The table shows the distribution of hours spent at work by the employees of a factory in a week. Time (in hours) 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 ...
Question 1 Report
The table shows the distribution of hours spent at work by the employees of a factory in a week.
| Time (in hours) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| No of persons | 8 | 11 | 23 | 25 | 8 | 5 |
(a) Draw an ogive for the distribution.
(b) Using your graph, estimate the (i) lower quartile (ii) median (iii) 40th percentile (iv) number of employees that spent at least 50 hours 30 minutes at work.
Answer Details
(a) Cumulative-frequency table
| Time (hours) | Frequency | Class boundaries | Cumulative frequency |
|---|---|---|---|
| 20–29 | 8 | 19.5–29.5 | 8 |
| 30–39 | 11 | 29.5–39.5 | 19 |
| 40–49 | 23 | 39.5–49.5 | 42 |
| 50–59 | 25 | 49.5–59.5 | 67 |
| 60–69 | 8 | 59.5–69.5 | 75 |
| 70–79 | 5 | 69.5–79.5 | 80 |
Plot cumulative frequency against the upper class boundary, beginning at .5, 0. The resulting less-than ogive is:
(b) Total number of employees, N=80.
- The lower quartile corresponds to cumulative frequency \(\frac{N}{4}=20\). From the ogive, \(Q_1\approx 40\) hours.
- The median corresponds to cumulative frequency \(\frac{N}{2}=40\). From the ogive, median \(\approx 49\) hours.
- The 40th percentile corresponds to cumulative frequency \(0.40\times80=32\). From the ogive, \(P_{40}\approx45.5\) hours.
- At 50 hours 30 minutes, that is, 50.5 hours, the cumulative frequency read from the ogive is approximately 45. Hence, the number who spent at least 50.5 hours at work is \[80-45=\boxed{35\text{ employees}}.\]
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