IGCSE · Practice questions · Histograms with unequal class widths

Height times width.

Extended. Six original Cambridge-style questions on histograms with unequal class widths.

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

Divide to draw, multiply to read.

Width = upper \(-\) lower boundary of each class.
Draw: frequency density = frequency \(\div\) width.
Read: frequency = frequency density \(\times\) width (the area).
Part of a class: use only the width inside your range.
Hidden scale: use a known class to find what one unit of height is worth.

Recall

[3 marks]

The times, \(t\) seconds, taken by 65 students to solve a puzzle are shown. Calculate the frequency density of each class.

Time	\(0 < t \le 20\)	\(20 < t \le 30\)	\(30 < t \le 35\)	\(35 < t \le 40\)	\(40 < t \le 60\)
Frequency	12	18	14	11	10

Widths: 20, 10, 5, 5, 20. ✓
Frequency densities: \(12 \div 20 = 0.6\), \(18 \div 10 = 1.8\), \(14 \div 5 = 2.8\), \(11 \div 5 = 2.2\), \(10 \div 20 = 0.5\). ✓

Analysis

[2 marks]

For the puzzle times in the previous question, which is the modal class? Explain why it is not \(20 < t \le 30\).

The modal class has the highest frequency DENSITY: \(30 < t \le 35\) (2.8). ✓
\(20 < t \le 30\) has the highest frequency, 18, but it is twice as wide, so its values are less crowded. ✓

Calculation

[3 marks]

Using the puzzle times, estimate the number of students who took between 25 and 38 seconds.

25 to 30 s: \(5 \times 1.8 = 9\). ✓
30 to 35 s: the whole class, 14. 35 to 38 s: \(3 \times 2.2 = 6.6\). ✓
Total \(= 9 + 14 + 6.6 = 29.6\), so about 30 students. ✓

Calculation

[2 marks]

A histogram has bars for \(0 < d \le 5\) with frequency density 4 and \(5 < d \le 15\) with frequency density 2.5. How many values are there in these two classes altogether?

\(4 \times 5 = 20\) and \(2.5 \times 10 = 25\). ✓
Total \(= 20 + 25 = 45\). ✓

Calculation

[3 marks]

Calculate an estimate of the mean time for the 65 students in the first question.

Midpoints: 10, 25, 32.5, 37.5, 50. ✓
\(12 \times 10 + 18 \times 25 + 14 \times 32.5 + 11 \times 37.5 + 10 \times 50 = 120 + 450 + 455 + 412.5 + 500 = 1937.5\). ✓
Mean \(\approx \dfrac{1937.5}{65} = 29.8\) seconds (3 s.f.) ✓

Analysis

[5 marks]

A histogram shows the ages of people at a concert, but the frequency density axis has no scale. The bar for \(20 < a \le 30\) represents 60 people and is 4 cm tall. The bar for \(30 < a \le 50\) is 3 cm tall, and the bar for \(50 < a \le 80\) is 0.8 cm tall.

(a) How many people are aged \(30 < a \le 50\)? (b) How many are aged \(50 < a \le 80\)? (c) Estimate how many people are aged between 40 and 60.

First bar: frequency density \(= 60 \div 10 = 6\), shown by 4 cm, so 1 cm represents 1.5. ✓
(a) Density \(= 3 \times 1.5 = 4.5\); frequency \(= 4.5 \times 20 = 90\) people. ✓
(b) Density \(= 0.8 \times 1.5 = 1.2\); frequency \(= 1.2 \times 30 = 36\) people. ✓
(c) 40 to 50: \(10 \times 4.5 = 45\). 50 to 60: \(10 \times 1.2 = 12\). Total \(\approx 45 + 12 = 57\) people. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.