| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify outliers using mean and standard deviation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for mean, standard deviation, outlier detection (using the 2 standard deviations rule), and linear transformations of data. All parts require only direct substitution into well-known formulas with no problem-solving or conceptual insight needed—easier than average A-level questions. |
| Spec | 2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \frac{3026}{56} = 54.0\) | B1 for mean | |
| \(S_{xx} = 178890 - \frac{3026^2}{56} = 15378\) | M1 for attempt at \(S_{xx}\) | |
| \(s = \sqrt{\frac{15378}{55}} = 16.7\) | A1 CAO | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} + 2s = 54.0 + 2\times16.7 = 87.4\), so 93 is an outlier | M1 for their \(\bar{x} + 2\times\) their \(s\), A1 FT for 87.4 and comment | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| New \(s = 1.2\times16.7 = 20.1\) | B1 FT, M1A1 FT | 3 marks |
# Question 3:
## Part (i)
Mean $= \frac{3026}{56} = 54.0$ | B1 for mean |
$S_{xx} = 178890 - \frac{3026^2}{56} = 15378$ | M1 for attempt at $S_{xx}$ |
$s = \sqrt{\frac{15378}{55}} = 16.7$ | A1 CAO | 3 marks
## Part (ii)
$\bar{x} + 2s = 54.0 + 2\times16.7 = 87.4$, so 93 is an outlier | M1 for their $\bar{x} + 2\times$ their $s$, A1 FT for 87.4 and comment | 2 marks
## Part (iii)
New mean $= 1.2\times54.0 - 10 = 54.8$
New $s = 1.2\times16.7 = 20.1$ | B1 FT, M1A1 FT | 3 marks
---3 The marks $x$ scored by a sample of 56 students in an examination are summarised by
$$n = 56 , \quad \Sigma x = 3026 , \quad \Sigma x ^ { 2 } = 178890 .$$
(i) Calculate the mean and standard deviation of the marks.\\
(ii) The highest mark scored by any of the 56 students in the examination was 93 . Show that this result may be considered to be an outlier.\\
(iii) The formula $y = 1.2 x - 10$ is used to scale the marks. Find the mean and standard deviation of the scaled marks.
\hfill \mbox{\textit{OCR MEI S1 2007 Q3 [8]}}