| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw cumulative frequency graph from frequency table (unequal class widths) |
| Difficulty | Easy -1.3 This is a straightforward statistics question requiring routine application of cumulative frequency techniques: constructing a cumulative frequency graph from grouped data and reading values from it. All steps are standard textbook procedures with no problem-solving or conceptual challenges beyond basic recall and careful plotting. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency |
| Number of rooms occupied | \(1 - 20\) | \(21 - 40\) | \(41 - 50\) | \(51 - 60\) | \(61 - 70\) | \(71 - 90\) |
| Frequency | 10 | 32 | 62 | 50 | 28 | 18 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | Attempt at cf table (up to 200) |
| M1 | Linear scale minimum 0 to 200 and 20 to 80, and labels | |
| M1 | Attempt to plot points at \((20.5, 10), (40.5, 42), (50.5, 104), (60.5, 154), (70.5, 182), (90.5, 200)\), accept \((20, 10), (40, 42)\) or \((21, 10), (41, 42)\) etc | |
| A1 [4] | All points correct and joined up, allow \((0, 0)\) or \((0.5, 0)\) | |
| (ii) Line on graph up from \(30\) \(200 - 20 = 180\) | M1, A1 [2] | Line or mark seen, can be implied if matches graph and in range. Accept \(174 - 180\) if reading from graph |
| OR using lin int \(10 + \frac{(30-20.5)}{20} \times 32 = 25.2 = 174.8\) | M1, A1 | Can have \(20\) or \(20.5\); Accept decimals \(174 - 175\) if using lin int |
| (iii) Line on graph across from \(150\) | M1 | Line or mark seen, can be implied if matches graph and in range. \(150\) seen and line between \(140\) and \(160\) |
| \(59\) rooms | A1 [2] | Accept \(58 - 60\) |
| OR lin int \(50.5 + \frac{46}{50} \times 10 = 59\) or \(60\) | M1, A1 | Can have \(50\) or \(50.5\); Must be integer |
**(i)** | M1 | Attempt at cf table (up to 200)
| M1 | Linear scale minimum 0 to 200 and 20 to 80, and labels
| M1 | Attempt to plot points at $(20.5, 10), (40.5, 42), (50.5, 104), (60.5, 154), (70.5, 182), (90.5, 200)$, accept $(20, 10), (40, 42)$ or $(21, 10), (41, 42)$ etc
| A1 [4] | All points correct and joined up, allow $(0, 0)$ or $(0.5, 0)$
**(ii)** Line on graph up from $30$ $200 - 20 = 180$ | M1, A1 [2] | Line or mark seen, can be implied if matches graph and in range. Accept $174 - 180$ if reading from graph
OR using lin int $10 + \frac{(30-20.5)}{20} \times 32 = 25.2 = 174.8$ | M1, A1 | Can have $20$ or $20.5$; Accept decimals $174 - 175$ if using lin int
**(iii)** Line on graph across from $150$ | M1 | Line or mark seen, can be implied if matches graph and in range. $150$ seen and line between $140$ and $160$
$59$ rooms | A1 [2] | Accept $58 - 60$
OR lin int $50.5 + \frac{46}{50} \times 10 = 59$ or $60$ | M1, A1 | Can have $50$ or $50.5$; Must be integer5 A hotel has 90 rooms. The table summarises information about the number of rooms occupied each day for a period of 200 days.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of rooms occupied & $1 - 20$ & $21 - 40$ & $41 - 50$ & $51 - 60$ & $61 - 70$ & $71 - 90$ \\
\hline
Frequency & 10 & 32 & 62 & 50 & 28 & 18 \\
\hline
\end{tabular}
\end{center}
(i) Draw a cumulative frequency graph on graph paper to illustrate this information.\\
(ii) Estimate the number of days when over 30 rooms were occupied.\\
(iii) On $75 \%$ of the days at most $n$ rooms were occupied. Estimate the value of $n$.
\hfill \mbox{\textit{CAIE S1 2011 Q5 [8]}}