| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.3 This is a straightforward Normal distribution question requiring standard techniques: z-score calculations, using tables, and raising probabilities to powers for independent events. Part (iii) involves solving simultaneous equations with inverse Normal values, which is slightly more demanding but still routine for S2. No novel problem-solving or proof required—slightly easier than average A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 450) = P\!\left(Z < \frac{450-435}{30}\right)\) | M1 | For standardising; M0 if 'continuity correction' applied |
| \(= P(Z < 0.5) = \Phi(0.5)\) | M1 | For correct structure |
| \(= 0.6915\) | A1 | CAO; Allow 0.692 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(400 < X < 450) = P\!\left(\frac{400-435}{30} < Z < \frac{450-435}{30}\right)\) | ||
| \(= P(-1.1667 < Z < 0.5)\) | ||
| \(= \Phi(0.5) - \Phi(-1.1667)\) | M1 | For correct structure |
| \(= 0.6915 - 0.1216\) | B1 | For use of difference column to obtain 0.8784, 0.8783, 0.1216 or 0.1217; Condone 0.8782 or 0.1218 |
| \(= 0.5699\) | A1 | FT "their \(0.6915\)" \(- 0.1216\) (or 0.1217) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{all 5 between 400 and 450}) = 0.5699^5\) | M1 | |
| \(= 0.0601\) | A1 | FT; Allow 0.060 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P\!\left(Z < \frac{350-\mu}{\sigma}\right) = 0.2\), \(\Phi^{-1}(0.2) = -0.8416\) | M1 | For equation as seen or equivalent with their \(-\)ive \(z\) value; If 'continuity corrections' applied allow M marks but do not award final A marks |
| \(\frac{350-\mu}{\sigma} = -0.8416\) | ||
| \(\Phi^{-1}(0.9) = 1.282\) | B1 | For 1.282 or \(-0.8416\) |
| \(\frac{390-\mu}{\sigma} = 1.282\) | M1 | For equation as seen or equivalent with their \(+\)ive \(z\) value |
| \(350 = \mu - 0.8416\sigma\) | ||
| \(390 = \mu + 1.282\sigma\) | ||
| \(2.1236\sigma = 40\), \(\sigma = 18.84\) | A1 | Allow 18.8; Answers to max 2 d.p. |
| \(\mu = 350 + (0.8416 \times 18.84) = 365.85\) | A1 | Allow 365.86, 366, 365.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Phi^{-1}(0.975) = 1.96\) | B1 | For using a suitable pair of \(z\) values e.g. \(\pm 1.96\) |
| \(a = 365.85 - (1.96 \times 18.84) = 328.9\) | M1 | For either equation provided that a suitable pair of \(z\)-values is used, e.g. \(+2.326\) and \(-1.751\); FT their \(\mu\) and \(\sigma\) to 2 d.p. (A0 if 'continuity correction' used) |
| A1 | Accept any correct values of \(a\) and \(b\) | |
| \(b = 365.85 + (1.96 \times 18.84) = 402.8\) | A1 | FT their \(\mu\) and \(\sigma\) to 2 d.p. (A0 if 'continuity correction' used) |
# Question 3:
## Part (i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 450) = P\!\left(Z < \frac{450-435}{30}\right)$ | M1 | For standardising; M0 if 'continuity correction' applied |
| $= P(Z < 0.5) = \Phi(0.5)$ | M1 | For correct structure |
| $= 0.6915$ | A1 | CAO; Allow 0.692 |
## Part (i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(400 < X < 450) = P\!\left(\frac{400-435}{30} < Z < \frac{450-435}{30}\right)$ | | |
| $= P(-1.1667 < Z < 0.5)$ | | |
| $= \Phi(0.5) - \Phi(-1.1667)$ | M1 | For correct structure |
| $= 0.6915 - 0.1216$ | B1 | For use of difference column to obtain 0.8784, 0.8783, 0.1216 or 0.1217; Condone 0.8782 or 0.1218 |
| $= 0.5699$ | A1 | FT "their $0.6915$" $- 0.1216$ (or 0.1217) |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{all 5 between 400 and 450}) = 0.5699^5$ | M1 | |
| $= 0.0601$ | A1 | FT; Allow 0.060 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P\!\left(Z < \frac{350-\mu}{\sigma}\right) = 0.2$, $\Phi^{-1}(0.2) = -0.8416$ | M1 | For equation as seen or equivalent with their $-$ive $z$ value; If 'continuity corrections' applied allow M marks but do not award final A marks |
| $\frac{350-\mu}{\sigma} = -0.8416$ | | |
| $\Phi^{-1}(0.9) = 1.282$ | B1 | For 1.282 or $-0.8416$ |
| $\frac{390-\mu}{\sigma} = 1.282$ | M1 | For equation as seen or equivalent with their $+$ive $z$ value |
| $350 = \mu - 0.8416\sigma$ | | |
| $390 = \mu + 1.282\sigma$ | | |
| $2.1236\sigma = 40$, $\sigma = 18.84$ | A1 | Allow 18.8; Answers to max 2 d.p. |
| $\mu = 350 + (0.8416 \times 18.84) = 365.85$ | A1 | Allow 365.86, 366, 365.9 |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Phi^{-1}(0.975) = 1.96$ | B1 | For using a suitable pair of $z$ values e.g. $\pm 1.96$ |
| $a = 365.85 - (1.96 \times 18.84) = 328.9$ | M1 | For either equation provided that a suitable pair of $z$-values is used, e.g. $+2.326$ and $-1.751$; FT their $\mu$ and $\sigma$ to 2 d.p. (A0 if 'continuity correction' used) |
| | A1 | Accept any correct values of $a$ and $b$ |
| $b = 365.85 + (1.96 \times 18.84) = 402.8$ | A1 | FT their $\mu$ and $\sigma$ to 2 d.p. (A0 if 'continuity correction' used) |3 The amount of data, $X$ megabytes, arriving at an internet server per second during the afternoon is modelled by the Normal distribution with mean 435 and standard deviation 30.
\begin{enumerate}[label=(\roman*)]
\item Find\\
(A) $\mathrm { P } ( X < 450 )$,\\
(B) $\mathrm { P } ( 400 < X < 450 )$.
\item Find the probability that, during 5 randomly selected seconds, the amounts of data arriving are all between 400 and 450 megabytes.
The amount of data, $Y$ megabytes, arriving at the server during the evening is modelled by the Normal distribution with mean $\mu$ and standard deviation $\sigma$.
\item Given that $\mathrm { P } ( Y < 350 ) = 0.2$ and $\mathrm { P } ( Y > 390 ) = 0.1$, find the values of $\mu$ and $\sigma$.
\item Find values of $a$ and $b$ for which $\mathrm { P } ( a < Y < b ) = 0.95$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2013 Q3 [17]}}