The table shows the heights in cm of some seedlings in a certain garden. Height (cm) 36-40 41-45 46-50 51-55 56-60 Frequency 3 9 21 12 5 (a) Draw the cumula...
Question 1 Report
The table shows the heights in cm of some seedlings in a certain garden.
| Height (cm) | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 |
| Frequency | 3 | 9 | 21 | 12 | 5 |
(a) Draw the cumulative frequency curve for the distribution.
(b) Using the curve in (a), find thesemi-interquartile range.
Answer Details
The continuous class boundaries and cumulative frequencies are:
| Height (cm) | Class boundaries (cm) | Frequency | Cumulative frequency |
|---|---|---|---|
| 36–40 | 35.5–40.5 | 3 | 3 |
| 41–45 | 40.5–45.5 | 9 | 12 |
| 46–50 | 45.5–50.5 | 21 | 33 |
| 51–55 | 50.5–55.5 | 12 | 45 |
| 56–60 | 55.5–60.5 | 5 | 50 |
(a) The cumulative frequency curve is shown below. The point [35.5, 0[0m is included before plotting the upper class boundaries against their cumulative frequencies.
(b) Total frequency, \(N=50\).
On the curve, the lower quartile corresponds to cumulative frequency
\[\frac{N}{4}=\frac{50}{4}=12.5.\]
Reading across from \(12.5\) to the curve and down to the height axis gives
\[Q_1\approx 45.6\text{ cm}.\]
The upper quartile corresponds to cumulative frequency
\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]
Reading across from \(37.5\) to the curve and down to the height axis gives
\[Q_3\approx 52.0\text{ cm}.\]
Hence,
\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]
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