Measures of Dispersion for Grouped Data
How to Estimate the mean of grouped data
Use this to estimate the mean when data is given in class intervals rather than as single values.
Before you start
- Finding the midpoint of an interval
- Multiplying a decimal by a whole number
- Reading a frequency table
- Reading Σ (sigma) as 'add up'
When to use it
Use this to estimate the mean when data is given in class intervals rather than as single values.
The steps
- Find the midpoint of each class interval.
- Multiply each midpoint by the frequency of its class.
- Add up these products to get Σfx.
- Add up the frequencies to get Σf.
- Estimate the mean as Σfx ÷ Σf.
Worked example
The masses of 20 parcels are grouped: 1–5 kg (4 parcels), 6–10 kg (6), 11–15 kg (7), 16–20 kg (3). Estimate the mean mass.
- Find the midpoint of each class: (1+5)/2 = 3, (6+10)/2 = 8, (11+15)/2 = 13, (16+20)/2 = 18.
- Multiply each midpoint by the frequency of its class: 3×4 = 12, 8×6 = 48, 13×7 = 91, 18×3 = 54.
- Add these products to get Σfx: 12 + 48 + 91 + 54 = 205.
- Add the frequencies to get Σf: 4 + 6 + 7 + 3 = 20.
- Estimate the mean as Σfx ÷ Σf: 205 ÷ 20 = 10.25.
A second example, with a twist
The classes 1–10, 11–20, … give decimal midpoints (5.5, 15.5, …) and the final mean does not divide exactly, so it must be rounded. The marks of 30 students are grouped: 1–10 (5 students), 11–20 (9), 21–30 (11), 31–40 (5).
Estimate the mean mark, correct to 2 decimal places.
- Find the midpoint of each class: (1+10)/2 = 5.5, (11+20)/2 = 15.5, (21+30)/2 = 25.5, (31+40)/2 = 35.5.
- Multiply each midpoint by the frequency of its class: 5.5×5 = 27.5, 15.5×9 = 139.5, 25.5×11 = 280.5, 35.5×5 = 177.5.
- Add these products to get Σfx: 27.5 + 139.5 + 280.5 + 177.5 = 625.
- Add the frequencies to get Σf: 5 + 9 + 11 + 5 = 30.
- Estimate the mean as Σfx ÷ Σf: 625 ÷ 30 = 20.83 (2 d.p.).
Formulae you may need
Formula pages
Practise this in a KBAT problem
Frequently asked questions
Why is this only an estimate of the mean?
Because grouped data hides the exact values, we replace every value in a class with its midpoint and assume the values are evenly spread. That assumption is rarely perfect, so Σfx ÷ Σf gives a close estimate rather than the true mean.
How do I find a class midpoint?
Add the lower and upper values of the class and divide by 2. For 11–20 that is (11 + 20) ÷ 2 = 15.5.
Do it for every class; the midpoint stands in for all the values in that class when you multiply by the frequency.
Do I divide by the number of classes?
No. Divide Σfx by Σf, which is the total frequency, how many data values there are altogether, not how many classes.
In the second example that is 30 students, so you divide by 30, not by the 4 class intervals.