| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Rounded or discrete from continuous |
| Difficulty | Standard +0.3 This is a standard Normal distribution question with continuity correction, requiring routine application of z-scores and inverse Normal calculations. All parts follow textbook procedures: (i-ii) apply continuity correction and use tables, (iii) use binomial probability, (iv) find inverse Normal for 90th percentile, (v) find mean given a percentile. No novel insight or complex multi-step reasoning required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y = 76) = P\!\left(\frac{75.5-76}{12} \leq Z \leq \frac{76.5-76}{12}\right)\) | B1 | For one correct continuity correction used |
| \(= P(-0.04166\ldots < Z < 0.04166\ldots)\) | M1 | For standardising |
| \(= \Phi(0.04166\ldots) - (1 - \Phi(0.04166\ldots))\) | ||
| \(= 2 \times \Phi(0.04166\ldots) - 1\) | M1 | For correctly structured probability calculation. |
| \(= 2 \times 0.5167 - 1\) | ||
| \(= 0.0334\) | A1 | CAO inc use of diff tables. Allow \(0.0330 - 0.0340\) www. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y \geq 80) = P\!\left(Z \geq \frac{79.5 - 76}{12}\right)\) | B1 | For correct cc used |
| \(= P(Z > 0.2917) = 1 - \Phi(0.2917)\) | M1 | For correct structure |
| \(= 1 - 0.6148 = 0.3852 = 0.385\) to 3 sig fig | A1 | CAO do not allow 0.386 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3 \times 0.3852 \times 0.6148^2 = 0.4368\) | M1 | \(3 \times \text{their } p \times (1 - \text{their } p)^2\) |
| A1 | FT their \(p\). Allow 2sf if working seen. | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(P(\text{Score} \geq k) = 0.1\) | ||
| \(\Phi^{-1}(0.9) = 1.282\) | B1 | For 1.282 |
| \(\frac{k - 76}{12} = 1.282\) | M1 | Allow \(k - 0.5\) used for \(k\). Positive \(z\) used. |
| \(k = 76 + (1.282 \times 12) = 91.38\) or \(k = 76 + 0.5 + (1.282 \times 12) = 91.88\) | A1 | For 91.38 or 91.88 |
| \(91.38 > 90.5\) or \(91.88 > 91\) so lowest reported mark \(= 92\) | M1 | Relevant comparison (e.g. diagram) |
| A1 | ||
| OR Trial and improvement method | M1 | M1 for attempt to find \(P(\text{Mark} \geq \text{integer})\) |
| \(P(\text{Mark} \geq 91) = P(\text{Score} \geq 90.5) = 0.1135\) | A1 | A1 for 0.1135 |
| \(P(\text{Mark} \geq 92) = P(\text{Score} \geq 91.5) = 0.0982\) | A1 | A1 for 0.0982 |
| \(P(\text{Mark} \geq 91) > 10\%\) and \(P(\text{Mark} \geq 92) < 10\%\) | M1 | M1 for comparisons |
| so lowest reported mark \(= 92\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y \leq 50) = 0.2\) | ||
| \(P\!\left(Z \leq \frac{50.5 - \mu}{12}\right) = 0.2\) | B1 | For 50.5 used |
| \(\frac{50.5 - \mu}{12} = \Phi^{-1}(0.2) = -0.8416\) | B1 | For \(-0.8416\). Condone \(-0.842\). Condone 0.8416 if numerator reversed. |
| M1 | For structure. | |
| \(\mu = 50.5 + (12 \times 0.8416) = 60.6\) | A1 | CAO |
| [4] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y = 76) = P\!\left(\frac{75.5-76}{12} \leq Z \leq \frac{76.5-76}{12}\right)$ | B1 | For one correct continuity correction used |
| $= P(-0.04166\ldots < Z < 0.04166\ldots)$ | M1 | For standardising |
