| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Forward transformation: find new statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of standard transformation formulas for mean and standard deviation. Part (i) uses direct formulas from summary statistics, while parts (ii) and (iii) require only recall of how linear transformations affect measures of location and spread (multiplication affects both mean and SD; addition affects only mean). No problem-solving or conceptual insight needed beyond textbook rules. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \frac{759.00}{60} = £12.65\) | B1 | Ignore units |
| \(S_{xx} = 11736.59 - \frac{759^2}{60} = 2135.24\) | M1 | For \(S_{xx}\); M1 for \(11736.59 - 60 \times \text{their mean}^2\); BUT NOTE M0 if their \(S_{xx} < 0\) |
| \(s = \sqrt{\frac{2135.24}{59}} = £6.02\) | A1 | CAO ignore units; Allow more accurate answers |
| [3] | CAO Do not allow \(\frac{759}{60}\) as final answer but allow \(12\frac{13}{20}\); For \(s^2\) of 36.2 (or better) allow M1A0 with or without working; For RMSD of 5.97 or 5.96 (or better) allow M1A0 provided working seen; For RMSD\(^2\) of 35.6 (or better) allow M1A0 provided working seen |
| Answer | Marks | Guidance |
|---|---|---|
| New mean \(= 12.65 \times 1.02 = £12.90\) | B1 | FT their mean; Awrt 12.90; Allow 12.9 |
| New sd \(= 6.02 \times 1.02 = £6.14\) | B1 | FT their sd |
| [2] | If candidate 'starts again' only award marks for CAO; Deduct at most 1 mark overall in whole question for overspecification of Mean and 1 mark overall for SD |
| Answer | Marks | Guidance |
|---|---|---|
| New mean \(= 12.65 + 0.25 = £12.90\) | B1 | FT their mean; Awrt 12.90 |
| New sd \(= £6.02\) | B1 | FT their sd (unless negative); Awrt 6.02; Allow sd unchanged (or similar) |
| [2] | If candidate 'starts again' only award marks for CAO |
# Question 1
## (i)
Mean $= \frac{759.00}{60} = £12.65$ | B1 | Ignore units
$S_{xx} = 11736.59 - \frac{759^2}{60} = 2135.24$ | M1 | For $S_{xx}$; M1 for $11736.59 - 60 \times \text{their mean}^2$; BUT NOTE M0 if their $S_{xx} < 0$
$s = \sqrt{\frac{2135.24}{59}} = £6.02$ | A1 | CAO ignore units; Allow more accurate answers
**[3]** | CAO Do not allow $\frac{759}{60}$ as final answer but allow $12\frac{13}{20}$; For $s^2$ of 36.2 (or better) allow M1A0 with or without working; For RMSD of 5.97 or 5.96 (or better) allow M1A0 provided working seen; For RMSD$^2$ of 35.6 (or better) allow M1A0 provided working seen
## (ii)
New mean $= 12.65 \times 1.02 = £12.90$ | B1 | FT their mean; Awrt 12.90; Allow 12.9
New sd $= 6.02 \times 1.02 = £6.14$ | B1 | FT their sd
**[2]** | If candidate 'starts again' only award marks for CAO; Deduct at most 1 mark overall in whole question for overspecification of Mean and 1 mark overall for SD
## (iii)
New mean $= 12.65 + 0.25 = £12.90$ | B1 | FT their mean; Awrt 12.90
New sd $= £6.02$ | B1 | FT their sd (unless negative); Awrt 6.02; Allow sd unchanged (or similar)
**[2]** | If candidate 'starts again' only award marks for CAO1 The hourly wages, $\pounds x$, of a random sample of 60 employees working for a company are summarised as follows.
$$n = 60 \quad \sum x = 759.00 \quad \sum x ^ { 2 } = 11736.59$$
(i) Calculate the mean and standard deviation of $x$.\\
(ii) The workers are offered a wage increase of $2 \%$. Use your answers to part (i) to deduce the new mean and standard deviation of the hourly wages after this increase.\\
(iii) As an alternative the workers are offered a wage increase of 25 p per hour. Write down the new mean and standard deviation of the hourly wages after this 25p increase.
\hfill \mbox{\textit{OCR MEI S1 Q1 [7]}}