| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Finding n from sample mean distribution |
| Difficulty | Standard +0.8 This question requires understanding that the sample mean distribution is fully determined by two pieces of information, then using z-scores to set up simultaneous equations. Part (ii) tests deeper conceptual understanding of degrees of freedom in statistical systems. While the mechanics are standard S2 level, the conceptual reasoning in part (ii) elevates this above routine exercises. |
| 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 | Marks | Guidance |
| \(\dfrac{\mu - 20}{\sigma/\sqrt{n}} = 1.0\); \(\quad \dfrac{35 - \mu}{\sigma/\sqrt{n}} = 2.0\) | M1, A1, B1 | Standardise either 20 or 35, equate to \(\Phi^{-1}\). Both equations completely correct. Both correct \(z\)-values seen (to 3 SF at least). With \(\sqrt{n}\) or \(n\) and \(z\), allow "\(1-\)", cc; including signs, but can have wrong \(z\); independent of previous marks |
| Solve to get \(\sigma = 5\sqrt{n}\) | M1, A1 | Correctly obtain \(\sigma = k\sqrt{n}\) or \(\sigma^2 = kn\). \(\sigma = 5\sqrt{n}\) or \(\sqrt{25n}\) only. Allow \(\sqrt{}\) errors, ALLOW from not \(\Phi^{-1}\) [only mark from 0.7998 & 0.8358] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binya is right | B1 | Binya stated. "Aidan" used: max B0B1M0 |
| \(\mu = 25\) | B1 | \(\mu = 25\) following no wrong working. But allow if \(\sqrt{n}\) omitted or wrong |
| \(1 - \Phi\!\left(\dfrac{32 - \mu}{5}\right) = 1 - \Phi(1.4)\) | M1 | Standardise with their \(\sigma/\sqrt{n}\) and their numerical \(\mu\). NB: use of 1.282 probably implies "Aidan" |
| \(= 1 - 0.9192 = \mathbf{0.0808}\) | A1 | Answer, a.r.t. 0.081, CWO |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\mu - 20}{\sigma/\sqrt{n}} = 1.0$; $\quad \dfrac{35 - \mu}{\sigma/\sqrt{n}} = 2.0$ | M1, A1, B1 | Standardise either 20 or 35, equate to $\Phi^{-1}$. Both equations completely correct. Both correct $z$-values seen (to 3 SF at least). With $\sqrt{n}$ or $n$ and $z$, allow "$1-$", cc; including signs, but can have wrong $z$; independent of previous marks |
| Solve to get $\sigma = 5\sqrt{n}$ | M1, A1 | Correctly obtain $\sigma = k\sqrt{n}$ or $\sigma^2 = kn$. $\sigma = 5\sqrt{n}$ or $\sqrt{25n}$ only. Allow $\sqrt{}$ errors, ALLOW from not $\Phi^{-1}$ [only mark from 0.7998 & 0.8358] |
**[5 marks]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binya is right | B1 | Binya stated. "Aidan" used: max B0B1M0 |
| $\mu = 25$ | B1 | $\mu = 25$ following no wrong working. But allow if $\sqrt{n}$ omitted or wrong |
| $1 - \Phi\!\left(\dfrac{32 - \mu}{5}\right) = 1 - \Phi(1.4)$ | M1 | Standardise with their $\sigma/\sqrt{n}$ and their numerical $\mu$. NB: use of 1.282 probably implies "Aidan" |
| $= 1 - 0.9192 = \mathbf{0.0808}$ | A1 | Answer, a.r.t. 0.081, CWO |
**[4 marks]**
---7 The continuous random variable $X$ has the distribution $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$. The mean of a random sample of $n$ observations of $X$ is denoted by $\bar { X }$. It is given that $\mathrm { P } ( \bar { X } < 35.0 ) = 0.9772$ and $\mathrm { P } ( \bar { X } < 20.0 ) = 0.1587$.\\
(i) Obtain a formula for $\sigma$ in terms of $n$.
Two students are discussing this question. Aidan says "If you were told another probability, for instance $\mathrm { P } ( \bar { X } > 32 ) = 0.1$, you could work out the value of $\sigma$." Binya says, "No, the value of $\mathrm { P } ( \bar { X } > 32 )$ is fixed by the information you know already."\\
(ii) State which of Aidan and Binya is right. If you think that Aidan is right, calculate the value of $\sigma$ given that $\mathrm { P } ( \bar { X } > 32 ) = 0.1$. If you think that Binya is right, calculate the value of $\mathrm { P } ( \bar { X } > 32 )$.
\hfill \mbox{\textit{OCR S2 2013 Q7 [9]}}