| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Use linear interpolation for median or quartiles |
| Difficulty | Moderate -0.8 This is a routine S1 question testing standard procedures: linear interpolation from grouped frequency data (a straightforward algorithm once cumulative frequencies are found) and using normal distribution tables with inverse lookup. Part (c) requires only basic comparison. The techniques are mechanical with no problem-solving insight needed, making it easier than average A-level maths questions. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| \(0 -\) | \(30 -\) | \(60 -\) | \(90 -\) | \(120 -\) | \(240 -\) | \(360 -\) | ||
| Number of users | 15 | 31 | 32 | 23 | 17 | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) cum. freqs: 15, 46, 78, 101, 118, 120 | M1 | |
| \(Q_1 = 30.25^{\text{th}} = 30 + 30(\frac{15.25}{32}) = 44.8\) [30th → 44.5] | M2 A3 | |
| \(Q_2 = 60.5^{\text{th}} = 60 + 30(\frac{14.5}{32}) = 73.6\) [60th → 73.1] | ||
| \(Q_3 = 90.75^{\text{th}} = 90 + 30(\frac{12.75}{32}) = 106.6\) [90th → 105.7] | ||
| (b) median = mean = 72 minutes | A1 | |
| \(P(Z < \frac{Q_1 - 72}{48}) = 0.25\) | M1 | |
| \(\therefore \frac{Q_1 - 72}{48} = -0.67; \quad Q_1 = 39.8\) (1dp) | M1 A1 | |
| \(P(Z < \frac{Q_3 - 72}{48}) = 0.75\) | M1 | |
| \(\therefore \frac{Q_3 - 72}{48} = 0.67; \quad Q_3 = 104.2\) (1dp) | M1 A1 | |
| (c) e.g. median and quartiles from model all slightly lower than in new results but reasonably close so fairly suitable model | B2 | (15 marks) |
(a) cum. freqs: 15, 46, 78, 101, 118, 120 | M1 |
$Q_1 = 30.25^{\text{th}} = 30 + 30(\frac{15.25}{32}) = 44.8$ [30th → 44.5] | M2 A3 |
$Q_2 = 60.5^{\text{th}} = 60 + 30(\frac{14.5}{32}) = 73.6$ [60th → 73.1] |
$Q_3 = 90.75^{\text{th}} = 90 + 30(\frac{12.75}{32}) = 106.6$ [90th → 105.7] |
(b) median = mean = 72 minutes | A1 |
$P(Z < \frac{Q_1 - 72}{48}) = 0.25$ | M1 |
$\therefore \frac{Q_1 - 72}{48} = -0.67; \quad Q_1 = 39.8$ (1dp) | M1 A1 |
$P(Z < \frac{Q_3 - 72}{48}) = 0.75$ | M1 |
$\therefore \frac{Q_3 - 72}{48} = 0.67; \quad Q_3 = 104.2$ (1dp) | M1 A1 |
(c) e.g. median and quartiles from model all slightly lower than in new results but reasonably close so fairly suitable model | B2 | (15 marks)7. A cyber-cafe recorded how long each user stayed during one day giving the following results.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
Length of stay \\
(minutes) \\
\end{tabular} & $0 -$ & $30 -$ & $60 -$ & $90 -$ & $120 -$ & $240 -$ & $360 -$ \\
\hline
Number of users & 15 & 31 & 32 & 23 & 17 & 2 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use linear interpolation to estimate the median and quartiles of these data.
The results of a previous study had led to the suggestion that the length of time each user stays can be modelled by a normal distribution with a mean of 72 minutes and a standard deviation of 48 minutes.
\item Find the median and quartiles that this model would predict.
\item Comment on the suitability of the suggested model in the light of the new results.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}