| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from discrete rounded data |
| Difficulty | Moderate -0.8 This is a straightforward S1 statistics question requiring basic probability calculation (selecting without replacement), mean from grouped data using given Σfx, and reverse calculation of a subset mean from an overall mean. All techniques are standard textbook exercises with no conceptual challenges or novel problem-solving required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
|
| ||||
| \(5 - 9\) | 2 | ||||
| \(10 - 14\) | 9 | ||||
| \(15 - 19\) | 20 | ||||
| \(20 - 24\) | 13 | ||||
| \(25 - 29\) | 8 | ||||
| \(30 - 34\) | 3 |
| Answer | Marks |
|---|---|
| \(P(\text{both longer than 24.5}) = \frac{11}{55} \times \frac{10}{54} = \frac{1}{27}\) or 0.037 or 0.037 | M1A1 |
| 2 fracs x w/o rep., awrt 0.037 |
| Answer | Marks |
|---|---|
| Estimate of mean time spent on their conversations is \(\bar{x} = \frac{1060}{55} = 19\frac{3}{11}\) or 19.27 or 19.3 | M1A1 |
| 1060/total, awrt 19.3 or 19mins 16s |
| Answer | Marks |
|---|---|
| \(\frac{1060 + \sum fy}{80} = 21\) | B1 |
| \(21 \times 80 = 1680\) | |
| \(\sum fy = 620\) | M1 |
| Subtracting 'their 1060' | |
| \(\bar{y} = \frac{620}{25} = 24.8\) | M1A1 |
| Dividing their 620 by 25 |
| Answer | Marks |
|---|---|
| Increase in mean value. Length of conversations increased considerably during 25 weeks relative to 55 weeks | B1 |
| context - ft only from comment above | B1 |
**Part (a):**
| $P(\text{both longer than 24.5}) = \frac{11}{55} \times \frac{10}{54} = \frac{1}{27}$ or 0.037 or 0.037 | M1A1 |
| 2 fracs x w/o rep., awrt 0.037 | |
**Part (b):**
| Estimate of mean time spent on their conversations is $\bar{x} = \frac{1060}{55} = 19\frac{3}{11}$ or 19.27 or 19.3 | M1A1 |
| 1060/total, awrt 19.3 or 19mins 16s | |
**Part (c):**
| $\frac{1060 + \sum fy}{80} = 21$ | B1 |
| $21 \times 80 = 1680$ | |
| $\sum fy = 620$ | M1 |
| Subtracting 'their 1060' | |
| $\bar{y} = \frac{620}{25} = 24.8$ | M1A1 |
| Dividing their 620 by 25 | |
**Part (d):**
| Increase in mean value. Length of conversations increased considerably during 25 weeks relative to 55 weeks | B1 |
| context - ft only from comment above | B1 |
**Total 10 marks**
---2. Sunita and Shelley talk to one another once a week on the telephone. Over many weeks they recorded, to the nearest minute, the number of minutes spent in conversation on each occasion. The following table summarises their results.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
\begin{tabular}{ c }
Time \\
(to the nearest minute) \\
\end{tabular} & \begin{tabular}{ c }
Number of \\
Conversations \\
\end{tabular} \\
\hline
$5 - 9$ & 2 \\
\hline
$10 - 14$ & 9 \\
\hline
$15 - 19$ & 20 \\
\hline
$20 - 24$ & 13 \\
\hline
$25 - 29$ & 8 \\
\hline
$30 - 34$ & 3 \\
\hline
\end{tabular}
\end{center}
Two of the conversations were chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that both of them were longer than 24.5 minutes.
The mid-point of each class was represented by $x$ and its corresponding frequency by $f$, giving $\Sigma f x = 1060$.
\item Calculate an estimate of the mean time spent on their conversations.
During the following 25 weeks they monitored their weekly conversations and found that at the end of the 80 weeks their overall mean length of conversation was 21 minutes.
\item Find the mean time spent in conversation during these 25 weeks.
\item Comment on these two mean values.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2006 Q2 [10]}}