| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Skewness identification and interpretation |
| Difficulty | Moderate -0.8 This is a straightforward statistics question requiring identification of skewness from a stem-and-leaf diagram (visual inspection), calculator use for mean/standard deviation, and application of the standard outlier formula (Q1 - 1.5×IQR and Q3 + 1.5×IQR). All parts are routine procedures with no problem-solving or conceptual challenge beyond basic S1 content. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| 3 | 4 | 6 | 7 | 8 | 9 | 9 | ||
| 4 | 0 | 2 | 2 | 3 | 4 | 6 | 8 | 9 |
| 5 | 0 | 1 | 3 | 5 | 8 | |||
| 6 | 2 | 4 | 5 | |||||
| 7 | 4 | 6 | ||||||
| 8 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Positive | [1] | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 5.064\) (allow 5.1 with working \(126.6/25\) or 5.06 without) | B1 | Also allow B1 for \(S_{xx} = 42.08\) or \(\sum x^2 = 683\) |
| SD \(= 1.324\) (allow 1.3 with working or 1.32 without) | B2 | Allow B1 for RMSD \(= 1.297\) or var \(= 1.753\) or MSD \(= 1.683\); SC1 for both mean \(= 50.64\) and SD \(= 13.24\) (even if over-specified) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} - 2s = 5.064 - 2 \times 1.324 = 2.416\) | B1FT | FT their mean and sd; For use of quartiles and IQR: \(Q_1 = 3.95\), \(Q_3 = 6.0\), IQR \(= 2.05\); \(3.95 - 1.5(2.05)\) gets M1 |
| \(\bar{x} + 2s = 5.064 + 2 \times 1.324 = 7.712\) | M1 | For \(\bar{x} + 2s\) but withhold final E mark if their limits mean there are no outliers; \(6.0 + 1.5(2.05)\) gets M1 |
| A1FT | For upper limit; Limits 0.875 and 9.075 | |
| So there is an outlier. | E1 | Incorrect statement such as 7.6 and 8.1 are outliers gets E0; Do not award E1 if calculation error in upper limit; FT from SC1 |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Positive | [1] | CAO |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 5.064$ (allow 5.1 with working $126.6/25$ or 5.06 without) | B1 | Also allow B1 for $S_{xx} = 42.08$ or $\sum x^2 = 683$ |
| SD $= 1.324$ (allow 1.3 with working or 1.32 without) | B2 | Allow B1 for RMSD $= 1.297$ or var $= 1.753$ or MSD $= 1.683$; SC1 for both mean $= 50.64$ and SD $= 13.24$ (even if over-specified) |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} - 2s = 5.064 - 2 \times 1.324 = 2.416$ | B1FT | FT their mean and sd; For use of quartiles and IQR: $Q_1 = 3.95$, $Q_3 = 6.0$, IQR $= 2.05$; $3.95 - 1.5(2.05)$ gets M1 |
| $\bar{x} + 2s = 5.064 + 2 \times 1.324 = 7.712$ | M1 | For $\bar{x} + 2s$ but withhold final E mark if their limits mean there are no outliers; $6.0 + 1.5(2.05)$ gets M1 |
| | A1FT | For upper limit; Limits 0.875 and 9.075 |
| So there is an outlier. | E1 | Incorrect statement such as 7.6 and 8.1 are outliers gets E0; Do not award E1 if calculation error in upper limit; FT from SC1 |
---3 The stem and leaf diagram illustrates the heights in metres of 25 young oak trees.
\begin{center}
\begin{tabular}{ l | l l l l l l l l }
3 & 4 & 6 & 7 & 8 & 9 & 9 & & \\
4 & 0 & 2 & 2 & 3 & 4 & 6 & 8 & 9 \\
5 & 0 & 1 & 3 & 5 & 8 & & & \\
6 & 2 & 4 & 5 & & & & & \\
7 & 4 & 6 & & & & & & \\
8 & 1 & & & & & & & \\
\end{tabular}
\end{center}
Key: 4 |2 represents 4.2\\
(i) State the type of skewness of the distribution.\\
(ii) Use your calculator to find the mean and standard deviation of these data.\\
(iii) Determine whether there are any outliers.
\hfill \mbox{\textit{OCR MEI S1 Q3 [8]}}