| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw cumulative frequency graph from cumulative frequency table |
| Difficulty | Easy -1.8 This is a straightforward cumulative frequency question requiring only routine plotting of given data points and reading values from the graph. Parts (iii) and (iv) involve basic arithmetic with the table values. No problem-solving insight or complex statistical reasoning is required—purely mechanical application of standard techniques taught early in S1. |
| 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 |
| Number of pupils in a school | \(\leqslant 100\) | \(\leqslant 150\) | \(\leqslant 200\) | \(\leqslant 250\) | \(\leqslant 350\) | \(\leqslant 450\) | \(\leqslant 600\) |
| Cumulative frequency | 200 | 800 | 1600 | 2100 | 4100 | 4700 | 5000 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | Sensible attempt at graph using u.c.b. |
| M1 (Indpt) | 2500 seen in median attempt on a CF graph. Can be implied | |
| A1 [3] | Correct answer \(\pm\) or \(- 5\) | |
| (ii) 20% less than 160 | M1 | Using 20% |
| A1 [2] | Correct answer \(\pm\) or \(- 5\) | |
| (iii) 2100 – 1600 = 500 | B1 [1] | |
| (iv) \((50.5 \times 200 + 125.5 \times 600 + 175.5 \times 800 + 225.5 \times 500 + 300.5 \times 2000 + 400.5 \times 600 + 525.5 \times 300) / 5000 = 268\) | M1 | Using an attempt at mid-points |
| M1 | Using an attempt at frequencies | |
| A1 | Correct mid-points or frequencies | |
| A1 [4] | Correct answer only |
**(i)** | M1 | Sensible attempt at graph using u.c.b.
| M1 (Indpt) | 2500 seen in median attempt on a CF graph. Can be implied
| A1 [3] | Correct answer $\pm$ or $- 5$
**(ii)** 20% less than 160 | M1 | Using 20%
| A1 [2] | Correct answer $\pm$ or $- 5$
**(iii)** 2100 – 1600 = 500 | B1 [1] |
**(iv)** $(50.5 \times 200 + 125.5 \times 600 + 175.5 \times 800 + 225.5 \times 500 + 300.5 \times 2000 + 400.5 \times 600 + 525.5 \times 300) / 5000 = 268$ | M1 | Using an attempt at mid-points
| M1 | Using an attempt at frequencies
| A1 | Correct mid-points or frequencies
| A1 [4] | Correct answer only6 There are 5000 schools in a certain country. The cumulative frequency table shows the number of pupils in a school and the corresponding number of schools.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of pupils in a school & $\leqslant 100$ & $\leqslant 150$ & $\leqslant 200$ & $\leqslant 250$ & $\leqslant 350$ & $\leqslant 450$ & $\leqslant 600$ \\
\hline
Cumulative frequency & 200 & 800 & 1600 & 2100 & 4100 & 4700 & 5000 \\
\hline
\end{tabular}
\end{center}
(i) Draw a cumulative frequency graph with a scale of 2 cm to 100 pupils on the horizontal axis and a scale of 2 cm to 1000 schools on the vertical axis. Use your graph to estimate the median number of pupils in a school.\\
(ii) $80 \%$ of the schools have more than $n$ pupils. Estimate the value of $n$ correct to the nearest ten.\\
(iii) Find how many schools have between 201 and 250 (inclusive) pupils.\\
(iv) Calculate an estimate of the mean number of pupils per school.
\hfill \mbox{\textit{CAIE S1 2011 Q6 [10]}}