The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
| Length (in mm) | Number of leaves |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
146.75 mm
Explanation
This question requires converting discontinuous classes to continuous classes before applying the median formula. The median class is identified where cumulative frequency first exceeds n/2 = 20. The formula Median = l + ((n/2 - cf)/f) × h is then applied with values from the median class 144.5 - 153.5.
Solution Steps
Step 1: Convert discontinuous classes to continuous classes. Since data is measured correct to nearest millimetre, adjust limits by 0.5: 117.5-126.5, 126.5-135.5, 135.5-144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5, 171.5-180.5
Step 2: Calculate cumulative frequencies: 3, 8, 17, 29, 34, 38, 40. Total n = 40.
Step 3: Find n/2 = 40/2 = 20. The median class is where cumulative frequency first exceeds 20, which is 144.5-153.5 (cf = 29).
Step 4: Identify values: l = 144.5, cf = 17 (preceding class), f = 12, h = 9
Step 5: Apply formula: Median = l + ((n/2 - cf)/f) × h = 144.5 + ((20 - 17)/12) × 9 = 144.5 + (3/12) × 9 = 144.5 + 2.25 = 146.75 mm