| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| 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 standard S1 statistics question testing routine procedures: histogram calculations using frequency density, linear interpolation for median, and mean/standard deviation from grouped data using given midpoints. All techniques are textbook exercises requiring formula application rather than problem-solving. Part (e) adds a simple weighted calculation but remains procedural. Easier than average A-level due to being pure recall and computation. |
| 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 deviation2.02j Clean data: missing data, errors |
| Length, \(L\) cm | Frequency, f | Class mid point, \(\boldsymbol { x } \mathbf { c m }\) |
| \(5 \leqslant L < 8\) | 5 | 6.5 |
| \(8 \leqslant L < 10\) | 13 | 9 |
| \(10 \leqslant L < 12\) | 16 | 11 |
| \(12 \leqslant L < 15\) | 25 | 13.5 |
| \(15 \leqslant L < 20\) | 30 | 17.5 |
| \(20 \leqslant L < 28\) | 11 | 24 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Width \(= \mathbf{2.5}\) (cm) | B1 | For width \(= 2.5\) cm |
| \(1.5\text{ cm}^2\) for freq of 5 so \(6\times 1.5 = 9\text{ cm}^2\) for freq of 30 or \(\text{fd} = \frac{5}{3}\), \(w\times h = 9\) | M1 | For sight of \(9\text{ cm}^2\) or \(w\times h = 9\) or \(\text{fd} = \frac{5}{3}\) |
| So \(h = 9\div 2.5\) or \(6\div\frac{5}{3} = \mathbf{3.6}\) (cm) | A1 | For height \(= 3.6\) cm |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Q_2 = [12] + \frac{16}{25}\times 3\), allow use of \((n+1)\) giving \([12]+\frac{16.5}{25}\times 3\) | M1 | For \(\frac{16}{25}\times 3\) or \(\frac{9}{25}\times 3\) or \(\frac{m-12}{15-m}=\frac{16}{9}\). For any correct equation leading to \(Q_2\) or correct fraction as part of \(Q_2\) |
| \(= 13.92 =\) awrt \(\mathbf{13.9}\) | A1 | For awrt 13.9 (use of \((n+1)\) giving 13.98 = awrt 14.0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum fx = 5\times 6.5 + 13\times 9 + 16\times 11 + 25\times 13.5 + 30\times 17.5 + 11\times 24 = 1452\) | M1 | For attempt at \(\sum fx\) with at least 3 correct terms or \(900 < \sum fx < 1800\). For info \(\sum fx = 32.5+117+176+337.5+525+264\) |
| \(\bar{x} = 14.52 =\) awrt \(\mathbf{14.5}\) | A1 | For awrt 14.5 (correct answer only 2/2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum fx^2 = 6.5^2\times 5 + 9^2\times 13 + 11^2\times 16 + 13.5^2\times 25 + 17.5^2\times 30 + 24^2\times 11 = 23280\) | M1 | For attempt at \(\sum fx^2\) with at least 3 correct terms or \(20000 < \sum fx^2 < 26000\). For info \(\sum fx^2 = 211.25+1053+1936+4556.25+9187.5+6336\) |
| \(\sigma_x = \sqrt{\frac{\text{"23280"}}{100} - (\text{"14.52"})^2}\) or \(\sqrt{21.9696}\) | M1 | For a correct expression including \(\sqrt{\ }\) (ft their \(\sum fx^2\) if clear it is \(\sum fx^2\)). Do not allow \((\sum fx)^2\) for \(\sum fx^2\) |
| \(\sigma_x = 4.687\ldots =\) awrt \(\mathbf{4.69}\) | A1 | For awrt 4.69 (allow \(s = 4.7107\ldots\) awrt 4.71) (correct answer only 3/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}\times 13 + 16 + 25 + 30 + \frac{1}{4}\times 11\) | M1 | For attempt at a correct expression (allow 1 error or omission) eg \(100-(5+\frac{13}{2})-\frac{33}{4}\) |
| So proportion is \(80.25\%\) or \(0.8025\) awrt \(\mathbf{0.803}\) | A1 | For awrt 80.3% or 0.803 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Profit} = 2.2\times\text{"0.8025"} + 0.8\times\frac{0.75\times 11}{100} - 1.2\times\left(1-\left[0.8025+\frac{0.75\times 11}{100}\right]\right)\) | M1 | For a correct expression ft their 0.8025. eg \(\left[2.2\times(100-11.5-8.25)+0.8\times 8.25-1.2\times 11.5\right]\div 100\). Condone \(\left[2.2\times\text{"80"}+0.8\times(8)-1.2\times(12)\right]\div 100\) |
| \(= 1.6935\) awrt \(\mathbf{1.7\ (p)}\) | A1 | For awrt 1.7. Allow £0.017 (this must have units) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Width $= \mathbf{2.5}$ **(cm)** | B1 | For width $= 2.5$ cm |
| $1.5\text{ cm}^2$ for freq of 5 so $6\times 1.5 = 9\text{ cm}^2$ for freq of 30 or $\text{fd} = \frac{5}{3}$, $w\times h = 9$ | M1 | For sight of $9\text{ cm}^2$ or $w\times h = 9$ or $\text{fd} = \frac{5}{3}$ |
