| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2001 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then find median/quartiles from cumulative frequency |
| Difficulty | Moderate -0.3 This is a standard S1 statistics question covering routine histogram construction, mean/standard deviation calculation, and basic interpretation. While it has multiple parts, each part follows textbook procedures with no novel problem-solving required. The interpolation for median and skewness coefficient are standard A-level techniques, making this slightly easier than average overall. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Delay (mins) | Number of motorists |
| \(4 - 6\) | 15 |
| \(7 - 8\) | 28 |
| 9 | 49 |
| 10 | 53 |
| \(11 - 12\) | 30 |
| \(13 - 15\) | 15 |
| \(16 - 20\) | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Histogram with frequency densities: \(5, 14, 49, 53, 15, 5, 2\) | B4 (4) | B1 scales & labels; M1 histogram shape; A1 shape; A1 accuracy |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The variable (minutes delayed) is continuous | B1 (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Median \(= 9.5 + \dfrac{(100-92)}{53}\times1 = 9.65\) (or \(9.66\)) | M1 A1 (2) | M1 must use \(100.5\); accept \(9.65\) or \(9.66\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum fx = 1991\), \(\sum fx^2 = 21366.5\) (using midpoints) | M1 M1 | M1 for \(\sum fx\); M1 for \(\sum fx^2\) using midpoints |
| Mean \(= \dfrac{1991}{200} = 9.955\) or \(9.96\) | M1 A1 | |
| \(s = \sqrt{\dfrac{21366.5}{200} - \left(\dfrac{1991}{200}\right)^2} = 2.78\) or \(2\text{ min }67\text{ sec}\) | M1 A1 (6) | Accept \(2.78\); (\(S_{n-1} = 2.79\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{3(9.955 - 9.65)}{2.78} = 0.329\) | M1 A1 (2) | Answer approximately \(0.3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For normal distribution skewness is zero. In this case skewness is \(0.329\), so normal may not be suitable | B1, B1 (2) |
## Question 5:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Histogram with frequency densities: $5, 14, 49, 53, 15, 5, 2$ | B4 (4) | B1 scales & labels; M1 histogram shape; A1 shape; A1 accuracy |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| The variable (minutes delayed) is continuous | B1 (1) | |
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median $= 9.5 + \dfrac{(100-92)}{53}\times1 = 9.65$ (or $9.66$) | M1 A1 (2) | M1 must use $100.5$; accept $9.65$ or $9.66$ |
**Part (d)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum fx = 1991$, $\sum fx^2 = 21366.5$ (using midpoints) | M1 M1 | M1 for $\sum fx$; M1 for $\sum fx^2$ using midpoints |
| Mean $= \dfrac{1991}{200} = 9.955$ or $9.96$ | M1 A1 | |
| $s = \sqrt{\dfrac{21366.5}{200} - \left(\dfrac{1991}{200}\right)^2} = 2.78$ or $2\text{ min }67\text{ sec}$ | M1 A1 (6) | Accept $2.78$; ($S_{n-1} = 2.79$) |
**Part (e)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{3(9.955 - 9.65)}{2.78} = 0.329$ | M1 A1 (2) | Answer approximately $0.3$ |
**Part (f)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| For normal distribution skewness is zero. In this case skewness is $0.329$, so normal may not be suitable | B1, B1 (2) | |
---5. The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 200 motorists were delayed by roadworks on a stretch of motorway.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Delay (mins) & Number of motorists \\
\hline
$4 - 6$ & 15 \\
\hline
$7 - 8$ & 28 \\
\hline
9 & 49 \\
\hline
10 & 53 \\
\hline
$11 - 12$ & 30 \\
\hline
$13 - 15$ & 15 \\
\hline
$16 - 20$ & 10 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Using graph paper represent these data by a histogram.
\item Give a reason to justify the use of a histogram to represent these data.
\item Use interpolation to estimate the median of this distribution.
\item Calculate an estimate of the mean and an estimate of the standard deviation of these data.
One coefficient of skewness is given by
$$\frac { 3 ( \text { mean - median } ) } { \text { standard deviation } } .$$
\item Evaluate this coefficient for the above data.
\item Explain why the normal distribution may not be suitable to model the number of minutes that motorists are delayed by these roadworks.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2001 Q5 [17]}}