| $= \Phi(0.04166\ldots) - (1 - \Phi(0.04166\ldots))$ | | |
| $= 2 \times \Phi(0.04166\ldots) - 1$ | M1 | For correctly structured probability calculation. |
| $= 2 \times 0.5167 - 1$ | | |
| $= 0.0334$ | A1 | CAO inc use of diff tables. Allow $0.0330 - 0.0340$ www. |
| | **[4]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y \geq 80) = P\!\left(Z \geq \frac{79.5 - 76}{12}\right)$ | B1 | For correct cc used |
| $= P(Z > 0.2917) = 1 - \Phi(0.2917)$ | M1 | For correct structure |
| $= 1 - 0.6148 = 0.3852 = 0.385$ to 3 sig fig | A1 | CAO do not allow 0.386 |
| | **[3]** | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 \times 0.3852 \times 0.6148^2 = 0.4368$ | M1 | $3 \times \text{their } p \times (1 - \text{their } p)^2$ |
| | A1 | FT their $p$. Allow 2sf if working seen. |
| | **[2]** | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $P(\text{Score} \geq k) = 0.1$ | | |
| $\Phi^{-1}(0.9) = 1.282$ | B1 | For 1.282 |
| $\frac{k - 76}{12} = 1.282$ | M1 | Allow $k - 0.5$ used for $k$. Positive $z$ used. |
| $k = 76 + (1.282 \times 12) = 91.38$ or $k = 76 + 0.5 + (1.282 \times 12) = 91.88$ | A1 | For 91.38 or 91.88 |
| $91.38 > 90.5$ or $91.88 > 91$ so lowest reported mark $= 92$ | M1 | Relevant comparison (e.g. diagram) |
| | A1 | |
| **OR** Trial and improvement method | M1 | M1 for attempt to find $P(\text{Mark} \geq \text{integer})$ |
| $P(\text{Mark} \geq 91) = P(\text{Score} \geq 90.5) = 0.1135$ | A1 | A1 for 0.1135 |
| $P(\text{Mark} \geq 92) = P(\text{Score} \geq 91.5) = 0.0982$ | A1 | A1 for 0.0982 |
| $P(\text{Mark} \geq 91) > 10\%$ and $P(\text{Mark} \geq 92) < 10\%$ | M1 | M1 for comparisons |
| so lowest reported mark $= 92$ | A1 | |
| | **[5]** | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y \leq 50) = 0.2$ | | |
| $P\!\left(Z \leq \frac{50.5 - \mu}{12}\right) = 0.2$ | B1 | For 50.5 used |
| $\frac{50.5 - \mu}{12} = \Phi^{-1}(0.2) = -0.8416$ | B1 | For $-0.8416$. Condone $-0.842$. Condone 0.8416 if numerator reversed. |
| | M1 | For structure. |
| $\mu = 50.5 + (12 \times 0.8416) = 60.6$ | A1 | CAO |
| | **[4]** | |3 The scores, $X$, in Paper 1 of an English examination have an underlying Normal distribution with mean 76 and standard deviation 12. The scores are reported as integer marks. So, for example, a score for which $75.5 \leqslant X < 76.5$ is reported as 76 marks.\\
(i) Find the probability that a candidate's reported mark is 76 .\\
(ii) Find the probability that a candidate's reported mark is at least 80 .\\
(iii) Three candidates are chosen at random. Find the probability that exactly one of these three candidates' reported marks is at least 80 .
The proportion of candidates who receive an A* grade (the highest grade) must not exceed $10 \%$ but should be as close as possible to $10 \%$.\\
(iv) Find the lowest reported mark that should be awarded an A* grade.
The scores in Paper 2 of the examination have an underlying Normal distribution with mean $\mu$ and standard deviation 12.\\
(v) Given that $20 \%$ of candidates receive a reported mark of 50 or less, find the value of $\mu$.
\hfill \mbox{\textit{OCR MEI S2 2013 Q3 [18]}}