| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Identify outliers using IQR rule |
| Difficulty | Moderate -0.8 This is a straightforward application of standard statistical procedures: calculating mean and standard deviation from a small dataset, then applying the IQR rule (a routine formula) to identify an outlier. The calculations are simple with only 8 values, and the discussion in part (iii) requires basic contextual reasoning rather than mathematical insight. This is easier than average A-level work. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Mean} = 83.95/8 = 10.49\) | B1 | |
| \(\text{Variance} = \dfrac{881.2119 - \dfrac{83.95^2}{8}}{7} = 0.03737\) | M1 | |
| Standard deviation \(= 0.193\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 2 standard deviations below mean \(= 10.49 - 2(0.193) = 10.104\) | M1 | Follow through if divisor \(n\) has been used above |
| but \(10.04 < 10.104\), so 10.04 is an outlier | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| This time is much faster than the others. This may be the result of wind assistance, faulty timing, false start and should be discarded. | E1 | Appreciating need for investigation |
| Opposite conclusion e.g. could be a genuinely fast time, can also receive full credit | E1 | Comment in context |
## Question 2:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Mean} = 83.95/8 = 10.49$ | B1 | |
| $\text{Variance} = \dfrac{881.2119 - \dfrac{83.95^2}{8}}{7} = 0.03737$ | M1 | |
| Standard deviation $= 0.193$ | A1 | |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| 2 standard deviations below mean $= 10.49 - 2(0.193) = 10.104$ | M1 | Follow through if divisor $n$ has been used above |
| but $10.04 < 10.104$, so 10.04 is an outlier | A1 | |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| This time is much faster than the others. This may be the result of wind assistance, faulty timing, false start and should be discarded. | E1 | Appreciating need for investigation |
| Opposite conclusion e.g. could be a genuinely fast time, can also receive full credit | E1 | Comment in context |
---2 A sprinter runs many 100 -metre trials, and the time, $x$ seconds, for each is recorded. A sample of eight of these times is taken, as follows.
$$\begin{array} { l l l l l l l l }
10.53 & 10.61 & 10.04 & 10.49 & 10.63 & 10.55 & 10.47 & 10.63
\end{array}$$
(i) Calculate the sample mean, $\bar { x }$, and sample standard deviation, $s$, of these times.\\
(ii) Show that the time of 10.04 seconds may be regarded as an outlier.\\
(iii) Discuss briefly whether or not the time of 10.04 seconds should be discarded.
\hfill \mbox{\textit{OCR MEI S1 2005 Q2 [7]}}