The table shows the heights in cm of some seedlings in a certain garden.
(a) Draw the cumulative frequency curve for the distribution.
(b) Using the curve in (a), find thesemi-interquartile range.
Cumulative frequency table (cf plotted against upper class boundaries). \(N = 50\).
| Height (cm) | Freq | Upper boundary | Cumulative freq |
| 36 - 40 | 3 | 40.5 | 3 |
| 41 - 45 | 9 | 45.5 | 12 |
| 46 - 50 | 21 | 50.5 | 33 |
| 51 - 55 | 12 | 55.5 | 45 |
| 56 - 60 | 5 | 60.5 | 50 |
(a) Plot (40.5, 3), (45.5, 12), (50.5, 33), (55.5, 45), (60.5, 50) and join with a smooth ogive.
(b) Semi-interquartile range.
\(Q_1\) at position \( \tfrac{50}{4} = 12.5\), in class 46 - 50 (\(L=45.5, \text{cf}=12, f=21\)):
\[ Q_1 = 45.5 + \frac{12.5 - 12}{21}\times 5 = 45.5 + 0.12 = 45.62 \]
\(Q_3\) at position \( \tfrac{3\times 50}{4} = 37.5\), in class 51 - 55 (\(L=50.5, \text{cf}=33, f=12\)):
\[ Q_3 = 50.5 + \frac{37.5 - 33}{12}\times 5 = 50.5 + 1.88 = 52.38 \]
\[ \text{Semi-interquartile range} = \frac{Q_3 - Q_1}{2} = \frac{52.38 - 45.62}{2} \approx \mathbf{3.4 \text{ cm}} \]