob} = 0.9578 - 0.7005 = 0.2573$ | M1 | For subtracting two probs consistent with their mean ie usually $\Phi_1 - \Phi_2$ or $(1-\Phi_1)-(1-\Phi_2)$ but could be of type $\Phi_1 - (1-\Phi_2)$ if their mean is in between 40 and 46 |
| $(0.2573)^4$ | M1 | For raising their answer above to a power 4 |
| $= 0.00438$ or $4.38 \times 10^{-3}$; accept $0.00449 \times 10^{-3}$; NB $0.0045$ gets A0 and RE #1 | A1ft **4** | For correct answer |
---5 The length of Paulo's lunch break follows a normal distribution with mean $\mu$ minutes and standard deviation 5 minutes. On one day in four, on average, his lunch break lasts for more than 52 minutes.\\
(i) Find the value of $\mu$.\\
(ii) Find the probability that Paulo's lunch break lasts for between 40 and 46 minutes on every one of the next four days.
\hfill \mbox{\textit{CAIE S1 2004 Q5 [7]}}