The table below shows the distribution of the waiting times for some customers in a certain petrol station. Waiting time (in mins) No of customers 1.5 - 1.9...
Question 1 Report
The table below shows the distribution of the waiting times for some customers in a certain petrol station.
| Waiting time (in mins) | No of customers |
| 1.5 - 1.9 | 3 |
| 2.0 - 2.4 | 10 |
| 2.5 - 2.9 | 18 |
| 3.0 - 3.4 | 10 |
| 3.5 - 3.9 | 7 |
| 4.0 - 4.4 | 2 |
(a) Write down the class boundaries of the distribution.
(b) Construct a cumulative frequency curve for the data;
(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.
Answer Details
(a) Class boundaries. Subtract 0.05 from each lower limit and add 0.05 to each upper limit.
| Class | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1.5-1.9 | 1.45-1.95 | 3 | 3 |
| 2.0-2.4 | 1.95-2.45 | 10 | 13 |
| 2.5-2.9 | 2.45-2.95 | 18 | 31 |
| 3.0-3.4 | 2.95-3.45 | 10 | 41 |
| 3.5-3.9 | 3.45-3.95 | 7 | 48 |
| 4.0-4.4 | 3.95-4.45 | 2 | 50 |
(b) Plot cumulative frequency against the upper boundaries: \((1.95,3),(2.45,13),(2.95,31),(3.45,41),(3.95,48),(4.45,50)\) and join with a smooth curve. \(N=50\).
(c)(i) Interquartile range. \(Q_1\) at \(\frac{N}{4}=12.5\) (in class 1.95-2.45):
\[ Q_1=1.95+\frac{12.5-3}{10}\times 0.5 = 1.95+0.475 = 2.43 \]\(Q_3\) at \(\frac{3N}{4}=37.5\) (in class 2.95-3.45):
\[ Q_3=2.95+\frac{37.5-31}{10}\times 0.5 = 2.95+0.325 = 3.28 \] \[ \text{IQR}=Q_3-Q_1=3.28-2.43 \approx 0.85 \text{ min} \](ii) Proportion waiting more than 3 minutes. Read the c.f. at \(3.0\): between \(2.95\,(31)\) and \(3.45\,(41)\),
\[ \text{c.f.}(3.0)=31+\frac{3.0-2.95}{0.5}\times 10 = 32 \]Number waiting more than 3 min \(=50-32=18\), so the proportion is
\[ \frac{18}{50}=0.36 \;\;(36\%) \]Download The App On Google Playstore
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