| So $h = 9\div 2.5$ or $6\div\frac{5}{3} = \mathbf{3.6}$ **(cm)** | A1 | For height $= 3.6$ cm |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q_2 = [12] + \frac{16}{25}\times 3$, allow use of $(n+1)$ giving $[12]+\frac{16.5}{25}\times 3$ | M1 | For $\frac{16}{25}\times 3$ or $\frac{9}{25}\times 3$ or $\frac{m-12}{15-m}=\frac{16}{9}$. For any correct equation leading to $Q_2$ or correct fraction as part of $Q_2$ |
| $= 13.92 =$ awrt $\mathbf{13.9}$ | A1 | For awrt 13.9 (use of $(n+1)$ giving 13.98 = awrt 14.0) |
## Part (c)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum fx = 5\times 6.5 + 13\times 9 + 16\times 11 + 25\times 13.5 + 30\times 17.5 + 11\times 24 = 1452$ | M1 | For attempt at $\sum fx$ with at least 3 correct terms or $900 < \sum fx < 1800$. **For info** $\sum fx = 32.5+117+176+337.5+525+264$ |
| $\bar{x} = 14.52 =$ awrt $\mathbf{14.5}$ | A1 | For awrt 14.5 (correct answer only 2/2) |
## Part (c)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum fx^2 = 6.5^2\times 5 + 9^2\times 13 + 11^2\times 16 + 13.5^2\times 25 + 17.5^2\times 30 + 24^2\times 11 = 23280$ | M1 | For attempt at $\sum fx^2$ with at least 3 correct terms or $20000 < \sum fx^2 < 26000$. **For info** $\sum fx^2 = 211.25+1053+1936+4556.25+9187.5+6336$ |
| $\sigma_x = \sqrt{\frac{\text{"23280"}}{100} - (\text{"14.52"})^2}$ or $\sqrt{21.9696}$ | M1 | For a correct expression including $\sqrt{\ }$ (ft their $\sum fx^2$ if clear it is $\sum fx^2$). Do not allow $(\sum fx)^2$ for $\sum fx^2$ |
| $\sigma_x = 4.687\ldots =$ awrt $\mathbf{4.69}$ | A1 | For awrt 4.69 (allow $s = 4.7107\ldots$ awrt 4.71) (correct answer only 3/3) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\times 13 + 16 + 25 + 30 + \frac{1}{4}\times 11$ | M1 | For attempt at a correct expression (allow 1 error or omission) eg $100-(5+\frac{13}{2})-\frac{33}{4}$ |
| So proportion is $80.25\%$ or $0.8025$ awrt $\mathbf{0.803}$ | A1 | For awrt 80.3% or 0.803 |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Profit} = 2.2\times\text{"0.8025"} + 0.8\times\frac{0.75\times 11}{100} - 1.2\times\left(1-\left[0.8025+\frac{0.75\times 11}{100}\right]\right)$ | M1 | For a correct expression ft their 0.8025. eg $\left[2.2\times(100-11.5-8.25)+0.8\times 8.25-1.2\times 11.5\right]\div 100$. Condone $\left[2.2\times\text{"80"}+0.8\times(8)-1.2\times(12)\right]\div 100$ |
| $= 1.6935$ awrt $\mathbf{1.7\ (p)}$ | A1 | For awrt 1.7. Allow £0.017 (this must have units) |
---\begin{enumerate}
\item A random sample of 100 carrots is taken from a farm and their lengths, $L \mathrm {~cm}$, recorded. The data are summarised in the following table.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Length, $L$ cm & Frequency, f & Class mid point, $\boldsymbol { x } \mathbf { c m }$ \\
\hline
$5 \leqslant L < 8$ & 5 & 6.5 \\
\hline
$8 \leqslant L < 10$ & 13 & 9 \\
\hline
$10 \leqslant L < 12$ & 16 & 11 \\
\hline
$12 \leqslant L < 15$ & 25 & 13.5 \\
\hline
$15 \leqslant L < 20$ & 30 & 17.5 \\
\hline
$20 \leqslant L < 28$ & 11 & 24 \\
\hline
\end{tabular}
\end{center}
A histogram is drawn to represent these data.\\
The bar representing the class $5 \leqslant L < 8$ is 1.5 cm wide and 1 cm high.\\
(a) Find the width and height of the bar representing the class $15 \leqslant L < 20$\\
(b) Use linear interpolation to estimate the median length of these carrots.\\
(c) Estimate\\
(i) the mean length of these carrots,\\
(ii) the standard deviation of the lengths of these carrots.
A supermarket will only buy carrots with length between 9 cm and 22 cm .\\
(d) Estimate the proportion of carrots from the farm that the supermarket will buy.
Any carrots that the supermarket does not buy are sold as animal feed.
The farm makes a profit of 2.2 pence on each carrot sold to the supermarket, a profit of 0.8 pence on each carrot longer than 22 cm and a loss of 1.2 pence on each carrot shorter than 9 cm .\\
(e) Find an estimate of the mean profit per carrot made by the farm.
\hfill \mbox{\textit{Edexcel S1 2021 Q3 [14]}}