| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward S1 question testing basic normal distribution calculations (z-scores and probability lookup), binomial probability with n=3, and interpretation of a box plot. All parts are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(7.8 < T < 11) - \Phi(1.143) - 0.5 = 0.8735 - 0.5 = 0.3735\) (accept ans rounded to 0.37 to 0.374) | M1 | For standardising, can be implied, no cc, no \(\sigma^2\) but accept \(\sqrt{\sigma}\) |
| A1 | For seeing 0.8735 | |
| M1 | For subtracting two probs, \(p_2 - p_1\) where \(p_2 > p_1\) | |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \((0.1265)^2 \times (0.8735) \times {}_3C_2\) \(= 0.0419\) | M1 | For any three term binomial-type expression with powers summing to 3 |
| A1ft | For correct answer ft on their 0.8735/0.1265 |
| Answer | Marks | Guidance |
|---|---|---|
| Does not agree with the hospital's figures | B1 | For any valid reason |
| B1dep | For stating it does not agree, with no invalid reasons |
**(i)** $z = \pm 1.143$
$P(7.8 < T < 11) - \Phi(1.143) - 0.5 = 0.8735 - 0.5 = 0.3735$ (accept ans rounded to 0.37 to 0.374) | M1 | For standardising, can be implied, no cc, no $\sigma^2$ but accept $\sqrt{\sigma}$
A1 | For seeing 0.8735
M1 | For subtracting two probs, $p_2 - p_1$ where $p_2 > p_1$
A1 | For correct answer
**4 marks total**
**(ii)** $(0.1265)^2 \times (0.8735) \times {}_3C_2$ $= 0.0419$ | M1 | For any three term binomial-type expression with powers summing to 3
A1ft | For correct answer ft on their 0.8735/0.1265
**2 marks total**
**(iii)** Not symmetric so not normal
Does not agree with the hospital's figures | B1 | For any valid reason
B1dep | For stating it does not agree, with no invalid reasons
**2 marks total**
---7 The length of time a person undergoing a routine operation stays in hospital can be modelled by a normal distribution with mean 7.8 days and standard deviation 2.8 days.\\
(i) Calculate the proportion of people who spend between 7.8 days and 11.0 days in hospital.\\
(ii) Calculate the probability that, of 3 people selected at random, exactly 2 spend longer than 11.0 days in hospital.\\
(iii) A health worker plotted a box-and-whisker plot of the times that 100 patients, chosen randomly, stayed in hospital. The result is shown below.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{26776153-9477-4155-b5e4-f35e6d33a5ff-3_447_917_767_657}
\captionsetup{labelformat=empty}
\caption{Days}
\end{center}
\end{figure}
State with a reason whether or not this agrees with the model used in parts (i) and (ii).
\hfill \mbox{\textit{CAIE S1 2003 Q7 [8]}}