Histograms – GCSE Maths

Introduction

  • A histogram is a graphical representation of numerical data that shows the frequency of values within specific intervals.
  • The data is grouped into continuous ranges (called bins), and bars are used to display how many data points fall into each range. Unlike bar graphs, the bars in a histogram touch, indicating continuous data.

For Example:

Image of a histogram for the topic histograms showing frequency density and class

What is Frequency Density?

  • A Frequency density is used when data is grouped into classes (intervals), and those classes have different widths. It tells us how packed the data is within each class.

Image of mathematical equation to calculate frequency density using frequency and classwidth

Terms related to the concept of Histograms:

  • Class Interval (or Bin)-

A range of values (like 0–10, 10–20) into which data is grouped.

  • Frequency-

The number of data points that fall into a particular class interval.

  • Frequency Density-

Used when class widths are unequal.

  • Class Width-

The difference between the upper and lower boundary of a class.

Example: For 10 –20, the class width is 10.

How to Work Out With Frequency Density?

  • Frequency density helps when your class intervals (group ranges) are not all equal. Here’s how you calculate it:

Steps to Work It Out:

  • Step#1: Identify the Frequency

This is how many values fall into each class interval.

  • Step#2: Find the Class Width

Class Width = (Upper Limit−Lower Limit)

  • Step#3: Use the Formula to find Frequency Density

Divide frequency by class width for each class

Example Table:Image of table for step by step solved example to correlate frequency with classwidth for histograms topic

certified Physics and Maths tutorSolved Example:

Example: Construct a histogram for the following data.

Table of data for step by step solved example of histogramsSolution:

Step #1: Identify the Frequency:

  • This is how many values fall into each class interval.

Step #2: Find the Class Width:

Class Width = (Upper Limit−Lower Limit)

Class Width-

Calculation for solved example

Let us take a note that there is no need to calculate Frequency density when class width of each and every interval comes out to be same.

Step #3: Use the Formula:

  • Divide frequency by class width for each class but as here the class width of all the class intervals is same. Thus, we will take frequency as frequency density.
  • Hence, by the help of the table given to us, let us construct the histogram of the given data. 

image of final histogram for step by step solved examples of topic histograms for gcse maths

certified Physics and Maths tutorSolved Example:

Example: Construct a histogram for the following data.image of interval and frequency dataset in the form of table for step by step solved examples of histograms Solution:

Step #1: Identify the Frequency:

  • This is how many values fall into each class interval.

Step #2: Find the Class Width:

Class Width = (Upper Limit−Lower Limit)

Step #3: Use the Formula:

  • Divide frequency by class width for each class in order to find Frequency Density.
  • Let us find the Frequency densities of the respective class intervals.

image of table showing interval, frequency density, classwidth and frequency for solved example

  • Hence, by the help of the table given to us, let us construct the histogram of the given data. 

image of solution of histogram for step by step solved examples

certified Physics and Maths tutorSolved Example:

Problem: A class of 30 students took a math exam (max score = 100). Construct a histogram for the following data and analyze the distribution:

[45, 72, 68, 90, 53, 81, 75, 62, 58, 77, 82, 64, 88, 92, 78, 60, 85, 70, 95, 50, 63, 71, 65, 55, 80, 73, 67, 79, 83, 69]

Solution: 

Step #1: We will firstly make the Frequency table:dataset in table form for step by step solved example

Step #2: Let us now draw the respective histogram by the help of frequency table.Solution of step by step solved example of histograms

certified Physics and Maths tutorSolved Example:

Problem: A store recorded daily sales (in $) for 20 days:Create a histogram and analyze spending trends.

[120, 85, 150, 200, 95, 130, 180, 75, 160, 110, 140, 90, 170, 105, 125, 115, 155, 80, 190, 100]

Solution: 

Step #1: We will firstly make the Frequency table:dataset in table form for step by step solved example

Step #2: Let us now draw the respective histogram by the help of frequency table.Solution of step by step solved example of histograms

certified Physics and Maths tutorSolved Example:

Problem: The histogram below shows information about the time tt seconds taken for students to complete a 400m race.

How many students ran 400m in under 90 seconds?final solved example histograms for gcse mathsSolution: 

Step #1: Locate the frequency density for the class interval(s).

  • The frequency density for the class intervalcalculation for final solved example histograms for gcse maths
  • The frequency density for the class intervalcalculation for final solved example histograms for gcse maths

Step #2: Determine the class width for the class interval(s).

  • The interval 30 < t < 60 has a class width

classwidth where w = 30

  • The interval 60 < t < 90 has a class width 

classwidth where w = 30 for step by step solved example of histograms

Step #3: Use the frequency density formula to determine the frequency.

To calculate the frequency in both of these class intervals, we must work out the frequency in each class separately, then add them together at the end.

The frequency density formula isformula for frequency density of histograms step by step solved exampleAs D=0.2 and W=30 for the interval 30 ≤ t < 60,, we can substitute these values into the formula and solve for Frequency.frequency density

As D=0.5 and W=30, for the interval 60 ≤ t < 90, we can substitute these values into the formula and solve for Frequency.

frequency density for step by step solved examples for histograms

The frequency of students who ran 400m in under 90 seconds is-

15 + 6 = 21 students.