| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Topic | Measures of Location and Spread |
| Type | Identify outliers using mean and standard deviation |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring only standard calculations (mean and standard deviation from a calculator) and application of the basic outlier rule (typically values more than 2 standard deviations from the mean). No problem-solving or conceptual insight is needed—just routine statistical procedures that are well below average A-level difficulty. |
| Spec | 2.02g Calculate mean and standard deviation2.02h Recognize outliers |
**Question 1(i)**
- Mean $= 1365/18 = 75.8(33\ldots)$ **B1**
- $\text{sd} = \sqrt{\dfrac{111381}{18} - 75.83...^2} = 20.9(07...)$ **B1** (Accept unbiased estimate $= 21.5(14\ldots)$)
**Question 1(ii)**
- Either threshold found correctly **M1**
- Lower limit $= 75.833 - 2 \times 20.907 = 34.0(17\ldots)$
Upper limit $= 75.833 + 2 \times 20.907 = 117(.64\ldots)$ **A1** (Both thresholds found correctly. FT *their* mean and sd.)
- $\therefore$ 19 is the only outlier. **A1** (C.a.o., but FT *their* mean and sd provided 19 only is identified as an outlier.)
**Total: 5 marks**1 Levels of nitrogen dioxide in the atmosphere are being monitored at the side of a road in a busy city centre. A sample of 18 measurements taken (in suitable units) is as follows.
$$\begin{array} { l l l l l l l l l l l l l l l l l l }
83 & 44 & 95 & 92 & 98 & 63 & 69 & 76 & 19 & 91 & 70 & 91 & 74 & 65 & 62 & 70 & 95 & 108
\end{array}$$
(i) Find the mean and standard deviation of the sample.\\
(ii) Hence identify, with justification, any possible outliers.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2017 Q1 [5]}}