| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single normal population sample mean |
| Difficulty | Moderate -0.8 This is a straightforward application of the sampling distribution of the mean with known population parameters. Part (a) requires standardizing using the standard error and looking up a normal table, while part (b) involves working backwards from a probability. Both are routine procedures with no problem-solving insight required, making this easier than average but not trivial due to the two-part structure. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{36-35}{8.1 \div \sqrt{140}}\) \([= 1.461]\) | M1 | Ignore inclusion of cc for M1. Must have \(\sqrt{140}\) |
| \(1 - \Phi(1.461)\) | M1 | For area consistent with their values |
| \(= 0.0720\) (3 sf) | A1 | Allow 0.072 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\Phi^{-1}(0.986)] = 2.197\) to \(2.198\) | B1 | Seen. Note: 2.2 and nothing better seen scores B0 |
| \(\pm\frac{a-35}{8.1 \div \sqrt{140}} = \pm 2.198\) | M1 | Must be a z value |
| \(a = 36.5\) (3 sf) | A1 | CWO. Note: use of 2.2 scores A1 so 2/3. But e.g. 2.196 gives 36.5 but scores B0 M1 A0 so 1/3 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{36-35}{8.1 \div \sqrt{140}}$ $[= 1.461]$ | M1 | Ignore inclusion of cc for M1. Must have $\sqrt{140}$ |
| $1 - \Phi(1.461)$ | M1 | For area consistent with their values |
| $= 0.0720$ (3 sf) | A1 | Allow 0.072 |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\Phi^{-1}(0.986)] = 2.197$ to $2.198$ | B1 | Seen. Note: 2.2 and nothing better seen scores **B0** |
| $\pm\frac{a-35}{8.1 \div \sqrt{140}} = \pm 2.198$ | M1 | Must be a z value |
| $a = 36.5$ (3 sf) | A1 | CWO. Note: use of 2.2 scores A1 so 2/3. But e.g. 2.196 gives 36.5 but scores **B0 M1 A0** so 1/3 |
---4 A population is normally distributed with mean 35 and standard deviation 8.1 . A random sample of size 140 is chosen from this population and the sample mean is denoted by $\bar { X }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( \bar { X } > 36 )$.
\item It is given that $\mathrm { P } ( \bar { X } < a ) = 0.986$. Find the value of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q4 [6]}}