| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw cumulative frequency graph from frequency table (unequal class widths) |
| Difficulty | Easy -1.2 This is a straightforward S1 statistics question requiring standard procedures: plotting cumulative frequency (routine), reading off a percentile value (direct application), and calculating standard deviation from grouped data using a formula (mechanical computation with given mean). All three parts are textbook exercises with no problem-solving or conceptual challenges. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Height (cm) | \(4 - 10\) | \(11 - 15\) | \(16 - 20\) | \(21 - 25\) | \(26 - 30\) |
| Frequency | 22 | 32 | 78 | 40 | 28 |
| Answer | Marks | Guidance |
|---|---|---|
| ht | \(<10.5\) | \(<15.5\) |
| CF | 22 | 54 |
| B1 | At least 4 CFs correct seen on graph | |
| B1 | Labels correct, i.e. all of ht, cm, cf | |
| M1 | Attempt at upper end points either 10 or 10.5 or 11 at least 4 upper end points | |
| A1 4 | All correct, i.e. points joined up from (3.5, 0) to (10.5, 22)...to (30.5, 200) Straight lines or curve | |
| (ii) 72% less, i.e. 144 less than ht \(h\). \(h = 22.5\) cm | M1, A1 2 | 144 used can be implied or single value in range 21 to 23 inclusive |
| Answer | Marks | Guidance |
|---|---|---|
| \[\text{sd} = 6.01\] | M1, B1, A1 3 | Using mid points attempt 7 ± 0.5in correct var formula incl − mean\(^2\) or At least 4 correct midpoints or Correct ans |
**(i)**
| ht | $<10.5$ | $<15.5$ | $<20.5$ | $<25.5$ | $<30.5$ |
|----|---------|---------|---------|---------|---------|
| CF | 22 | 54 | 132 | 172 | 200 |
| B1 | At least 4 CFs correct seen on graph
| B1 | Labels correct, i.e. all of ht, cm, cf
| M1 | Attempt at upper end points either 10 or 10.5 or 11 at least 4 upper end points
| A1 4 | All correct, i.e. points joined up from (3.5, 0) to (10.5, 22)...to (30.5, 200) Straight lines or curve
**(ii)** 72% less, i.e. 144 less than ht $h$. $h = 22.5$ cm | M1, A1 2 | 144 used can be implied or single value in range 21 to 23 inclusive
**(iii)** $$\text{var} = (7^2 \times 22 + 13^2 \times 32 + 18^2 \times 78 + 23^2 \times 40 + 28^2 \times 28)/200 - 18.39^2$$
$$= 74870/200 - 18.39^2 = 374.35 - 18.39^2 = 36.1579$$
$$\text{sd} = 6.01$$ | M1, B1, A1 3 | Using mid points attempt 7 ± 0.5in correct var formula incl − mean$^2$ or At least 4 correct midpoints or Correct ans
---On a certain day in spring, the heights of 200 daffodils are measured, correct to the nearest centimetre. The frequency distribution is given below.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Height (cm) & $4 - 10$ & $11 - 15$ & $16 - 20$ & $21 - 25$ & $26 - 30$ \\
\hline
Frequency & 22 & 32 & 78 & 40 & 28 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Draw a cumulative frequency graph to illustrate the data. [4]
\item 28\% of these daffodils are of height $h$ cm or more. Estimate $h$. [2]
\item You are given that the estimate of the mean height of these daffodils, calculated from the table, is 18.39 cm. Calculate an estimate of the standard deviation of the heights of these daffodils. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2014 Q6 [9]}}