01 · A density
A class [5,15) has frequency 30. Find its width and density.
Hint
Width is 15 − 5.
Worked solution
Width = 10; density = 30/10 = 3 observations per measurement unit.
Understand · explore · practise
Histograms and frequency density
Understand why histogram area represents frequency, calculate frequency density for unequal classes and distinguish bar height from count. Original adjustable model and worked questions.
Before you startGrouped frequency tables, class boundaries and rectangle area.
01 / Let area represent count
A histogram height must account for class width.
Frequency density = frequency / class width.
With a true density axis: frequency = class width × frequency density.
If two classes have different widths, equal frequencies need not produce equal heights. Area, not height alone, represents how many observations the bar contains.
Keep track of areaExplore
Explore one isolated class [0,w). Its frequency is f. The graph uses true frequency density, so width × height = frequency.
Density = 20/10 = 2; area = 10×2 = 20.
Holding the frequency fixed while widening the class lowers the density. This is a geometric comparison, not a claim that an actual dataset can be regrouped without checking its counts.
02 / Find true class widths
Use upper boundary minus lower boundary.
Class [10,30) has frequency 16.Worked example
Width = 30 − 10 = 20
Subtract boundaries, not category ranks.
Density = 16/20 = 0.8
The count is spread over twenty measurement units.
Bar area in axis units = 20×0.8 = 16
This recovers the class frequency.
02 · Rounded class labels
Lengths rounded to the nearest centimetre are grouped as 20–29 cm. Frequency is 15. Find true width and density.
Hint
Use boundaries 19.5 and 29.5.
Worked solution
Width = 10 cm, so density = 1.5 observations per cm. Using 29 − 20 = 9 ignores the true boundaries.
03 / Compare unequal-width bars
The tallest bar need not contain the most observations.
Watch: double the width, halve the height
Pause, replay or seek freely. The notes explain the same idea and stay in view.
A: [0,5), frequency 10. B: [5,20), frequency 15.Worked example
A density = 10/5 = 2
A is the taller bar.
B density = 15/15 = 1
B is three times as wide.
B has the larger frequency, despite its smaller height
Its area is 15 rather than 10.
03 · Equal heights
Classes [0,10) and [10,30) both have density 2. Find their frequencies.
Hint
Multiply each density by its own width.
Worked solution
Frequencies 20 and 40. Equal height does not imply equal frequency when widths differ.
04 · Equal counts
Classes have widths 4 and 10, both with frequency 20. Find their densities.
Hint
Divide the same count by each width.
Worked solution
Densities 5 and 2. The narrower class has the taller bar; both areas are 20.
04 / Recover missing frequencies
Multiply width by density.
05 · Read a bar
A bar spans 30 to 50 and reaches density 0.6. Find its frequency.
Hint
Area = width × height.
Worked solution
Frequency = 20×0.6 = 12, provided the axis is true frequency density.
06 · Recover a width
A class has frequency 24 and density 3. Find its width.
Hint
Rearrange density = frequency/width.
Worked solution
Width = 24/3 = 8 measurement units.
05 / Read the vertical scale
A physical height is not automatically a density.
A labelled frequency-density axis lets you use f = width×density directly. If heights are given only in centimetres or the vertical scale is omitted, areas are proportional to frequencies but a scale factor must first be established from known information.
07 · Units
A histogram’s horizontal variable is time in minutes. What are the units of true frequency density?
Hint
Density is count divided by the horizontal measurement unit.
Worked solution
Observations per minute. Multiplying by a width in minutes gives a count.
08 · Physical height
A printed bar is 3 cm high and covers a time interval of 10 minutes. Is its frequency necessarily 30?
Hint
Centimetres on paper are not observations per minute.
Worked solution
No. You need the vertical density scale or a known area-to-frequency calibration. Multiplying mixed units without that scale is invalid.
06 / Build the histogram consistently
Use adjoining intervals with clear axes.
Draw each rectangle over its true class boundaries with the calculated density height. Adjacent numerical intervals touch. A gap over an interval represents zero frequency or explicitly missing coverage; do not insert decorative gaps as in a category bar chart.
09 · Empty interval
An occupied class is followed by a class of width 5 with frequency 0. What density and bar height does the empty class have?
Hint
Zero count divided by positive width.
Worked solution
Density and height are zero. Preserve its horizontal interval on the axis; do not remove that part of the measurement scale.
10 · Category bars
Can categories such as favourite colours be treated as adjoining histogram intervals with meaningful widths?
Hint
There is no continuous numerical interval width for a colour category.
Worked solution
No. A category bar chart is appropriate for such counts. Histogram areas rely on numerical class intervals, not arbitrary category widths.
07 / Distinguish concentration from frequency
Name what a comparison measures.
11 · Tallest class
In the unequal-width example, which class has the largest density and which has the largest frequency?
Hint
A has density 2 and count 10; B has density 1 and count 15.
Worked solution
A has the highest density; B has the highest class frequency. If asked for a modal class, check whether the question means greatest frequency or greatest histogram density, and state your interpretation.
12 · Unbounded class
The final class is “50 or more”. Can its density be calculated from its count alone?
Hint
The class width is not finite and specified.
Worked solution
No. Obtain a finite upper boundary or another justified model before drawing that density bar. Do not invent a convenient width.
08 / Check areas, not just heights
A width calculation comes before a density calculation.
Use true boundaries, divide frequency by width, label the vertical scale and check that each area recovers its count. For unequal classes, separate the largest frequency from the greatest concentration per unit.
Section 1 of 8 · Let area represent count