IGCSE · Practice questions · Averages from frequency tables

Multiply, add, divide.

Six original Cambridge-style questions on averages from ungrouped and grouped frequency tables.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Add an fx column, divide by the total frequency.

Mean: \(\dfrac{\sum fx}{\sum f}\).
Median: position \(\dfrac{n+1}{2}\), then count down the running totals.
Mode: the value (or class) with the highest frequency.
Grouped (Extended): use midpoints, and say 'estimate'.
Backwards: total = mean \(\times\) number; add totals, never means.

Recall

[3 marks]

The table shows the numbers of goals scored by a team in 20 matches.

Goals	0	1	2	3	4	5
Frequency	4	6	5	3	1	1

Find (a) the mode (b) the median (c) the range.

(a) Highest frequency 6, so the mode is 1 goal. ✓
(b) \(n = 20\): median halfway between the 10th and 11th values. Running totals 4, 10, 15: the 10th is 1 and the 11th is 2, so the median is 1.5. ✓
(c) Range \(= 5 - 0 = 5\). ✓

Calculation

[3 marks]

Calculate the mean number of goals for the table in the previous question.

\(\sum fx = 0 + 6 + 10 + 9 + 4 + 5 = 34\). ✓
Mean \(= \dfrac{34}{20} = 1.7\) goals. ✓

Calculation

[4 marks]

(Extended) The masses, \(m\) kg, of 40 parcels are shown.

Mass	\(0 < m \le 20\)	\(20 < m \le 40\)	\(40 < m \le 60\)	\(60 < m \le 100\)
Frequency	6	14	16	4

Calculate an estimate of the mean mass.

Midpoints: 10, 30, 50, 80 (the last class is 40 kg wide). ✓
\(6 \times 10 + 14 \times 30 + 16 \times 50 + 4 \times 80 = 60 + 420 + 800 + 320 = 1600\). ✓
Estimated mean \(= \dfrac{1600}{40} = 40\) kg. ✓

Analysis

[2 marks]

(Extended) For the parcels in the previous question, write down (a) the modal class (b) the class that contains the median.

(a) Highest frequency 16: \(40 < m \le 60\). (Its width, 20 kg, matches the first three classes.) ✓
(b) Median position \(\dfrac{40+1}{2} = 20.5\). Running totals 6, 20, 36, so the 20.5th value is in \(40 < m \le 60\). ✓

Proof

[3 marks]

\(x\)	2	3	4	5
\(f\)	6	\(x\)	8	4

The mean of this distribution is 3.4. Show that the missing frequency is 7.

\(\sum fx = 12 + 3x + 32 + 20 = 64 + 3x\) and \(\sum f = 18 + x\). ✓
\(64 + 3x = 3.4(18 + x) = 61.2 + 3.4x\). ✓
\(2.8 = 0.4x\), so \(x = 7\). Check: \(\dfrac{85}{25} = 3.4\). ✓

Analysis

[3 marks]

A class has 30 students. The 18 girls do a mean of 6.5 hours of homework a week. The mean for the whole class is 6.1 hours. Find the mean for the boys.

Class total \(= 30 \times 6.1 = 183\) hours. Girls' total \(= 18 \times 6.5 = 117\) hours. ✓
Boys' total \(= 183 - 117 = 66\) hours, shared by \(30 - 18 = 12\) boys. ✓
Boys' mean \(= \dfrac{66}{12} = 5.5\) hours. ✓

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