| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | January |
| Marks | 18 |
| 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 histogram and summary statistics question requiring routine application of well-practiced techniques: calculating frequency densities, drawing a histogram, linear interpolation for median/quartiles, and computing mean/SD from grouped data. While multi-part with several calculations, each step follows textbook procedures without requiring problem-solving insight or novel approaches. |
| 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 deviation2.02i Select/critique data presentation |
| Sales | Number of days |
| \(1 - 200\) | 166 |
| \(201 - 400\) | 100 |
| \(401 - 700\) | 59 |
| \(701 - 1000\) | 30 |
| \(1001 - 1500\) | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sales: \(1\)–\(200\), fd \(= 0.830\); \(201\)–\(400\), fd \(= 0.500\); \(401\)–\(700\), fd \(= 0.197\); \(701\)–\(1000\), fd \(= 0.100\); \(1001\)–\(1500\), fd \(= 0.010\) | M1 A1 | Frequency densities (can be scored on graph) |
| Graph | 3 marks for graph |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Q_2 = 200.5 + \frac{(180-166)}{100} \times 200 = \underline{228.5}\) | M1 A1 | \(228/229/230\) |
| \(Q_1 = 0.5 + \frac{90}{166} \times 200 = \underline{108.933\ldots}\) | A1 | \(109\) AWRT |
| \(Q_3 = 400.5 + \frac{(270-266)}{59} \times 300 = \underline{420.838}\) | A1 | AWRT \(421/425\) |
| \(\text{IQR} = 420.830\ldots - 108.933\ldots = \underline{311.905\ldots}\) | B1 \(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum fx = 110980\); \(\sum fx^2 = 58105890\) | M1 | Attempt at \(\sum fx\) or \(\sum fy\) |
| \(\sum fy = 748\); \(\sum fy^2 = 3943.5\) where \(y = \frac{x - 100.5}{100}\) | M1 | Attempt at \(\sum fx^2\) or \(\sum fy^2\) |
| \(\mu = 308.277\overline{7}\) | M1 A1 | \(308\) AWRT |
| \(\sigma = 257.6238\) | M1 A1 | \(258\) AWRT |
| No working shown: SR B1 B1 only for \(\mu\), \(\sigma\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Median & IQR | B1 | |
| Sensible reason e.g. Assuming other years are skewed. | B1 dep | (2) |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sales: $1$–$200$, fd $= 0.830$; $201$–$400$, fd $= 0.500$; $401$–$700$, fd $= 0.197$; $701$–$1000$, fd $= 0.100$; $1001$–$1500$, fd $= 0.010$ | M1 A1 | Frequency densities (can be scored on graph) |
| Graph | | 3 marks for graph | **(5)** |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Q_2 = 200.5 + \frac{(180-166)}{100} \times 200 = \underline{228.5}$ | M1 A1 | $228/229/230$ |
| $Q_1 = 0.5 + \frac{90}{166} \times 200 = \underline{108.933\ldots}$ | A1 | $109$ AWRT |
| $Q_3 = 400.5 + \frac{(270-266)}{59} \times 300 = \underline{420.838}$ | A1 | AWRT $421/425$ |
| $\text{IQR} = 420.830\ldots - 108.933\ldots = \underline{311.905\ldots}$ | B1 $\checkmark$ | | **(5)** |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum fx = 110980$; $\sum fx^2 = 58105890$ | M1 | Attempt at $\sum fx$ or $\sum fy$ |
| $\sum fy = 748$; $\sum fy^2 = 3943.5$ where $y = \frac{x - 100.5}{100}$ | M1 | Attempt at $\sum fx^2$ or $\sum fy^2$ |
| $\mu = 308.277\overline{7}$ | M1 A1 | $308$ AWRT |
| $\sigma = 257.6238$ | M1 A1 | $258$ AWRT |
| No working shown: SR B1 B1 only for $\mu$, $\sigma$ | | | **(6)** |
# Question (d) [Statistics - Measures]:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median & IQR | B1 | |
| Sensible reason e.g. Assuming other years are skewed. | B1 dep | (2) |
---5. The values of daily sales, to the nearest $\pounds$, taken at a newsagents last year are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Sales & Number of days \\
\hline
$1 - 200$ & 166 \\
\hline
$201 - 400$ & 100 \\
\hline
$401 - 700$ & 59 \\
\hline
$701 - 1000$ & 30 \\
\hline
$1001 - 1500$ & 5 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a histogram to represent these data.
\item Use interpolation to estimate the median and inter-quartile range of daily sales.
\item Estimate the mean and the standard deviation of these data.
The newsagent wants to compare last year's sales with other years.
\item State whether the newsagent should use the median and the inter-quartile range or the mean and the standard deviation to compare daily sales. Give a reason for your answer.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q5 [18]}}