| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate frequency density from frequency |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard procedures: frequency density calculation (dividing frequency by class width), linear interpolation for quartiles, and mean/standard deviation from summary statistics. All techniques are direct applications of formulas with no problem-solving or insight required. The multi-part structure and calculation volume don't add conceptual difficulty—it's purely procedural work that any well-prepared S1 student should handle comfortably. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02i Select/critique data presentation |
| Distance (km) | Number of examiners |
| 41-45 | 4 |
| 46-50 | 19 |
| 51-60 | 53 |
| 61-70 | 37 |
| 71-90 | 15 |
| 91-150 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Distance is continuous | B1 | Answer: "continuous" |
| Answer | Marks | Guidance |
|---|---|---|
| \(F.D = \text{freq/class width} \Rightarrow 0.8, 3.8, 5.3, 3.7, 0.75, 0.1\) | M1 A1 | Or the same multiple of |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_2 = 50.5 + \frac{(67-23)}{53} \times 10 = 58.8\) | M1 A1 | awrt 58.8/58.9 |
| \(Q_1 = 52.48\); \(Q_3 = 67.12\) | A1 A1 | awrt 52.5/52.6, 67.1/67.3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{8379.5}{134} = 62.5335...\) | B1 | awrt 62.5 |
| \(s = \sqrt{\frac{557489.75}{134} - \left(\frac{8379.5}{134}\right)^2}\) | M1 A1\(\sqrt{}\) | |
| \(s = 15.8089...\) (\(S_{n-1} = 15.86825...\)) | A1 | awrt 15.8 (15.9) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1} = \frac{67.12 - 2 \times 58.8 + 52.48}{67.12 - 52.48} = 0.1366 \Rightarrow\) +ve skew | M1 A1\(\sqrt{}\) | Subst their \(Q_1, Q_2\) & \(Q_3\); need to show working for A1\(\sqrt{}\) and have reasonable values for quartiles |
| awrt 0.14 | A1; B1 |
| Answer | Marks |
|---|---|
| For +ve skew Mean \(>\) Median & \(62.53 > 58.80\), or \(Q_3 - Q_2\ (8.32) > Q_2 - Q_1\ (6.32)\), therefore +ve skew | B1 |
# Question 2:
## Part (a)
Distance is continuous | B1 | Answer: "continuous"
## Part (b)
$F.D = \text{freq/class width} \Rightarrow 0.8, 3.8, 5.3, 3.7, 0.75, 0.1$ | M1 A1 | Or the same multiple of
## Part (c)
$Q_2 = 50.5 + \frac{(67-23)}{53} \times 10 = 58.8$ | M1 A1 | awrt 58.8/58.9
$Q_1 = 52.48$; $Q_3 = 67.12$ | A1 A1 | awrt 52.5/52.6, 67.1/67.3
Special case: no working B1 B1 B1 ($\equiv$ A's on the epen)
## Part (d)
$\bar{x} = \frac{8379.5}{134} = 62.5335...$ | B1 | awrt 62.5
$s = \sqrt{\frac{557489.75}{134} - \left(\frac{8379.5}{134}\right)^2}$ | M1 A1$\sqrt{}$ |
$s = 15.8089...$ ($S_{n-1} = 15.86825...$) | A1 | awrt 15.8 (15.9)
Special case: answer only B1 B1 ($\equiv$ A's on the epen)
## Part (e)
$\frac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1} = \frac{67.12 - 2 \times 58.8 + 52.48}{67.12 - 52.48} = 0.1366 \Rightarrow$ +ve skew | M1 A1$\sqrt{}$ | Subst their $Q_1, Q_2$ & $Q_3$; need to show working for A1$\sqrt{}$ and have reasonable values for quartiles
awrt 0.14 | A1; B1 |
## Part (f)
For +ve skew Mean $>$ Median & $62.53 > 58.80$, or $Q_3 - Q_2\ (8.32) > Q_2 - Q_1\ (6.32)$, therefore +ve skew | B1 |
---2. The following table summarises the distances, to the nearest km , that 134 examiners travelled to attend a meeting in London.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Distance (km) & Number of examiners \\
\hline
41-45 & 4 \\
\hline
46-50 & 19 \\
\hline
51-60 & 53 \\
\hline
61-70 & 37 \\
\hline
71-90 & 15 \\
\hline
91-150 & 6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Give a reason to justify the use of a histogram to represent these data.
\item Calculate the frequency densities needed to draw a histogram for these data.\\
(DO NOT DRAW THE HISTOGRAM)
\item Use interpolation to estimate the median $Q _ { 2 }$, the lower quartile $Q _ { 1 }$, and the upper quartile $Q _ { 3 }$ of these data.
The mid-point of each class is represented by $x$ and the corresponding frequency by $f$. Calculations then give the following values
$$\Sigma f _ { x } = 8379.5 \quad \text { and } \quad \Sigma f _ { x ^ { 2 } } = 557489.75$$
\item Calculate an estimate of the mean and an estimate of the standard deviation for these data.
One coefficient of skewness is given by
$$\frac { Q _ { 3 } - 2 Q _ { 2 } + Q _ { 1 } } { Q _ { 3 } - Q _ { 1 } }$$
\item Evaluate this coefficient and comment on the skewness of these data.
\item Give another justification of your comment in part (e).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2005 Q2 [16]}}