| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Variance estimation from probability |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem requiring students to work backwards from a probability to find variance. The calculation involves standard normal tables and the formula Var(VĚ„) = Var(V)/n, but the steps are routine and clearly signposted. Slightly above average difficulty due to the reverse-working aspect and needing to explain CLT justification, but still a standard S2 question. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{71.2 - 72.0}{\sigma/\sqrt{40}}\right) = -0.3853\) | M1 | Standardise with \(\Phi^{-1}\) & \(\sqrt{40}\), allow cc, \(\sqrt{}\) errors e.g. \(\sigma^2\). RHS must be \(\Phi^{-1}\), i.e. *not* 0.7411 or 0.2589 or 0.6368 or 0.35 |
| A1 | Square roots and sign correct, no cc, no "\(1-\)" error. "\(1-\)" error or \(\times 40/39\): M1A0 | |
| \(z\) in range \((\pm)\ [0.385, 0.386]\) seen | B1 | \([0.674\) may be from "\(1 - 0.35 = 0.75\)"] |
| \([\sigma = 13.13]\ \text{Var}(V) = \mathbf{172.4}\) | A1 | *Final* answer in range \([172, 173]\), or \(13.1^2\) cwo. Needs variance, not SD. NB: Look out for \(-13.1 \to 172\), M1A0B1A0 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Parent distribution not known | B1 | Or clear equivalent. Not "*sample* not normal". Don't bother about order of these statements. |
| \(n\) is large | B1 | Or clear equiv, e.g. sample size \(> 30\). Extras: max 1. "\(n\) large, \(n > n_0\)": B1 if \(n_0 \geq 30\). If numerical must be 30. Ignore "continuous". |
| [2] |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{71.2 - 72.0}{\sigma/\sqrt{40}}\right) = -0.3853$ | M1 | Standardise with $\Phi^{-1}$ & $\sqrt{40}$, allow cc, $\sqrt{}$ errors e.g. $\sigma^2$. RHS must be $\Phi^{-1}$, i.e. *not* 0.7411 or 0.2589 or 0.6368 or 0.35 |
| | A1 | Square roots and sign correct, no cc, no "$1-$" error. "$1-$" error or $\times 40/39$: M1A0 |
| $z$ in range $(\pm)\ [0.385, 0.386]$ seen | B1 | $[0.674$ may be from "$1 - 0.35 = 0.75$"] |
| $[\sigma = 13.13]\ \text{Var}(V) = \mathbf{172.4}$ | A1 | *Final* answer in range $[172, 173]$, or $13.1^2$ cwo. Needs variance, not SD. NB: Look out for $-13.1 \to 172$, M1A0B1A0 |
| | **[4]** | |
---
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Parent distribution not known | B1 | Or clear equivalent. Not "*sample* not normal". Don't bother about order of these statements. |
| $n$ is large | B1 | Or clear equiv, e.g. sample size $> 30$. Extras: max 1. "$n$ large, $n > n_0$": B1 if $n_0 \geq 30$. If numerical must be 30. Ignore "continuous". |
| | **[2]** | |
---2 (i) For the continuous random variable $V$, it is known that $\mathrm { E } ( V ) = 72.0$. The mean of a random sample of 40 observations of $V$ is denoted by $\bar { V }$. Given that $\mathrm { P } ( \bar { V } < 71.2 ) = 0.35$, estimate the value of $\operatorname { Var } ( V )$.\\
(ii) Explain why you need to use the Central Limit Theorem in part (i), and why its use is justified.
\hfill \mbox{\textit{OCR S2 2012 Q2 [6]}}