All subjects · Mathematics · WST01 (S1)

Grouped data: estimating the mean and median

Once data are grouped the exact values are lost, so you estimate the mean using midpoints and the median using linear interpolation.

Estimated mean

Use the midpoint of each class as its representative value: x̄ ≈ Σfx / Σf, where x is the midpoint. It is only an estimate because the values in each class are not all at the midpoint.

Time, t (minutes)	Frequency	Cumulative frequency
0 ≤ t < 10	12	12
10 ≤ t < 15	18	30
15 ≤ t < 20	25	55
20 ≤ t < 30	20	75
30 ≤ t < 50	10	85

Median by interpolation

Assume values are spread evenly through each class.

1. Find the position n/2 (Pearson accepts (n + 1)/2 as well).

2. Use cumulative frequencies to find the class containing it.

3. Median = lower boundary + (position - cumulative frequency before the class) ÷ class frequency × class width.

For n = 85: position 42.5 lies in 15 ≤ t < 20 (cumulative 30 before it, frequency 25). Median = 15 + (42.5 - 30)/25 × 5 = 17.5.

Interpreting

State the estimate in context ('the median time is about 17.5 minutes'). The mode for grouped data is the modal class, the class with the highest frequency density (not simply the highest frequency when widths differ).

Worked example

Estimate the mean time from the table.

Midpoints: 5, 12.5, 17.5, 25, 40.
Σfx = 1622.5; Σf = 85.
x̄ ≈ 1622.5 ÷ 85

Answer: x̄ ≈ 19.1 minutes

Key idea: Estimated mean = Σfx/Σf with midpoints; median = L + (n/2 - cf before)/f × width.

Check yourself

Estimate the mean time. Give your answer to 3 significant figures.
Show answer
19.1 minutes. Σfx = 1622.5 using midpoints, so x̄ ≈ 1622.5/85 = 19.1.
Use linear interpolation to estimate the median time.
Show answer
17.5 minutes. Position 42.5; class 15 ≤ t < 20; 15 + (42.5 - 30)/25 × 5 = 17.5.
Find the midpoint of the class 30 ≤ t < 50.
Show answer
40. (30 + 50)/2 = 40.
Grouped data have 60 values. The class 20 ≤ x < 30 has frequency 15, and 24 values lie below 20. Use linear interpolation, with position n/2, to estimate the median.
Show answer
24. Position 30: 20 + (30 - 24)/15 × 10 = 24.
In the table, which is the modal class?
- 15 ≤ t < 20
- 20 ≤ t < 30
- 30 ≤ t < 50
- 0 ≤ t < 10
Show answer
15 ≤ t < 20. Frequency densities are 1.2, 3.6, 5, 2, 0.5: the highest is 5 for 15 ≤ t < 20.
Why is the mean from a grouped frequency table only an estimate?
- The frequencies might be wrong
- The classes have different widths
- The exact values are unknown, so midpoints are used
- The mean must always be a whole number
Show answer
The exact values are unknown, so midpoints are used. Grouping loses the raw values; midpoints stand in for them.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.