Statistics with Grouped Data
Senior 2 · Mathematics
Statistics with Grouped Data
Lesson 4 of 5
25 min
The mean is not always the most useful average. A shop selling school shoes cares about the most common size: the mode. A head teacher setting a start time might want the time by which half the students have arrived: the median. With grouped data we first find the class that contains each one, then estimate the value inside it.
The modal class is the class with the highest frequency. For the walking times, it is 25–34 (frequency 13).
To estimate the mode inside it, we lean towards the neighbouring class with the bigger frequency:
Mode ≈ L + (d1 ÷ (d1 + d2)) × c
Worked example 1
Walking times: L = 24.5, d1 = 13 − 12 = 1, d2 = 13 − 7 = 6, c = 10.
Mode ≈ 24.5 + (1 ÷ 7) × 10 = 24.5 + 1.43 = 25.9 minutes (to 1 decimal place).
This makes sense: the class before (12) is much fuller than the class after (7), so the mode sits near the bottom of the modal class.
The median is the middle value when the data are in order. With N values, it is the value at position N ÷ 2 in the cumulative frequency. For N = 40, that is the 20th value.
The cumulative frequencies are 5, 17, 30, 37, 40. The running total passes 20 in the class 25–34 (17 before it, 30 by its end), so the median class is 25–34. Then
Median ≈ L + ((N ÷ 2 − F) ÷ f) × c
Worked example 2
L = 24.5, N ÷ 2 = 20, F = 17, f = 13, c = 10.
Median ≈ 24.5 + ((20 − 17) ÷ 13) × 10 = 24.5 + 2.31 = 26.8 minutes.
The idea: we need 3 more students out of the 13 in the class, so we go 313 of the way through the 10-minute class.
For comparison, the raw data's median is 26.5 minutes and the grouped mean is 27.25 minutes. The three averages are close but not equal.
Milk yields (litres) of 20 cows: 0 ≤ x < 4: 3; 4 ≤ x < 8: 8; 8 ≤ x < 12: 6; 12 ≤ x < 16: 3. Here the limits are the boundaries.
Tip
The median class is found from the cumulative frequency, not by picking the middle row of the table. Find N ÷ 2, run down the cumulative column, and stop at the first class whose running total reaches it. And use the lower boundary (24.5), not the lower limit (25), for L.
Marks out of 50 of 30 students: 0–9: 2; 10–19: 5; 20–29: 9; 30–39: 10; 40–49: 4.
Which is the modal class?
Solution: 30–39, with frequency 10.
Estimate the mode.
Solution: L = 29.5, d1 = 10 − 9 = 1, d2 = 10 − 4 = 6, c = 10. Mode ≈ 29.5 + (1 ÷ 7) × 10 = 30.9 marks.
Which is the median class?
Solution: N ÷ 2 = 15. Cumulative frequencies 2, 7, 16, …: the total first reaches 15 in 20–29.
Estimate the median.
Solution: L = 19.5, F = 7, f = 9, c = 10. Median ≈ 19.5 + ((15 − 7) ÷ 9) × 10 = 19.5 + 8.89 = 28.4 marks.
Remember
Sources and credits
The modal-class and median-class ideas are adapted from the source above. The estimation formulas, the Rwandan examples, practice items and exercises are written by Ganzaa.
