| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | State distribution of sample mean |
| Difficulty | Moderate -0.8 This question tests basic recall of sampling distribution formulas (E(X̄) = μ, Var(X̄) = σ²/n) and a straightforward normal probability calculation with a sample mean. Part (iii) requires understanding when CLT is needed (not here, since population is already normal). All parts are routine textbook exercises requiring no problem-solving or novel insight. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mu\) | B1 | (as an expression) |
| \(\dfrac{\sigma^2}{n}\) | B1 [2] | (as an expression); SC If B0B0 scored, \(N\!\left(\mu, \dfrac{\sigma^2}{n}\right)\) scores B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{176-177.8}{\dfrac{6.1}{\sqrt{12}}} = -1.022\) | M1 | Standardise with \(\sqrt{12}\); accept 'totals' method, no mixed methods; \((2112-2133.6)/\sqrt{6.1^2 \times 12}\) |
| \(\Phi(-1.022) = 1 - \Phi(1.022)\) | M1 | Correct area (consistent with working) |
| \(= 0.153\) (3 sfs) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No; \(X\) not normally distributed or pop not normally distributed, or hts normal distributed, or original dist normal | B1 [1] | Need 'No' stated or implied AND correct reason; NB 'No, because small sample' scores B0; NB 'it is normally distributed' scores B0 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu$ | B1 | (as an expression) |
| $\dfrac{\sigma^2}{n}$ | B1 [2] | (as an expression); SC If B0B0 scored, $N\!\left(\mu, \dfrac{\sigma^2}{n}\right)$ scores B1 |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{176-177.8}{\dfrac{6.1}{\sqrt{12}}} = -1.022$ | M1 | Standardise with $\sqrt{12}$; accept 'totals' method, no mixed methods; $(2112-2133.6)/\sqrt{6.1^2 \times 12}$ |
| $\Phi(-1.022) = 1 - \Phi(1.022)$ | M1 | Correct area (consistent with working) |
| $= 0.153$ (3 sfs) | A1 [3] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No; $X$ not normally distributed or pop not normally distributed, or hts normal distributed, or original dist normal | B1 [1] | Need 'No' stated or implied AND correct reason; NB 'No, because small sample' scores B0; NB 'it is normally distributed' scores B0 |
---2 (i) A random variable $X$ has mean $\mu$ and variance $\sigma ^ { 2 }$. The mean of a random sample of $n$ values of $X$ is denoted by $\bar { X }$. Give expressions for $\mathrm { E } ( \bar { X } )$ and $\operatorname { Var } ( \bar { X } )$.\\
(ii) The heights, in centimetres, of adult males in Brancot are normally distributed with mean 177.8 and standard deviation 6.1. Find the probability that the mean height of a random sample of 12 adult males from Brancot is less than 176 cm .\\
(iii) State, with a reason, whether it was necessary to use the Central Limit Theorem in the calculation in part (ii).
\hfill \mbox{\textit{CAIE S2 2012 Q2 [6]}}