| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from frequency table |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard procedures: histogram bar calculations using frequency density, linear interpolation for median, and mean/standard deviation from grouped data using given formulas. All techniques are direct applications of textbook methods with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts and calculations involved. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Delay (minutes) | Number of motorists (f) | Delay midpoint (x) |
| 3-6 | 38 | 4.5 |
| 7-8 | 25 | 7.5 |
| 9-10 | 18 | 9.5 |
| 11-15 | 12 | 13 |
| 16-20 | 7 | 18 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((3-6)\) mins has width 4, is 2 cm; \((11-15)\) mins has width 5, so \(\mathbf{2.5}\) cm | B1 | For width of \(2.5\) cm, allow \(\frac{5}{2}\) |
| \((3-6)\) mins has frequency 38, area \(19\ \text{cm}^2\), so 2 people per cm² | M1 | For 2 people per cm², or correct numerical equation for \(h\), or width \(\times\) height \(= 6\) |
| \((11-15)\) mins has area \(2.5 \times h\ \text{cm}^2\), so \(h = \dfrac{12}{2 \times 2.5} = \mathbf{2.4}\) cm | A1 | For height of 2.4 cm. If just 2.4 and 2.5 seen, must be clear which is \(h\) and which is \(w\). Allow \(\frac{12}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Q_2 = (6.5) + \dfrac{12}{25} \times 2\) or \((8.5) - \dfrac{13}{25} \times 2\) | M1 | For correct expression with sign (ignoring end point). Condone 12.5 for use of \((n+1)\) |
| \(= \text{awrt}\ \mathbf{7.46}\) | A1 | For awrt 7.46 (or 7.5 if using \((n+1)\), but must see evidence of \((n+1)\) approach) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum fx = 38 \times 4.5 + \ldots + 7 \times 18 = 811.5\) and \(\bar{x} = \dfrac{811.5}{100}\), \(= \text{awrt}\ \mathbf{8.12}\) | M1, A1 | M1 for attempt at \(\sum fx\) (full expression or \(650 < \sum fx < 950\)) and division by 100. \(\sum fx\) may be in table. A1 for 8.115 or awrt 8.12 (allow 8.11) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sigma = \sqrt{\dfrac{8096.25}{100} - \bar{x}^2} = \sqrt{80.9625 - \text{"65.85..."}} = \sqrt{15.1(0)\ldots} = \text{awrt}\ \mathbf{3.89}\) | M1, A1 | M1 for correct expression (ft their mean) including \(\sqrt{\phantom{x}}\). Allow \(s\) leading to \(\sqrt{15.26\ldots}\). A1 for awrt 3.89. Allow use of \(s =\) awrt 3.91. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Skewness \(= \dfrac{3(\text{"8.12"} - \text{"7.46"})}{\text{"3.89"}} = 0.5055\ldots = \text{awrt}\ \mathbf{0.47 \sim 0.51}\) | B1 | For correct expression using their values (\(\sigma\) must be \(> 0\)) or awrt \(0.47 \sim 0.51\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Skewness for Monday and Friday are different (o.e.) | B1 | For comment that skewness is different. (Only commenting on "correlation" is B0.) If ans. to (e) \(> 0\), allow B1 for e.g. "skewness on Fri is \(< 0\)" ["on Fri" may be implied] |
| Suggests more longer delays on Friday (o.e.) | B1 | For comment about length of delay, e.g. "more long ones (on Fri.)" or "longer delays on Fri." |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3-6)$ mins has width 4, is 2 cm; $(11-15)$ mins has width 5, so $\mathbf{2.5}$ cm | B1 | For width of $2.5$ cm, allow $\frac{5}{2}$ |
| $(3-6)$ mins has frequency 38, area $19\ \text{cm}^2$, so 2 people per cm² | M1 | For 2 people per cm², or correct numerical equation for $h$, or width $\times$ height $= 6$ |
