| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 6 |
| 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 formulas for mean and standard deviation from summary statistics, followed by routine application of linear transformation rules (mean and SD under y = ax + b). Both parts require only direct substitution into well-known formulas with no problem-solving or conceptual challenges. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02j Clean data: missing data, errors |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \frac{59972}{40} = 1499\) | B1 | CAO Ignore units. NB Allow 1500. NB Answer must be decimal |
| \(S_{xx} = 96767028 - \frac{59972^2}{40} = 6851008\) | M1 | For \(S_{xx}\). M1 for \(96767028 - 40 \times \text{their mean}^2\) BUT NOTE M0 if their \(S_{xx} < 0\) |
| \(s = \sqrt{\frac{6851008}{39}} = \sqrt{175667} = 419\) | A1 | CAO ignore units. NB Full answer is \(419.1263\) (but only allow to 4sf due to over-specification rule). For \(s^2\) of \(176000\) (or better) allow M1A0 with or without working. For RMSD of \(414\) (or better) allow M1A0 provided working seen. For RMSD\(^2\) of \(171000\) (or better) allow M1A0 provided working seen. |
| Answer | Marks | Guidance |
|---|---|---|
| New mean \(= (0.163 \times 1499) + 14.5 = £258.84\) | B1 | FT their mean provided answer is positive. Allow \(£259\) or \(£259.00\) from \(1500\) or \(£258.89\) from \(1499.3\). Condone \(258.8\) and \(258.9\). Accept answers rounded to \(3\)sf or more eg \(£258.80\), \(£258.90\). Do not penalise lack of units in mean. (No penalty for giving to 5sf as this is an exact sum of money) |
| New sd \(= 0.163 \times 419 = £68.30\) | M1 | FT their sd for M1 and A1 |
| A1 | Allow \(£68.29\) to \(£68.32\). Allow \(68.3\). Or for \(0.163 \times 419.1\) oe. Do not penalise lack of units in sd. |
# Question 1
## (i)
Mean $= \frac{59972}{40} = 1499$ | B1 | CAO Ignore units. NB Allow 1500. NB Answer must be decimal
$S_{xx} = 96767028 - \frac{59972^2}{40} = 6851008$ | M1 | For $S_{xx}$. M1 for $96767028 - 40 \times \text{their mean}^2$ BUT NOTE M0 if their $S_{xx} < 0$
$s = \sqrt{\frac{6851008}{39}} = \sqrt{175667} = 419$ | A1 | CAO ignore units. NB Full answer is $419.1263$ (but only allow to 4sf due to over-specification rule). For $s^2$ of $176000$ (or better) allow M1A0 with or without working. For RMSD of $414$ (or better) allow M1A0 provided working seen. For RMSD$^2$ of $171000$ (or better) allow M1A0 provided working seen.
**Condone full answer of $1499.3$ (despite over-specification rule)**
**For use of $1499$: $S_{xx} = 6886988$, $s^2 = 176589$, $s = 420.225$, RMSD $= 414.9$**
**For use of $1500$: $S_{xx} = 6767028$, $s^2 = 173513.5$, $s = 416.549$, RMSD $= 411.3$**
**Give same credit to answers as for correct answers**
## (ii)
New mean $= (0.163 \times 1499) + 14.5 = £258.84$ | B1 | FT their mean provided answer is positive. Allow $£259$ or $£259.00$ from $1500$ or $£258.89$ from $1499.3$. Condone $258.8$ and $258.9$. Accept answers rounded to $3$sf or more eg $£258.80$, $£258.90$. Do not penalise lack of units in mean. (No penalty for giving to 5sf as this is an exact sum of money)
New sd $= 0.163 \times 419 = £68.30$ | M1 | FT their sd for M1 and A1
| A1 | Allow $£68.29$ to $£68.32$. Allow $68.3$. Or for $0.163 \times 419.1$ oe. Do not penalise lack of units in sd.
**If candidate 'starts again' only award marks for CAO**
**Deduct at most 1 mark overall in whole question for over-specification of either mean or SD or both**1 The amounts of electricity, $x \mathrm { kWh }$ (kilowatt hours), used by 40 households in a three-month period are summarised as follows.
$$n = 40 \quad \sum x = 59972 \quad \sum x ^ { 2 } = 96767028$$
(i) Calculate the mean and standard deviation of $x$.\\
(ii) The formula $y = 0.163 x + 14.5$ gives the cost in pounds of the electricity used by each household. Use your answers to part (i) to deduce the mean and standard deviation of the costs of the electricity used by these 40 households.
\hfill \mbox{\textit{OCR MEI S1 Q1 [6]}}