Handling Data 2
3 marks · NormalA botanist records the heights (h, in cm) of a random sample of ferns as shown in the grouped frequency table below:
| Height range (cm) | Frequency |
|---|---|
| 10 ≤ h < 15 | 4 |
| 15 ≤ h < 20 | 6 |
| 20 ≤ h < 25 | 11 |
| 25 ≤ h < 30 | 5 |
| 30 ≤ h < 35 | 2 |
Estimate the mean height of the ferns. Show your working.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
21.2 cm (±0.1)
Must show: midpoints in working
Common mistakes: Using lower bounds or upper bounds instead of mid-points
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Uses midpoints for each class and multiplies by frequency (follow-through)
M1 · 1 mark · Accurate addition and division by total frequency (Σfx/Σf) (follow-through)
A1 · 1 mark · Mean correctly estimated, answer to 1 d.p.
Worked solution
First, calculate mid-point of each class:
| Class | Mid-point (x) | Frequency (f) | f×x |
|---|---|---|---|
| 10–15 | 12.5 | 4 | 50.0 |
| 15–20 | 17.5 | 6 | 105.0 |
| 20–25 | 22.5 | 11 | 247.5 |
| 25–30 | 27.5 | 5 | 137.5 |
| 30–35 | 32.5 | 2 | 65.0 |
Sum frequencies: 4+6+11+5+2=28
Sum f×x: 50+105+247.5+137.5+65=605
Estimated mean = 605 / 28 = 21.6 (rounded to 1 d.p.: 21.6 cm)