| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Direct cumulative frequency graph reading |
| Difficulty | Easy -1.2 This is a straightforward cumulative frequency question requiring only direct reading from a graph (median at 300th value, quartiles at 150th and 450th values), basic outlier calculation using 1.5×IQR rule, frequency table completion by subtraction, and standard formula applications for mean calculation and linear transformations of standard deviation. All techniques are routine S1 procedures with no problem-solving or novel insight required. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers5.02c Linear coding: effects on mean and variance |
| \(3.0 \leqslant t \leqslant 3.4\) | \(3.4 < t \leqslant 3.8\) | \(3.8 < t \leqslant 4.2\) | \(4.2 < t \leqslant 4.6\) | \(4.6 < t \leqslant 5.0\) | ||
| Frequency | 243 | 157 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Median \(= 4.06 - 4.075\) (inclusive) | B1 cao | |
| \(Q_1 = 3.8\) | B1 | For \(Q_1\) (cao) |
| \(Q_3 = 4.3\) | B1 | For \(Q_3\) (cao) |
| Inter-quartile range \(= 4.3 - 3.8 = 0.5\) | B1 ft | IQR must be using t-values not locations to earn this mark |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Lower limit: their \(3.8 - 1.5 \times\) their \(0.5 = (3.05)\) | B1 ft | Must have \(-1.5\) |
| Upper limit: their \(4.3 + 1.5 \times\) their \(0.5 = (5.05)\) | B1 ft | Must have \(+1.5\) |
| Very few if any temperatures below 3.05 (but not zero) | E1 ft | dep on \(-1.5\) and \(Q_1\) |
| None above 5.05 | E1 ft | dep on \(+1.5\) and \(Q_3\); must be using t-values NOT locations |
| 'So few, if any outliers' scores SC1 | SC1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Valid argument such as 'Probably not, because there is nothing to suggest they are not genuine data items; they do not appear to form a separate pool of data.' Accept: exclude outlier – 'measuring equipment was wrong' or 'there was a power cut' or ref to hot/cold day | E1 | |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Missing frequencies 25, 125, 50 | B1, B1, B1 | All cao |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean \(= (3.2 \times 25 + 3.6 \times 125 + 4.0 \times 243 + 4.4 \times 157 + 4.8 \times 50)/600\) | M1 | For at least 4 midpoints correct and being used in attempt to find \(\Sigma ft\) |
| \(= 2432.8/600 = 4.05(47)\) | A1 cao | awfw (4.05 – 4.055) ISW or rounding |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| New mean \(= 1.8 \times\) their \(4.05(47) + 32 = 39.29(84)\) to 39.3 | B1 FT | |
| New \(s = 1.8 \times 0.379 = 0.682\) | M1 | For \(1.8 \times 0.379\) |
| A1 CAO | awfw (0.68 – 0.6822) | |
| TOTAL | 17 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Median $= 4.06 - 4.075$ (inclusive) | B1 cao | |
| $Q_1 = 3.8$ | B1 | For $Q_1$ (cao) |
| $Q_3 = 4.3$ | B1 | For $Q_3$ (cao) |
| Inter-quartile range $= 4.3 - 3.8 = 0.5$ | B1 ft | IQR must be using t-values not locations to earn this mark |
| **Total** | **4** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Lower limit: their $3.8 - 1.5 \times$ their $0.5 = (3.05)$ | B1 ft | Must have $-1.5$ |
| Upper limit: their $4.3 + 1.5 \times$ their $0.5 = (5.05)$ | B1 ft | Must have $+1.5$ |
| Very few if any temperatures below 3.05 (but not zero) | E1 ft | dep on $-1.5$ and $Q_1$ |
| None above 5.05 | E1 ft | dep on $+1.5$ and $Q_3$; must be using t-values NOT locations |
| 'So few, if any outliers' scores SC1 | SC1 | |
| **Total** | **4** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Valid argument such as 'Probably not, because there is nothing to suggest they are not genuine data items; they do not appear to form a separate pool of data.' Accept: exclude outlier – 'measuring equipment was wrong' or 'there was a power cut' or ref to hot/cold day | E1 | |
| **Total** | **1** | |
### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Missing frequencies 25, 125, 50 | B1, B1, B1 | All cao |
| **Total** | **3** | |
### Part (v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= (3.2 \times 25 + 3.6 \times 125 + 4.0 \times 243 + 4.4 \times 157 + 4.8 \times 50)/600$ | M1 | For at least 4 midpoints correct and being used in attempt to find $\Sigma ft$ |
| $= 2432.8/600 = 4.05(47)$ | A1 cao | awfw (4.05 – 4.055) ISW or rounding |
| **Total** | **2** | |
### Part (vi):
| Answer/Working | Marks | Guidance |
|---|---|---|
| New mean $= 1.8 \times$ their $4.05(47) + 32 = 39.29(84)$ to 39.3 | B1 FT | |
| New $s = 1.8 \times 0.379 = 0.682$ | M1 | For $1.8 \times 0.379$ |
| | A1 CAO | awfw (0.68 – 0.6822) |
| **TOTAL** | **17** | |
---6 The temperature of a supermarket fridge is regularly checked to ensure that it is working correctly. Over a period of three months the temperature (measured in degrees Celsius) is checked 600 times. These temperatures are displayed in the cumulative frequency diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{7b92607f-1bf9-45f0-997b-fe76c88b5fcd-4_1054_1649_539_248}\\
(i) Use the diagram to estimate the median and interquartile range of the data.\\
(ii) Use your answers to part (i) to show that there are very few, if any, outliers in the sample.\\
(iii) Suppose that an outlier is identified in these data. Discuss whether it should be excluded from any further analysis.\\
(iv) Copy and complete the frequency table below for these data.
\begin{center}
\begin{tabular}{ | l | l | l | c | c | c | }
\hline
\begin{tabular}{ l }
Temperature \\
$( t$ degrees Celsius $)$ \\
\end{tabular} & $3.0 \leqslant t \leqslant 3.4$ & $3.4 < t \leqslant 3.8$ & $3.8 < t \leqslant 4.2$ & $4.2 < t \leqslant 4.6$ & $4.6 < t \leqslant 5.0$ \\
\hline
Frequency & & & 243 & 157 & \\
\hline
\end{tabular}
\end{center}
(v) Use your table to calculate an estimate of the mean.\\
(vi) The standard deviation of the temperatures in degrees Celsius is 0.379 . The temperatures are converted from degrees Celsius into degrees Fahrenheit using the formula $F = 1.8 C + 32$. Hence estimate the mean and find the standard deviation of the temperatures in degrees Fahrenheit.
\hfill \mbox{\textit{OCR MEI S1 2009 Q6 [17]}}