| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | October |
| Marks | 11 |
| 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 histogram question requiring standard calculations: frequency density from given area, proportional reasoning for bar heights, and linear interpolation for quartiles. All techniques are textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| 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 deviation |
| Stem length (cm) | Frequency |
| \(40 \leqslant x < 42\) | 12 |
| \(42 \leqslant x < 45\) | 18 |
| \(45 \leqslant x < 50\) | 23 |
| \(50 \leqslant x < 55\) | 35 |
| \(55 \leqslant x < 58\) | 24 |
| \(58 \leqslant x < 60\) | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{Area} = \frac{16.5}{12} \times 18\) | M1 | Allow equivalent e.g. \(16.5 \times \frac{3}{2}\) |
| \(= 24.75 \text{ cm}^2\) | A1 | Allow 24.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| fd method: \(\frac{24}{58-55}[=8]\) and \(\frac{35}{55-50}[=7]\) or Area method: \(\frac{16.5}{12}\times 24[=33]\) and \(\frac{16.5}{12}\times 35[=48.125]\) or \(\frac{16.5}{12}\times\frac{24}{3}[=11]\) and \(\frac{16.5}{12}\times\frac{35}{5}[=9.625]\) | M1 | Correct method for finding fd or area for highest and 2nd highest bars. Allow if 8 and 7 seen or 33 and 48.125 seen or 9.625 rather than 48.125 and/or 11 rather than 33 |
| \([h=]\frac{10}{8}\times 7\) or \([h=]\frac{48.125\times 10\times 3}{5\times 33}\) or \([h=]9.625\times\frac{10}{11}\) | dM1 | Dependent on previous M. Fully correct expression for \(h\) or fully correct equation e.g. \(\frac{33}{10\times 3}=\frac{48.125}{5h}\) |
| \(= 8.75 \text{ cm}\) | A1 | 8.75 oe. NB answer of 8.75 seen as final answer 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([Q_2=]50+\frac{7}{35}\times 5\) or \([Q_2=]55-\frac{28}{35}\times 5\) | M1 | For \(50+\frac{7}{35}\times k\) or \(55-\frac{28}{35}\times k\) or \(\frac{Q_2-50}{k}=\frac{60-53}{88-53}\) or \(\frac{55-Q_2}{k}=\frac{88-60}{88-53}\) where \(4, k, 6\). Condone use of \(n+1\) |
| \(= 51 \text{ cm}\) | A1 | 51 (condone for use of \(n+1\) awrt 51.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([Q_3=]55+\frac{2}{24}\times 3\) or \([Q_3=]58-\frac{22}{24}\times 3\) and \(55.25-45\) | M1 | \(55+\frac{2}{24}\times t\) or \(58-\frac{22}{24}\times t\) or \(\frac{Q_3-55}{t}=\frac{90-88}{112-88}\) or \(\frac{58-Q_3}{t}=\frac{112-90}{112-88}\) where \(2, t, 4\). and using "their \(Q_3\)" \(-45\). Condone use of \(n+1\) |
| \(= 10.25 \text{ cm}\) | A1 | 10.25 oe e.g. 41/4, allow 10.3 from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{55.25-2(51)+45}{55.25-45}\) | M1 | Substitution of their values from (c) seen or awrt \(-0.17\) or \(-7/41\) |
| \([=-0.17073\ldots < 0]\) negative [skew] | A1ft | Dependent on M1. Correct description of skewness consistent with their values from (c). Only allow no skew or symmetrical if their value should be 0. Ignore correlation. |
