| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from discrete rounded data |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question requiring standard calculations of mean, median, and variance from grouped frequency data. The methods are straightforward textbook procedures (using midpoints, linear interpolation for median, variance formula), and part (b) requires only basic interpretation. No problem-solving insight or novel techniques needed. |
| Spec | 2.01a Population and sample: terminology2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Body weight (kg) | 75-79 | 80-84 | 85-89 | 90-94 | 95-99 | 100-104 | 105-109 |
| Number of Players (2010) | 1 | 2 | 2 | 4 | 3 | 2 | 1 |
| Estimate in 1990 | Estimate in 2010 | |
| Mean | 83.0 | \(a\) |
| Median | 82.0 | \(b\) |
| Variance | 44.0 | \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = \frac{77\times1 + 82\times2 + ...}{15} = \frac{1385}{15} = \frac{277}{3} = 92.\dot{3}\) awrt 92.3 | M1A1 | 1st M1 for attempt to use correct midpoints in expression for mean. Accept \(\frac{\sum fx}{15}\) with at least 3 correct \(fx\) products seen and intention to add, or \(1300 < \sum fx < 1400\). 1st A1 for awrt 92.3 (don't insist on 3 sf). NB midpoints are: 77, 82, 87, 92, 97, 102, 107 |
| \(b = \left[89.5+\right]\frac{7.5-5}{9-5}(94.5-89.5) = 92.625\) awrt 92.6 | M1A1 | 2nd M1 for \(\frac{7.5-5}{9-5}(94.5-89.5)\) or \(\frac{8-5}{9-5}(94.5-89.5)\). 2nd A1 for awrt 92.6. For \(n+1\) case gives 93.25, allow awrt 93.3. Don't insist on 3 sf. Correct answer must not follow from incorrect expression. |
| \(c = \frac{1\times77^2 + 2\times82^2 + ....}{15} - 92.\dot{3}^2 = 64.88...\) awrt 64.9 | M1A1 | 3rd M1 for fully correct expression ft their \(a\), e.g. \(\frac{128855}{15} - a^2\) or \(\frac{25771}{3} - a^2\). 3rd A1 for awrt 64.9. Accept \(s^2 = 69.5238...\) or awrt 69.5. Don't insist on 3 sf. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Median in 2010 (92.6 kg) > Median in 1990 (82.0 kg); Mean in 2010 (92.3 kg) > Mean in 1990 (83.0 kg); Rugby coach's claim supported. | B1, dB1 | 1st B1 for suitable reason i.e. identifying an increase in mean or median. Ignore any comment about variance. 2nd dB1 dependent on suitable reason for stating coach's claim is supported. Allow if both \(a > 83.0\) and \(b > 82.0\). If NOT both \(a > 83.0\) and \(b > 82.0\), allow ft provided M1 scored for both \(a\) and \(b\) in part (a). |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = \frac{77\times1 + 82\times2 + ...}{15} = \frac{1385}{15} = \frac{277}{3} = 92.\dot{3}$ awrt 92.3 | M1A1 | 1st M1 for attempt to use correct midpoints in expression for mean. Accept $\frac{\sum fx}{15}$ with at least 3 correct $fx$ products seen and intention to add, or $1300 < \sum fx < 1400$. 1st A1 for awrt 92.3 (don't insist on 3 sf). NB midpoints are: 77, 82, 87, 92, 97, 102, 107 |
| $b = \left[89.5+\right]\frac{7.5-5}{9-5}(94.5-89.5) = 92.625$ awrt 92.6 | M1A1 | 2nd M1 for $\frac{7.5-5}{9-5}(94.5-89.5)$ or $\frac{8-5}{9-5}(94.5-89.5)$. 2nd A1 for awrt 92.6. For $n+1$ case gives 93.25, allow awrt 93.3. Don't insist on 3 sf. Correct answer must not follow from incorrect expression. |
| $c = \frac{1\times77^2 + 2\times82^2 + ....}{15} - 92.\dot{3}^2 = 64.88...$ awrt 64.9 | M1A1 | 3rd M1 for fully correct expression ft their $a$, e.g. $\frac{128855}{15} - a^2$ or $\frac{25771}{3} - a^2$. 3rd A1 for awrt 64.9. Accept $s^2 = 69.5238...$ or awrt 69.5. Don't insist on 3 sf. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Median in 2010 (92.6 kg) > Median in 1990 (82.0 kg); Mean in 2010 (92.3 kg) > Mean in 1990 (83.0 kg); Rugby coach's claim supported. | B1, dB1 | 1st B1 for suitable reason i.e. identifying an increase in mean or median. Ignore any comment about variance. 2nd dB1 dependent on suitable reason for stating coach's claim is supported. Allow if both $a > 83.0$ and $b > 82.0$. If NOT both $a > 83.0$ and $b > 82.0$, allow ft provided M1 scored for both $a$ and $b$ in part (a). |2. A rugby club coach uses club records to take a random sample of 15 players from 1990 and an independent random sample of 15 players from 2010. The body weight of each player was recorded to the nearest kg and the results from 2010 are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
Body weight (kg) & 75-79 & 80-84 & 85-89 & 90-94 & 95-99 & 100-104 & 105-109 \\
\hline
Number of Players (2010) & 1 & 2 & 2 & 4 & 3 & 2 & 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the estimated values in kg of the summary statistics $a$, $b$ and $c$ in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Estimate in 1990 & Estimate in 2010 \\
\hline
Mean & 83.0 & $a$ \\
\hline
Median & 82.0 & $b$ \\
\hline
Variance & 44.0 & $c$ \\
\hline
\end{tabular}
\end{center}
Give your answers to 3 significant figures.
The rugby coach claims that players' body weight increased between 1990 and 2010.
\item Using the table in part (a), comment on the rugby coach's claim.
\includegraphics[max width=\textwidth, alt={}, center]{a839a89a-17f0-473b-ac10-bcec3dbe97f7-05_104_97_2613_1784}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q2 [8]}}