| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Vertical line chart construction |
| Difficulty | Easy -1.8 This is a straightforward statistics question requiring basic skills: drawing a simple vertical line chart from a frequency table, calculating mean and RMSD using standard formulas with small integers, and applying a linear transformation (present = 30 - absent). All techniques are routine recall with no problem-solving or insight required, making it easier than the typical A-level question. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Number of absentees | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(> 6\) |
| Frequency | 8 | 15 | 11 | 8 | 3 | 4 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correctly labelled bar chart | G1, G1 | G1 labelled linear scales on both axes; G1 heights |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= \frac{99}{50} = 1.98\) | B1 | |
| \(S_{xx} = 315 - \frac{99^2}{50} = 118.98\) | M1 | M1 for attempt at \(S_{xx}\) |
| \(rmsd = \sqrt{\frac{118.98}{50}} = 1.54\) | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New mean \(= 30 - 1.98 = 28.02\) | B1 FT | FT their mean |
| New \(rmsd = 1.54\) (unchanged) | B1 FT | FT their rmsd |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correctly labelled bar chart | G1, G1 | G1 labelled linear scales on both axes; G1 heights |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= \frac{99}{50} = 1.98$ | B1 | |
| $S_{xx} = 315 - \frac{99^2}{50} = 118.98$ | M1 | M1 for attempt at $S_{xx}$ |
| $rmsd = \sqrt{\frac{118.98}{50}} = 1.54$ | A1 CAO | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New mean $= 30 - 1.98 = 28.02$ | B1 FT | FT their mean |
| New $rmsd = 1.54$ (unchanged) | B1 FT | FT their rmsd |
---2 The numbers of absentees per day from Mrs Smith's reception class over a period of 50 days are summarised below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Number of absentees & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $> 6$ \\
\hline
Frequency & 8 & 15 & 11 & 8 & 3 & 4 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
(i) Illustrate these data by means of a vertical line chart.\\
(ii) Calculate the mean and root mean square deviation of these data.\\
(iii) There are 30 children in Mrs Smith's class altogether. Find the mean and root mean square deviation of the number of children who are present during the 50 days.
\hfill \mbox{\textit{OCR MEI S1 2007 Q2 [7]}}