# Question 1:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{Area} = \frac{16.5}{12} \times 18$ | M1 | Allow equivalent e.g. $16.5 \times \frac{3}{2}$ |
| $= 24.75 \text{ cm}^2$ | A1 | Allow 24.8 |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| fd method: $\frac{24}{58-55}[=8]$ **and** $\frac{35}{55-50}[=7]$ **or** Area method: $\frac{16.5}{12}\times 24[=33]$ **and** $\frac{16.5}{12}\times 35[=48.125]$ or $\frac{16.5}{12}\times\frac{24}{3}[=11]$ **and** $\frac{16.5}{12}\times\frac{35}{5}[=9.625]$ | M1 | Correct method for finding fd or area for highest **and** 2nd highest bars. Allow if 8 and 7 seen or 33 and 48.125 seen or 9.625 rather than 48.125 and/or 11 rather than 33 |
| $[h=]\frac{10}{8}\times 7$ **or** $[h=]\frac{48.125\times 10\times 3}{5\times 33}$ **or** $[h=]9.625\times\frac{10}{11}$ | dM1 | Dependent on previous M. Fully correct expression for $h$ or fully correct equation e.g. $\frac{33}{10\times 3}=\frac{48.125}{5h}$ |
| $= 8.75 \text{ cm}$ | A1 | 8.75 oe. **NB** answer of 8.75 seen as final answer 3/3 |
## Part (c)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $[Q_2=]50+\frac{7}{35}\times 5$ **or** $[Q_2=]55-\frac{28}{35}\times 5$ | M1 | For $50+\frac{7}{35}\times k$ or $55-\frac{28}{35}\times k$ or $\frac{Q_2-50}{k}=\frac{60-53}{88-53}$ or $\frac{55-Q_2}{k}=\frac{88-60}{88-53}$ where $4, k, 6$. Condone use of $n+1$ |
| $= 51 \text{ cm}$ | A1 | 51 (condone for use of $n+1$ awrt 51.1) |
## Part (c)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $[Q_3=]55+\frac{2}{24}\times 3$ **or** $[Q_3=]58-\frac{22}{24}\times 3$ **and** $55.25-45$ | M1 | $55+\frac{2}{24}\times t$ or $58-\frac{22}{24}\times t$ or $\frac{Q_3-55}{t}=\frac{90-88}{112-88}$ or $\frac{58-Q_3}{t}=\frac{112-90}{112-88}$ where $2, t, 4$. **and** using "their $Q_3$" $-45$. Condone use of $n+1$ |
| $= 10.25 \text{ cm}$ | A1 | 10.25 oe e.g. 41/4, allow 10.3 from correct working |
## Part (d)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{55.25-2(51)+45}{55.25-45}$ | M1 | Substitution of their values from (c) seen **or** awrt $-0.17$ or $-7/41$ |
| $[=-0.17073\ldots < 0]$ negative [skew] | A1ft | Dependent on M1. Correct description of skewness consistent with their values from (c). Only allow no skew or symmetrical if their value should be 0. Ignore correlation. |
---\begin{enumerate}
\item The stem lengths of a sample of 120 tulips are recorded in the grouped frequency table below.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|}
\hline
Stem length (cm) & Frequency \\
\hline
$40 \leqslant x < 42$ & 12 \\
\hline
$42 \leqslant x < 45$ & 18 \\
\hline
$45 \leqslant x < 50$ & 23 \\
\hline
$50 \leqslant x < 55$ & 35 \\
\hline
$55 \leqslant x < 58$ & 24 \\
\hline
$58 \leqslant x < 60$ & 8 \\
\hline
\end{tabular}
\end{center}
A histogram is drawn to represent these data.\\
The area of the bar representing the $40 \leqslant x < 42$ class is $16.5 \mathrm {~cm} ^ { 2 }$\\
(a) Calculate the exact area of the bar representing the $42 \leqslant x < 45$ class.
The height of the tallest bar in the histogram is 10 cm .\\
(b) Find the exact height of the second tallest bar.\\
$Q _ { 1 }$ for these data is 45 cm .\\
(c) Use linear interpolation to find an estimate for\\
(i) $Q _ { 2 }$\\
(ii) the interquartile range.
One measure of skewness is given by
$$\frac { Q _ { 3 } - 2 Q _ { 2 } + Q _ { 1 } } { Q _ { 3 } - Q _ { 1 } }$$
(d) By calculating this measure, describe the skewness of these data.
\hfill \mbox{\textit{Edexcel S1 2022 Q1 [11]}}