| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate single values from cumulative frequency graph |
| Difficulty | Easy -1.2 This is a straightforward S1 statistics question requiring basic reading of a box plot to identify quartiles (parts a,b), applying a given formula for outliers (part c), reading values from the diagram (part d,e), and a standard binomial probability calculation (part f). All steps are routine recall and direct application with no problem-solving insight required. |
| Spec | 2.02f Measures of average and spread2.02h Recognize outliers2.03a Mutually exclusive and independent events |
| Answer | Marks | Guidance |
|---|---|---|
| \(30\) (pass for) | B1 (1) | Labelled or 1st answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(46\) (merit for) | B1 (1) | Labelled or 2nd answer |
| Answer | Marks | Guidance |
|---|---|---|
| \([1.5(Q_3 - Q_1) = 1.5 \times 16 = 24]\) so \(c = 70\) and \(d = 6\) | B1, B1 (2) | 1st B1 for \(c = 70\); 2nd B1 for \(d = 6\). Allow B1B1 for unlabelled 70 followed by 6. Award B1B0 for \(c = 6\) and \(d = 70\) or 6 and 70 in wrong order |
| Answer | Marks | Guidance |
|---|---|---|
| \(68, 72, 79\) | B2/1/0 (2) | B2 for all 3 correct values and no extra value; B1 for two correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 < d\) therefore 5 is an outlier | M1 | For identifying or stating that 5 is the only outlier or lower whisker ending at 6 or 10 |
| One outlier correctly marked at 5 (whisker must stop above 5) | A1 | For only one outlier correctly marked at 5 |
| Single lower whisker stopping at 10 | A1 (3) | Two whiskers is A0. Condone 15 marked on otherwise correct whisker. If outlier at 5 and lower whisker ends at 6 award M1A1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times \left(\frac{1}{4}\right)^2 \times 3\) | M1M1 | 1st M1 for \(\frac{1}{2} \times \left(\frac{1}{4}\right)^2\); 2nd M1 for expression of form \(pq^2 \times 3\) where \(p\) and \(q\) are probabilities \((p \neq q)\) |
| \(= \frac{3}{32}\) | A1 (3) | For \(\frac{3}{32}\) or exact equivalent. Allow 0.0937 or 0.0938 following correct expression. NB \(\frac{3}{4} \times \left(\frac{1}{4}\right)^2 \times 3 = \frac{9}{64}\) is common incorrect answer scoring M0M1A0. Warning: \(2 \times (0.25)^2 \times (0.75)\) or \(2 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4}\) gives correct answer but is M0M0A0 |
## Question 2:
### Part (a)
| $30$ (pass for) | B1 (1) | Labelled or 1st answer |
### Part (b)
| $46$ (merit for) | B1 (1) | Labelled or 2nd answer |
### Part (c)
| $[1.5(Q_3 - Q_1) = 1.5 \times 16 = 24]$ so $c = 70$ and $d = 6$ | B1, B1 (2) | 1st B1 for $c = 70$; 2nd B1 for $d = 6$. Allow B1B1 for unlabelled 70 followed by 6. Award B1B0 for $c = 6$ and $d = 70$ or 6 and 70 in wrong order |
### Part (d)
| $68, 72, 79$ | B2/1/0 (2) | B2 for all 3 correct values and no extra value; B1 for two correct |
### Part (e)
| $5 < d$ therefore 5 is an outlier | M1 | For identifying or stating that 5 is the only outlier or lower whisker ending at 6 or 10 |
| One outlier correctly marked at 5 (whisker must stop above 5) | A1 | For only one outlier correctly marked at 5 |
| Single lower whisker stopping at 10 | A1 (3) | Two whiskers is A0. Condone 15 marked on otherwise correct whisker. If outlier at 5 and lower whisker ends at 6 award M1A1A0 |
### Part (f)
| $\frac{1}{2} \times \left(\frac{1}{4}\right)^2 \times 3$ | M1M1 | 1st M1 for $\frac{1}{2} \times \left(\frac{1}{4}\right)^2$; 2nd M1 for expression of form $pq^2 \times 3$ where $p$ and $q$ are probabilities $(p \neq q)$ |
| $= \frac{3}{32}$ | A1 (3) | For $\frac{3}{32}$ or exact equivalent. Allow 0.0937 or 0.0938 following correct expression. NB $\frac{3}{4} \times \left(\frac{1}{4}\right)^2 \times 3 = \frac{9}{64}$ is common incorrect answer scoring M0M1A0. Warning: $2 \times (0.25)^2 \times (0.75)$ or $2 \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4}$ gives correct answer but is M0M0A0 |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{70137e9a-0a6b-48b5-8dd4-c436cb063351-04_284_1244_260_388}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of a box and whisker plot for the marks in an examination with a large number of candidates. Part of the lower whisker has been torn off.
\begin{enumerate}[label=(\alph*)]
\item Given that $75 \%$ of the candidates passed the examination, state the lowest mark for the award of a pass.
\item Given that the top $25 \%$ of the candidates achieved a merit grade, state the lowest mark for the award of a merit grade.
An outlier is defined as any value greater than $c$ or any value less than $d$ where
$$\begin{aligned}
& c = Q _ { 3 } + 1.5 \left( Q _ { 3 } - Q _ { 1 } \right) \\
& d = Q _ { 1 } - 1.5 \left( Q _ { 3 } - Q _ { 1 } \right)
\end{aligned}$$
\item Find the value of $c$ and the value of $d$.
\item Write down the 3 highest marks scored in the examination.
The 3 lowest marks in the examination were 5, 10 and 15
\item On the diagram on page 7, complete the box and whisker plot.
Three candidates are selected at random from those who took this examination.
\item Find the probability that all 3 of these candidates passed the examination but only 2 achieved a merit grade.
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{70137e9a-0a6b-48b5-8dd4-c436cb063351-05_285_1628_2343_166}
Turn over for a spare diagram if you need to redraw your plot.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q2 [12]}}