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
To find the median of grouped data, we first need to convert the discontinuous classes to continuous classes by adjusting the class limits. Then we find the cumulative frequencies and identify the median class where the n/2th observation lies. Finally, we apply the median formula for grouped data.
Solution Steps
Step 1: Convert discontinuous classes to continuous classes
Since the data is measured correct to the nearest millimetre, we adjust the class limits by 0.5. Continuous classes: 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: Construct cumulative frequency table
Length (in mm) Frequency (f) Cumulative Frequency (cf) 117.5 - 126.5 3 3 126.5 - 135.5 5 8 135.5 - 144.5 9 17 144.5 - 153.5 12 29 153.5 - 162.5 5 34 162.5 - 171.5 4 38 171.5 - 180.5 2 40 Step 3: Find n/2
n = 40
n/2 = 40/2 = 20
Step 4: Identify the median class
The median class is the class whose cumulative frequency is greater than n/2 = 20. Since cf = 29 > 20, the median class is 144.5 - 153.5.
Step 5: Apply the median formula
Median = l + [(n/2 - cf)/f] × h
Where:
- l = lower limit of median class = 144.5
- n = 40
- cf = cumulative frequency of class preceding median class = 17
- f = frequency of median class = 12
- h = class size = 153.5 - 144.5 = 9
Median = 144.5 + [(20 - 17)/12] × 9
= 144.5 + [3/12] × 9
= 144.5 + (1/4) × 9
= 144.5 + 2.25
= 146.75 mm