| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate using histogram bar dimensions |
| Difficulty | Moderate -0.3 This is a standard S1 histogram and summary statistics question requiring routine techniques: frequency density calculations for histogram dimensions, linear interpolation for median, and application of given formulas for mean/standard deviation. All methods are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation |
| Time (seconds) | Number of customers, \(f\) |
| 0-30 | 2 |
| 30-60 | 10 |
| 60-70 | 17 |
| 70-80 | 25 |
| 80-100 | 25 |
| 100-150 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) 70–80 group width 0.5 (cm). \(1.5\text{ cm}^2\) is 10 customers or \(3.75\text{ cm}^2\) is 25 customers or \(0.5c = 3.75\) or \(\frac{2.5}{\frac{1}{3}}\). 70–80 group height 7.5 (cm) | B1 M1 A1 (3) | B1 for 0.5. M1 for one of the given statements or any method where "their width" \(\times\) "their height" \(= 3.75\). Correct height scores M1A1 independent of width so B0M1A1 is possible |
| (b) \(\text{Median} = (70) + \frac{13.5}{25}\times10\) allow \((n+1) = (70)+\frac{14}{25}\times10\) \(= \mathbf{75.4}\) (or if using \((n+1)\) allow 75.6) | M1 A1 (2) | M1 for a correct fraction: \(+\frac{k}{25}\times10\) where \(k=13.5\) or 14 for \((n+1)\) case. NB may work down so look out for \((80)-\frac{11.5}{25}\times10\). Beware: \(69.5+\frac{13.5}{25}\times11=75.44\) (but M0) |
| (c) \(\left[\text{Mean} = \frac{6460}{85}\right] = \mathbf{76}\) \(\sigma = \sqrt{\frac{529400}{85}-76^2} = 21.2658\ldots\) \((s=21.3920)\) awrt 21.3 | B1 M1 A1 (3) | M1 for correct expression with square root, ft their mean. A1 for awrt 21.3 or, if clearly using \(s\), allow awrt 21.4. Must be evaluated — no surds |
| (d) \(\text{Coeff of skewness} = \frac{3(76-75.4)}{21.2658\ldots} = 0.08464\ldots\) awrt 0.08 (awrt 0.06 for 75.6). There is (very slight) positive skew or the data is almost symmetrical (or both) | M1 A1 B1ft (3) | M1 sub. their values into formula, allow use of \(s\) but their \(\sigma\) or \(s\) must be \(>0\). A1 for awrt 0.08 but accept 0.085. No fraction. B1ft for correct comment compatible with their coefficient. Allow "symmetrical" for \( |
# Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** 70–80 group width **0.5** (cm). $1.5\text{ cm}^2$ is 10 customers or $3.75\text{ cm}^2$ is 25 customers or $0.5c = 3.75$ or $\frac{2.5}{\frac{1}{3}}$. 70–80 group height **7.5** (cm) | B1 M1 A1 **(3)** | B1 for 0.5. M1 for one of the given statements or any method where "their width" $\times$ "their height" $= 3.75$. Correct height scores M1A1 independent of width so B0M1A1 is possible |
| **(b)** $\text{Median} = (70) + \frac{13.5}{25}\times10$ allow $(n+1) = (70)+\frac{14}{25}\times10$ $= \mathbf{75.4}$ (or if using $(n+1)$ allow 75.6) | M1 A1 **(2)** | M1 for a correct fraction: $+\frac{k}{25}\times10$ where $k=13.5$ or 14 for $(n+1)$ case. NB may work down so look out for $(80)-\frac{11.5}{25}\times10$. Beware: $69.5+\frac{13.5}{25}\times11=75.44$ (but M0) |
| **(c)** $\left[\text{Mean} = \frac{6460}{85}\right] = \mathbf{76}$ $\sigma = \sqrt{\frac{529400}{85}-76^2} = 21.2658\ldots$ $(s=21.3920)$ awrt **21.3** | B1 M1 A1 **(3)** | M1 for correct expression with square root, ft their mean. A1 for awrt 21.3 or, if clearly using $s$, allow awrt 21.4. Must be evaluated — no surds |
| **(d)** $\text{Coeff of skewness} = \frac{3(76-75.4)}{21.2658\ldots} = 0.08464\ldots$ awrt **0.08** (awrt 0.06 for 75.6). There is (very slight) positive skew or the data is almost symmetrical (or both) | M1 A1 B1ft **(3)** | M1 sub. their values into formula, allow use of $s$ but their $\sigma$ or $s$ must be $>0$. A1 for awrt 0.08 but accept 0.085. No fraction. B1ft for correct comment compatible with their coefficient. Allow "symmetrical" for $|\text{coeff}|\leq 0.25$. They may say "slightly skew" so omit "positive" but do not allow "negative" if coef is $+$ve. Condone "strongly" positive skew. Any mention of "correlation" is B0 |6. The times, in seconds, spent in a queue at a supermarket by 85 randomly selected customers, are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Time (seconds) & Number of customers, $f$ \\
\hline
0-30 & 2 \\
\hline
30-60 & 10 \\
\hline
60-70 & 17 \\
\hline
70-80 & 25 \\
\hline
80-100 & 25 \\
\hline
100-150 & 6 \\
\hline
\end{tabular}
\end{center}
A histogram was drawn to represent these data. The $30 - 60$ group was represented by a bar of width 1.5 cm and height 1 cm .
\begin{enumerate}[label=(\alph*)]
\item Find the width and the height of the $70 - 80$ group.
\item Use linear interpolation to estimate the median of this distribution.
Given that $x$ denotes the midpoint of each group in the table and
$$\sum f x = 6460 \quad \sum f x ^ { 2 } = 529400$$
\item calculate an estimate for
\begin{enumerate}[label=(\roman*)]
\item the mean,
\item the standard deviation,\\
for the above data.
One measure of skewness is given by
$$\text { coefficient of skewness } = \frac { 3 ( \text { mean } - \text { median } ) } { \text { standard deviation } }$$
\end{enumerate}\item Evaluate this coefficient and comment on the skewness of these data.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q6 [11]}}