| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then perform other calculations |
| Difficulty | Moderate -0.3 This is a standard S1 histogram question requiring interpretation of frequency density, area calculations, and median estimation. Part (a) is basic recall, parts (b)-(c) involve routine histogram calculations using given information, and part (d) requires a simple comment on distribution shape. All techniques are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Answer | Marks | Guidance |
|---|---|---|
| (a) [Time is] continuous | B1 | Allow not discrete |
| (b) \(16 \text{ photographers} = 160 \text{ small squares or } 64 \text{ photographers} = 640 \text{ small squares}\) | M1 | For establishing a ratio between photographers and area or calculating frequency density (may be implied by 2nd M1) |
| or \(\frac{16}{160} = 0.1\) or \(\frac{64}{640} = 0.1\) or \(\frac{160}{16} = 10\) or \(\frac{640}{64} = 10\) | ||
| Frequency density = 1.6 or Correct scale on the frequency density axis | M1 | For a correct ratio or expression using areas for photographers from 12 to 24 |
| \(\frac{x}{240} = \frac{16}{160}\) oe or \(\frac{(20-12)}{240} \times 16 + \frac{(24-20)}{10} \times 5 \times 14\) | ||
| \(= 24\) | A1 | Cao |
| (c) Using \(n\) | Using \(n + 1\) | M1 |
| \([Q_2] = 20 + \frac{(32-21)}{35-21} \times (25-20)\) oe | \([Q_2] = 20 + \frac{(32.5-21)}{35-21} \times (25-20)\) oe | |
| or \(25 - \frac{35-32}{35-21} \times (25-20)\) oe | or \(25 - \frac{35-32.5}{35-21} \times (25-20)\) oe | |
| = awrt 23.9 | = awrt 24.1 | A1 |
| (d) Mean = Median or Mean = Median e.g. Appropriate decision. Consistent with expectation for a normal distribution. | M1 A1ft | For a correct comment about mean and median ft their median |
| Allow mean is close to median to imply mean = median | ||
| Ignore any comments made about the shape of the histogram | ||
| For a correct compatible comment about Charlie's decision ft their median | ||
| If Mean = Median or Mean = Median, then the decision should be that a normal distribution is suitable [due to symmetry] | ||
| If Mean < Median or Mean > Median or mean ≠ median, then the decision should be that a normal distribution is not suitable [due to the skew in the data] |
**(a)** [Time is] continuous | B1 | Allow not discrete
**(b)** $16 \text{ photographers} = 160 \text{ small squares or } 64 \text{ photographers} = 640 \text{ small squares}$ | M1 | For establishing a ratio between photographers and area or calculating frequency density (may be implied by 2nd M1)
or $\frac{16}{160} = 0.1$ or $\frac{64}{640} = 0.1$ or $\frac{160}{16} = 10$ or $\frac{640}{64} = 10$ | |
Frequency density = 1.6 or Correct scale on the frequency density axis | M1 | For a correct ratio or expression using areas for photographers from 12 to 24
$\frac{x}{240} = \frac{16}{160}$ oe or $\frac{(20-12)}{240} \times 16 + \frac{(24-20)}{10} \times 5 \times 14$ | |
$= 24$ | A1 | Cao
**(c)** Using $n$ | Using $n + 1$ | M1 | For a correct method to find median using either $n$ or $n + 1$
$[Q_2] = 20 + \frac{(32-21)}{35-21} \times (25-20)$ oe | $[Q_2] = 20 + \frac{(32.5-21)}{35-21} \times (25-20)$ oe | |
or $25 - \frac{35-32}{35-21} \times (25-20)$ oe | or $25 - \frac{35-32.5}{35-21} \times (25-20)$ oe | |
= awrt 23.9 | = awrt 24.1 | A1 | awrt 23.9 or awrt 24.1 if using $n + 1$
**(d)** Mean = Median or Mean = Median e.g. Appropriate decision. Consistent with expectation for a normal distribution. | M1 A1ft | For a correct comment about mean and median ft their median
| | Allow mean is close to median to imply mean = median
| | Ignore any comments made about the shape of the histogram
| | For a correct compatible comment about Charlie's decision ft their median
| | If Mean = Median or Mean = Median, then the decision should be that a normal distribution is suitable [due to symmetry]
| | If Mean < Median or Mean > Median or mean ≠ median, then the decision should be that a normal distribution is not suitable [due to the skew in the data]
**Total: 8 marks**5.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fe416f2e-bc81-444b-a0ca-f0eae9a8b149-16_990_1473_246_296}
\end{center}
The histogram shows the number of hours worked in a given week by a group of 64 freelance photographers.
\begin{enumerate}[label=(\alph*)]
\item Give a reason to justify the use of a histogram to represent these data.
Given that 16 of these freelance photographers spent between 10 and 20 hours working in this week,
\item estimate the number that spent between 12 and 24 hours working in this week.
\item Find an estimate for the median time spent working in this week by these 64 freelance photographers.
Charlie decides to model these data using a normal distribution. Charlie calculates an estimate of the mean to be 23.9 hours to one decimal place.
\item Comment on Charlie's decision to use a normal distribution.
Give a justification for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2024 Q5}}