| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Compare using diagrams only |
| Difficulty | Moderate -0.8 This is a straightforward cumulative frequency question requiring basic interpretation of tables and standard calculations. Part (i) involves simple lookup of medians (150th value), part (ii) is direct subtraction of cumulative frequencies, and part (iii) is a routine mean calculation from grouped data using midpoints. All techniques are standard S1 procedures with no problem-solving or novel insight required. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Mark | \(< 10\) | \(< 20\) | \(< 35\) | \(< 50\) | \(< 70\) | \(< 100\) |
| Cumulative frequency, \(A\) | 25 | 68 | 159 | 234 | 260 | 300 |
| Cumulative frequency, \(B\) | 10 | 46 | 72 | 144 | 198 | 300 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) median\(_A < 35\) or \(20 \leq\) median\(_A < 35\) or median\(_B = 33.0/33.1/33.5/33.6\) or median\(_B \geq 50\) or \(50 \leq\) median\(_B < 70\) or median\(_B = 51.7/51.9/52.2/52.4\) median\(_B > \) median\(_A\) | B1 | Correct numerical statement re median\(_A\) or median\(_B\) |
| B1 [2] | Correct numerical statement re other median and a conclusion | |
| OR \(A\) has 66 card \(50 <\) mark \(< 100\), so med\(_A < 50\) or \(B\) has 156 card \(50 <\) mark \(< 100\), so med\(_B > 50\) median\(_B >\) median\(_A\) | B1, B1 | As before; As before |
| (ii) \(159 - 68 = 91\) | B1 [1] | Correct final answer |
| (iii) mean\(= \left(\frac{4.5 \times 25 + 14.5 \times 43 + 27 \times 91}{+ ..... + 84.5 \times 40}\right) / 300\) | M1 | Using an attempt at mid-points, not end points or class widths |
| M1 | Using an attempt at frequencies, not cum freqs | |
| M1 | Sum of 6 prods, correct freqs, divided by 300 | |
| \(= 11270 / 300 = 37.6\) | A1 [4] | Correct answer |
**(i)** median$_A < 35$ or $20 \leq$ median$_A < 35$ or median$_B = 33.0/33.1/33.5/33.6$ or median$_B \geq 50$ or $50 \leq$ median$_B < 70$ or median$_B = 51.7/51.9/52.2/52.4$ median$_B > $ median$_A$ | B1 | Correct numerical statement re median$_A$ or median$_B$
| B1 [2] | Correct numerical statement re other median and a conclusion
OR $A$ has 66 card $50 <$ mark $< 100$, so med$_A < 50$ or $B$ has 156 card $50 <$ mark $< 100$, so med$_B > 50$ median$_B >$ median$_A$ | B1, B1 | As before; As before
**(ii)** $159 - 68 = 91$ | B1 [1] | Correct final answer
**(iii)** mean$= \left(\frac{4.5 \times 25 + 14.5 \times 43 + 27 \times 91}{+ ..... + 84.5 \times 40}\right) / 300$ | M1 | Using an attempt at mid-points, not end points or class widths
| M1 | Using an attempt at frequencies, not cum freqs
| M1 | Sum of 6 prods, correct freqs, divided by 300
$= 11270 / 300 = 37.6$ | A1 [4] | Correct answer
---3 The following cumulative frequency table shows the examination marks for 300 candidates in country $A$ and 300 candidates in country $B$.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Mark & $< 10$ & $< 20$ & $< 35$ & $< 50$ & $< 70$ & $< 100$ \\
\hline
Cumulative frequency, $A$ & 25 & 68 & 159 & 234 & 260 & 300 \\
\hline
Cumulative frequency, $B$ & 10 & 46 & 72 & 144 & 198 & 300 \\
\hline
\end{tabular}
\end{center}
(i) Without drawing a graph, show that the median for country $B$ is higher than the median for country $A$.\\
(ii) Find the number of candidates in country $A$ who scored between 20 and 34 marks inclusive.\\
(iii) Calculate an estimate of the mean mark for candidates in country $A$.
\hfill \mbox{\textit{CAIE S1 2011 Q3 [7]}}