| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Easy -1.8 This is a straightforward S1 question requiring basic calculations from a frequency table: mean (sum fx/sum f), median (middle value), and mode (highest frequency). Part (b) requires minimal interpretation. All techniques are routine recall with no problem-solving or conceptual challenge. |
| Spec | 2.02f Measures of average and spread |
| Number of births | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Number of weeks | 1 | 2 | 9 | 13 | 7 | 13 | 6 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean = \(\frac{\sum fx}{\sum x} = \frac{247}{52} = 4.75\) or \(4\frac{3}{4}\) | B2 | CAO (4.75 = 5 ⟹ ISW) |
| If B0 but evidence of \(\frac{\sum fx}{52}\) | M1 | |
| Median (26, 26½) = 5 | B2 (B1) | CAO; Stated identification of 26 or 26½; Need to see attempt at ≥ 4 F-values |
| If B0 but evidence of cumulative frequencies \(F: (0) 1 3 12 25 32 45 51 52\) or If data assumed continuous so use of \(4+\frac{x}{7}\) where \(0 < x < 2\) | M1 | (4 < median < 4.29) |
| Mode(s) = 4 and 6 | B1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Mode(s) More than one mode/value; Two modes/values; No unique mode/value | B1 | |
| B1dep | Dep only on previous B1 scored | |
| Notes: If data treated as two separate sets, then only marks available are B1 B1dep in (b); If averages confused then mark (a) as stated; eg median = 4 and 6 ⟹ B0 in (a) and in (b) "median, as two values" ⟹ B0 B0 | 2 |
**1(a)**
| Mean = $\frac{\sum fx}{\sum x} = \frac{247}{52} = 4.75$ or $4\frac{3}{4}$ | B2 | CAO (4.75 = 5 ⟹ ISW) |
| If B0 but evidence of $\frac{\sum fx}{52}$ | M1 | |
| Median (26, 26½) = 5 | B2 (B1) | CAO; Stated identification of 26 or 26½; Need to see attempt at ≥ 4 F-values |
| If B0 but evidence of cumulative frequencies $F: (0) 1 3 12 25 32 45 51 52$ or If data assumed continuous so use of $4+\frac{x}{7}$ where $0 < x < 2$ | M1 | (4 < median < 4.29) |
| Mode(s) = 4 and 6 | B1 | 5 | CAO both (so mode = 5 ⟹ B0) |
**1(b)**
| Mode(s) More than one mode/value; Two modes/values; No unique mode/value | B1 | | CAO; Or equivalent; eg not unique |
| | B1dep | | Dep only on previous B1 scored |
| Notes: If data treated as two separate sets, then only marks available are B1 B1dep in (b); If averages confused then mark (a) as stated; eg median = 4 and 6 ⟹ B0 in (a) and in (b) "median, as two values" ⟹ B0 B0 | | 2 | |
**Total for Q1: 7 marks**
---1 Ms N Parker always reads the columns of announcements in her local weekly newspaper. During each week of 2008, she notes the number of births announced. Her results are summarised in the table.
\begin{center}
\begin{tabular}{ | l | l | l | l | c | l | c | l | l | }
\hline
Number of births & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Number of weeks & 1 & 2 & 9 & 13 & 7 & 13 & 6 & 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean, median and modes of these data.
\item State, with a reason, which of the three measures of average in part (a) you consider to be the least appropriate for summarising the number of births.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2009 Q1 [7]}}