| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| 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. The calculations are routine for S1 level. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \frac{3026}{56} = 54.0\) | 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 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} + 2s = 54.0 + 2 \times 16.7 = 87.4\), so 93 is an outlier | M1 | for their \(\bar{x} + 2\times\) their \(s\) |
| A1 FT | for 87.4 and comment |
| Answer | Marks |
|---|---|
| New mean \(= 1.2 \times 54.0 - 10 = 54.8\) | B1 FT |
| New \(s = 1.2 \times 16.7 = 20.1\) | M1A1 FT |
## Question 2:
### Part (i)
| $\text{Mean} = \frac{3026}{56} = 54.0$ | 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 | |
### Part (ii)
| $\bar{x} + 2s = 54.0 + 2 \times 16.7 = 87.4$, so 93 is an outlier | M1 | for their $\bar{x} + 2\times$ their $s$ |
| | A1 FT | for 87.4 and comment |
### Part (iii)
| New mean $= 1.2 \times 54.0 - 10 = 54.8$ | B1 FT | |
| New $s = 1.2 \times 16.7 = 20.1$ | M1A1 FT | |
---2 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 Q2 [8]}}