| $(11-15)$ mins has area $2.5 \times h\ \text{cm}^2$, so $h = \dfrac{12}{2 \times 2.5} = \mathbf{2.4}$ cm | A1 | For height of 2.4 cm. If just 2.4 and 2.5 seen, must be clear which is $h$ and which is $w$. Allow $\frac{12}{5}$ |
**(3 marks)**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q_2 = (6.5) + \dfrac{12}{25} \times 2$ or $(8.5) - \dfrac{13}{25} \times 2$ | M1 | For correct expression with sign (ignoring end point). Condone 12.5 for use of $(n+1)$ |
| $= \text{awrt}\ \mathbf{7.46}$ | A1 | For awrt 7.46 (or 7.5 if using $(n+1)$, but must see evidence of $(n+1)$ approach) |
**(2 marks)**
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum fx = 38 \times 4.5 + \ldots + 7 \times 18 = 811.5$ and $\bar{x} = \dfrac{811.5}{100}$, $= \text{awrt}\ \mathbf{8.12}$ | M1, A1 | M1 for attempt at $\sum fx$ (full expression or $650 < \sum fx < 950$) and division by 100. $\sum fx$ may be in table. A1 for 8.115 or awrt 8.12 (allow 8.11) |
**(2 marks)**
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sigma = \sqrt{\dfrac{8096.25}{100} - \bar{x}^2} = \sqrt{80.9625 - \text{"65.85..."}} = \sqrt{15.1(0)\ldots} = \text{awrt}\ \mathbf{3.89}$ | M1, A1 | M1 for correct expression (ft their mean) including $\sqrt{\phantom{x}}$. Allow $s$ leading to $\sqrt{15.26\ldots}$. A1 for awrt 3.89. Allow use of $s =$ awrt 3.91. |
**(2 marks)**
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Skewness $= \dfrac{3(\text{"8.12"} - \text{"7.46"})}{\text{"3.89"}} = 0.5055\ldots = \text{awrt}\ \mathbf{0.47 \sim 0.51}$ | B1 | For correct expression using their values ($\sigma$ must be $> 0$) or awrt $0.47 \sim 0.51$ |
**(1 mark)**
## Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Skewness for Monday and Friday are different (o.e.) | B1 | For comment that skewness is different. (Only commenting on "correlation" is B0.) If ans. to (e) $> 0$, allow B1 for e.g. "skewness on Fri is $< 0$" ["on Fri" may be implied] |
| Suggests more longer delays on Friday (o.e.) | B1 | For comment about length of delay, e.g. "more long ones (on Fri.)" or "longer delays on Fri." |
**(2 marks) [Total: 12]**2. The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 100 motorists were delayed by roadworks on a stretch of motorway one Monday.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Delay (minutes) & Number of motorists (f) & Delay midpoint (x) \\
\hline
3-6 & 38 & 4.5 \\
\hline
7-8 & 25 & 7.5 \\
\hline
9-10 & 18 & 9.5 \\
\hline
11-15 & 12 & 13 \\
\hline
16-20 & 7 & 18 \\
\hline
\end{tabular}
\end{center}
(You may use $\sum \mathrm { f } x ^ { 2 } = 8096.25$ )
A histogram has been drawn to represent these data.
The bar representing a delay of (3-6) minutes has a width of 2 cm and a height of 9.5 cm .
\begin{enumerate}[label=(\alph*)]
\item Calculate the width and the height of the bar representing a delay of (11-15) minutes.
\item Use linear interpolation to estimate the median delay.
\item Calculate an estimate of the mean delay.
\item Calculate an estimate of the standard deviation of the delays.
One coefficient of skewness is given by $\frac { 3 ( \text { mean } - \text { median } ) } { \text { standard deviation } }$
\item Evaluate this coefficient for the above data, giving your answer to 2 significant figures.
On the following Friday, the coefficient of skewness for the delays on this stretch of motorway was - 0.22
\item State, giving a reason, how the delays on this stretch of motorway on Friday are different from the delays on Monday.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q2 [12]}}