| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Standard +0.3 This question tests standard CLT application with routine calculations of sample statistics and a straightforward normal approximation. Part (iii) requires understanding when CLT is needed, but the conceptual demand is modest—students simply need to recognize that CLT justifies using normal distribution when the population distribution is unknown. The calculations are mechanical and the reasoning is textbook-standard for S2 level. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{est}(\mu) = 2005/200 = 10.025\) | B1 | |
| \(\text{est}(\sigma^2) = \dfrac{1}{99}\left(20175 - \dfrac{2005^2}{200}\right) = 0.376\) (3 sf) | M1, A1 [3] | Correct subst in correct formula |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{10 - 10.025}{\sqrt{\dfrac{0.376256}{50}}} = -0.288\) | M1 | Allow without \(\sqrt{}\), but \(\div\sqrt{50}\) essential |
| \(1 - \Phi(0.288) = 0.387\) (3 sf) | M1, A1 [3] | Use of 'biased' variance can still score fully in (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Yes; (assumed distr of \(\bar{X}\) normal) although distr of \(X\) unknown | B1, B1 [2] |
## Question 4:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{est}(\mu) = 2005/200 = 10.025$ | B1 | |
| $\text{est}(\sigma^2) = \dfrac{1}{99}\left(20175 - \dfrac{2005^2}{200}\right) = 0.376$ (3 sf) | M1, A1 [3] | Correct subst in correct formula |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{10 - 10.025}{\sqrt{\dfrac{0.376256}{50}}} = -0.288$ | M1 | Allow without $\sqrt{}$, but $\div\sqrt{50}$ essential |
| $1 - \Phi(0.288) = 0.387$ (3 sf) | M1, A1 [3] | Use of 'biased' variance can still score fully in (ii) |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Yes; (assumed distr of $\bar{X}$ normal) although distr of $X$ unknown | B1, B1 [2] | |
---4 The lengths, $x \mathrm {~m}$, of a random sample of 200 balls of string are found and the results are summarised by $\Sigma x = 2005$ and $\Sigma x ^ { 2 } = 20175$.\\
(i) Calculate unbiased estimates of the population mean and variance of the lengths.\\
(ii) Use the values from part (i) to estimate the probability that the mean length of a random sample of 50 balls of string is less than 10 m .\\
(iii) Explain whether or not it was necessary to use the Central Limit theorem in your calculation in part (ii).
\hfill \mbox{\textit{CAIE S2 2013 Q4 [8]}}