| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Direct frequency calculation from histogram |
| Difficulty | Moderate -0.8 This is a standard S1 histogram question requiring routine frequency density calculations and basic statistical measures. While multi-part with 5 sections, each part follows textbook procedures: reading frequencies from histogram bars, calculating mean from grouped data, finding median position, and making standard comments about skewness. No novel problem-solving or insight required—purely methodical application of learned techniques. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| One large square \(= \frac{450}{\text{"22.5"}}\) or one small square \(= \frac{450}{\text{"562.5"}}\) | M1 | For attempt to count squares (accept "22.5" in [22,23] and "562.5" in [550,575]) and use 450 to obtain measure of scale |
| One large square \(= 20\) cars or one small square \(= 0.8\) cars or 1 car \(= 1.25\) squares | A1 | For a correct calc. for 20 or 0.8 or 1.25 etc |
| No. \(> 35\) mph is: \(4.5 \times \text{"20"}\) or \(112.5 \times \text{"0.8"}\) | dM1 | Dependent on 1st M1; for correctly counting squares for \(> 35\) mph and forming suitable expression |
| \(= 90\) (cars) | A1 | For 90 with no incorrect working seen; e.g. \(\frac{4.5}{22.5} \times 450\) scores M1A1M1 and A1 when \(= 90\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{30 \times 12.5 + 240 \times 25 + 90 \times 32.5 + 30 \times 37.5 + 60 \times 42.5}{450} = \frac{12975}{450}\) | M1 | 1st M1 for clear sensible use of midpoints; at least 3 of (12.5, 25, 32.5, 37.5, 42.5) seen |
| M1 | 2nd M1 for expression for \(\bar{x}\) with at least 3 correct terms on numerator and compatible denominator | |
| \(= 28.83...\) or \(\frac{173}{6}\) awrt \(\mathbf{28.8}\) | A1 | For awrt 28.8 (answer only is 3/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Q_2 = 20 + \frac{195}{240} \times 10\) | M1 | For a full expression for median using their frequencies; do not accept boundaries of 19.5 or 20.5 (M0A0) |
| \(= 28.125\) awrt \(\mathbf{28.1}\) | A1 | For awrt 28.1 (answer only is 2/2); use of \((n+1)\) accept 28.15 but not 28.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Q_2 < \bar{x}\), so positive skew | B1ft | For correct statement about their \(Q_2\) and \(\bar{x}\); condone \(Q_2 \approx \bar{x}\) only if \( |
| Compatible description of skewness | dB1ft | Dependent on 1st B1; if \(Q_1 = 23.4\) and \(Q_3 = 33.7 \sim 33.8\) seen, allow comparison of quartiles for 1st B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [If chose skew in (d)] median (\(Q_2\)); since the data is skewed, or median not affected by extreme values | B1 | 1st B1 for correct choice based on skewness comment in (d); if no choice made in (d) only \(Q_2\) |
| Compatible supporting comment | dB1 | 2nd dB1 for a suitable compatible comment |
| [If chose symmetric in (d)] mean (\(\bar{x}\)); since it uses all the data |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| One large square $= \frac{450}{\text{"22.5"}}$ or one small square $= \frac{450}{\text{"562.5"}}$ | M1 | For attempt to count squares (accept "22.5" in [22,23] and "562.5" in [550,575]) and use 450 to obtain measure of scale |
| One large square $= 20$ cars or one small square $= 0.8$ cars or 1 car $= 1.25$ squares | A1 | For a correct calc. for 20 or 0.8 or 1.25 etc |
| No. $> 35$ mph is: $4.5 \times \text{"20"}$ or $112.5 \times \text{"0.8"}$ | dM1 | Dependent on 1st M1; for correctly counting squares for $> 35$ mph and forming suitable expression |
| $= 90$ (cars) | A1 | For 90 with no incorrect working seen; e.g. $\frac{4.5}{22.5} \times 450$ scores M1A1M1 and A1 when $= 90$ seen |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{30 \times 12.5 + 240 \times 25 + 90 \times 32.5 + 30 \times 37.5 + 60 \times 42.5}{450} = \frac{12975}{450}$ | M1 | 1st M1 for clear sensible use of midpoints; at least 3 of (12.5, 25, 32.5, 37.5, 42.5) seen |
| | M1 | 2nd M1 for expression for $\bar{x}$ with at least 3 correct terms on numerator and compatible denominator |
| $= 28.83...$ or $\frac{173}{6}$ awrt $\mathbf{28.8}$ | A1 | For awrt 28.8 (answer only is 3/3) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Q_2 = 20 + \frac{195}{240} \times 10$ | M1 | For a full expression for median using their frequencies; do not accept boundaries of 19.5 or 20.5 (M0A0) |
| $= 28.125$ awrt $\mathbf{28.1}$ | A1 | For awrt 28.1 (answer only is 2/2); use of $(n+1)$ accept 28.15 but not 28.2 |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Q_2 < \bar{x}$, so positive skew | B1ft | For correct statement about their $Q_2$ and $\bar{x}$; condone $Q_2 \approx \bar{x}$ only if $|Q_2 - \bar{x}| < 1$; do not accept argument based on shape of graph alone |
| Compatible description of skewness | dB1ft | Dependent on 1st B1; if $Q_1 = 23.4$ and $Q_3 = 33.7 \sim 33.8$ seen, allow comparison of quartiles for 1st B1 |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| [If chose skew in (d)] **median** ($Q_2$); since the data is skewed, or median not affected by extreme values | B1 | 1st B1 for correct choice based on skewness comment in (d); if no choice made in (d) only $Q_2$ |
| Compatible supporting comment | dB1 | 2nd dB1 for a suitable compatible comment |
| [If chose symmetric in (d)] **mean** ($\bar{x}$); since it uses all the data | | |
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5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0593544d-392d-465b-b922-c9cb1435abb5-08_1031_1239_116_354}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A policeman records the speed of the traffic on a busy road with a 30 mph speed limit. He records the speeds of a sample of 450 cars. The histogram in Figure 2 represents the results.
\begin{enumerate}[label=(\alph*)]
\item Calculate the number of cars that were exceeding the speed limit by at least 5 mph in the sample.
\item Estimate the value of the mean speed of the cars in the sample.
\item Estimate, to 1 decimal place, the value of the median speed of the cars in the sample.
\item Comment on the shape of the distribution. Give a reason for your answer.
\item State, with a reason, whether the estimate of the mean or the median is a better representation of the average speed of the traffic on the road.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2012 Q5 [13]}}