10 min readPathway sheetA 3-mark or 5-mark question; finding missing frequencies from the median is a favourite

Median of grouped data

Make a cumulative frequency column and find the median class, where n/2 falls. Then Median = l + ((n/2 − cf) ÷ f) × h.

FormulaMedian = l + ((n/2 − cf) ÷ f) × h

Median classThe first class whose cumulative frequency is more than n/2

Symbolsn: total frequency; l: lower limit of the median class; cf: cumulative frequency of the class before it; f: frequency of the median class; h: class size

Cumulative frequencyA running total of the frequencies, class by class

Pathway: median of grouped data

1
Add the frequencies to get n. Work out n/2.
2
Make a cumulative frequency (cf) column: keep a running total.
3
Find the median class: the first class whose cf is more than n/2.
4
Note l, h and f of the median class, and the cf of the class before it.
5
Median = l + ((n/2 − cf) ÷ f) × h.
6
Check that the median lies inside the median class.

Worked example 1: weights of students

Problem

The weights (in kg) of 50 students are: 30-35: 4, 35-40: 10, 40-45: 18, 45-50: 12, 50-55: 6. Find the median weight.

Working

n = 50, so n/2 = 25.
Cumulative frequencies: 4, 14, 32, 44, 50.
The first cf more than 25 is 32, so the median class is 40-45.
l = 40, h = 5, f = 18, cf of the class before = 14.
Formula: Median = l + ((n/2 − cf) ÷ f) × h.
Median = 40 + ((25 − 14) ÷ 18) × 5 = 40 + 55 ÷ 18 = 40 + 3.06 = 43.06.

Answer

Median weight ≈ 43.06 kg.

Cumulative frequency table for example 1

Weight (kg)	Frequency f	Cumulative frequency cf
30-35	4	4
35-40	10	4 + 10 = 14
40-45 (median class)	18	14 + 18 = 32
45-50	12	32 + 12 = 44
50-55	6	44 + 6 = 50

Worked example 2: two missing frequencies

Problem

The median of this distribution of 40 values is 25. Find x and y. Classes: 0-10: 4, 10-20: x, 20-30: 12, 30-40: y, 40-50: 6.

Working

Total: 4 + x + 12 + y + 6 = 40, so x + y = 18.
The median 25 lies in 20-30, so that is the median class.
n/2 = 20, l = 20, h = 10, f = 12, cf = 4 + x.
Formula: Median = l + ((n/2 − cf) ÷ f) × h, so 25 = 20 + ((20 − 4 − x) ÷ 12) × 10.
5 = (16 − x) × 10 ÷ 12, so 60 = 160 − 10x, giving x = 10.
y = 18 − 10 = 8.
Check: cf = 4, 14, 26, 34, 40; median = 20 + ((20 − 14) ÷ 12) × 10 = 25.

Answer

x = 10, y = 8.

Points examiners look for

cf in the formula is the cumulative frequency of the class before the median class.
f is the frequency of the median class itself.
If classes are like 10-19, 20-29, first make them continuous: 9.5-19.5, 19.5-29.5. With a gap of 1, take 0.5 off each lower limit and add 0.5 to each upper limit.
Always show the cf column. It carries marks.

In real life

Survey reports often give the median age in a district, the age of the middle person. A few very old people do not pull the median up, but they would pull the mean.

Exam angle

2 marks: “For a grouped table, n = 60. How do you decide the median class?” n/2 = 30. The median class is the first class whose cumulative frequency is more than 30.

3 marks: “Find the median of: 0-10: 5, 10-20: 9, 20-30: 14, 30-40: 8, 40-50: 4.” n = 40, so n/2 = 20. cf: 5, 14, 28, 36, 40, so the median class is 20-30. Median = 20 + ((20 − 14) ÷ 14) × 10 = 20 + 4.29 = 24.29.

5 marks: “The median of the distribution below is 35 and the total frequency is 50. Find a and b. Classes: 0-10: 3, 10-20: a, 20-30: 10, 30-40: 12, 40-50: b, 50-60: 5.” a + b = 20. Median class 30-40: l = 30, f = 12, cf = 13 + a, h = 10. 35 = 30 + ((25 − 13 − a) ÷ 12) × 10 gives a = 6, so b = 14. Check: cf 3, 9, 19, 31, 45, 50; median = 30 + (6 ÷ 12) × 10 = 35.

Remember it

Half, ladder, lock: find n/2, climb the cf column, and lock the first class that passes it.