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Statistics - Histograms and frequency polygons

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Histograms are used to represent continuous data where the area of the bar (not just the height) represents the frequency.

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Unlike bar charts, histograms have no gaps between bars because the data is continuous.

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For histograms with unequal class widths, the vertical axis must represent Frequency Density.

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Frequency Polygons are line graphs used to represent the distribution of data; they are created by joining the midpoints of the tops of histogram bars.

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To draw a frequency polygon without a histogram, plot the frequency against the midpoint of each class interval and join the points with straight lines.

📐Formulae

Frequency Density=FrequencyClass WidthFrequency\ Density = \frac{Frequency}{Class\ Width}

Frequency=Frequency Density×Class WidthFrequency = Frequency\ Density \times Class\ Width

Class Width=Upper Boundary−Lower BoundaryClass\ Width = Upper\ Boundary - Lower\ Boundary

Midpoint=Lower Boundary+Upper Boundary2Midpoint = \frac{Lower\ Boundary + Upper\ Boundary}{2}

💡Examples

Problem 1:

A group of 50 students measured the time (tt minutes) they spent on homework. A class interval is 20<t≤3020 < t \le 30 with a frequency of 15. Calculate the frequency density for this interval.

Solution:

Class Width=30−20=10Class\ Width = 30 - 20 = 10. Frequency Density=1510=1.5Frequency\ Density = \frac{15}{10} = 1.5.

Explanation:

To find the height of the bar on a histogram, divide the frequency by the width of the interval.

Problem 2:

In a histogram, a bar representing the interval 10<x≤2510 < x \le 25 has a height of 4 units on the Frequency Density axis. Find the frequency of this interval.

Solution:

Class Width=25−10=15Class\ Width = 25 - 10 = 15. Frequency=15×4=60Frequency = 15 \times 4 = 60.

Explanation:

The frequency is the area of the bar, which is calculated as Class Width multiplied by Frequency Density.

Problem 3:

Given the class interval 40<x≤6040 < x \le 60 with frequency 12, identify the coordinates needed to plot this on a frequency polygon.

Solution:

Midpoint=40+602=50Midpoint = \frac{40 + 60}{2} = 50. Coordinate=(50,12)Coordinate = (50, 12).

Explanation:

Frequency polygons are plotted using the midpoint of the class interval on the x-axis and the actual frequency on the y-axis.