01 · Equal heights
A bar of width 5 represents 12 observations. A second bar has width 15 and the same height. Find its frequency.
Hint
Its area is three times larger.
Worked solution
36 observations.
Understand · explore · practise
Scaled histograms and missing information
Calibrate histogram areas from a known frequency, distinguish physical drawing units from data units and solve missing heights, widths and counts.
Before you startFrequency density, ratios, rectangle areas and linear equations.
01 / Start with a known bar
A known count fixes the proportionality.
Frequency = k × displayed area.
Find k from a bar whose frequency is known, using consistent area units.
In a histogram, areas represent counts. An unnumbered or scaled height axis does not stop us using area ratios, but an absolute frequency needs a known count or equivalent calibration. The horizontal and vertical scales must be linear and common to all bars.
Calibrate the drawingExplore
The known bar covers [0,10) and represents 20 observations. The second covers [10,30). Its height is chosen relative to the known bar.
Width ratio 2; height ratio 0.5; area ratio 1; second frequency 20.
This comparison uses a common linear vertical scale with a zero baseline. No height is itself a frequency.
02 / Use an area ratio
Compare width and height together.
A known bar has frequency 20. Another is twice as wide and 1.5 times as tall.Worked example
Area ratio = 2×1.5 = 3
Both dimensions matter.
Unknown frequency = 20×3 = 60
The same area factor applies to frequencies.
A height ratio alone would incorrectly give 30
That ignores the width change.
02 · Two ratios
A known frequency of 30 is drawn with width 6 and height 4. A second bar has width 9 and height 2. Find its frequency.
Hint
Use the area ratio (9×2)/(6×4).
Worked solution
30×18/24 = 22.5. As an exact observed frequency this is impossible; the dimensions/counts cannot all be exact. If measured from a drawing, treat it as an estimate and use the question’s rounding instruction.
03 / Keep drawing units consistent
Square centimetres are areas on paper.
Watch: calibrate one area, then another
Pause, replay or seek freely. The notes explain the same idea and stay in view.
A rectangle of area 8 cm² represents 24 observations. Another has area 15 cm².Worked example
k = 24/8 = 3 observations per cm²
This is the drawing’s area conversion.
Second frequency = 3×15 = 45
Use the same diagram and physical units.
Do not call 3 a frequency density in data units
The horizontal paper scale also matters.
03 · Physical area
A bar is 2 cm wide and 3 cm high and represents 18 observations. Another is 5 cm wide and 2 cm high. Find its frequency.
Hint
Calibrate 6 cm² first.
Worked solution
k = 18/6 = 3 observations/cm²; the 10 cm² bar represents 30.
04 · Unit change
A drawing uses k = 3 observations/cm². What is the conversion per mm²?
Hint
One cm² equals 100 mm².
Worked solution
0.03 observations/mm². Converting just one length factor is wrong.
04 / Recover the vertical scale
Connect displayed height to true frequency density.
A class has width 10 and frequency 30. Its displayed height is 1.5 cm.Worked example
True density = 30/10 = 3
This uses the variable’s units.
Density per cm of height = 3/1.5 = 2
Height h cm represents density 2h.
A class of width 8 drawn 2 cm high has frequency 8×4 = 32
Do not multiply by the known class width.
05 · Vertical calibration
A width-4 class with frequency 20 is drawn 2.5 cm high. Find the frequency of a width-6 class drawn 3 cm high.
Hint
The known density is 5, so each cm represents density 2.
Worked solution
Unknown density 6; frequency 6×6 = 36.
06 · Horizontal scale
One cm horizontally represents 5 minutes; one cm vertically represents density 2 observations/minute. How many observations does 1 cm² represent?
Hint
Multiply the two scale factors.
Worked solution
5×2 = 10 observations. A 3 cm by 2 cm bar represents 60.
05 / Solve a missing dimension
Area fixes a product, not each factor alone.
07 · Missing height
A known bar of width 5 and height 4 represents 40 observations. Another width-10 bar represents 30. Find its height on the same display scale.
Hint
k = 40/(5×4) = 2.
Worked solution
30 = 2×10×h, so h = 1.5.
08 · Missing width
Using the calibration in question 7, a height-3 bar represents 24 observations. Find its width.
Hint
24 = 2×w×3.
Worked solution
w = 4. If the bar begins at measurement 12, its upper boundary is 16.
06 / Use differences carefully
A height difference is a density difference.
Classes of widths 5 and 10 each contain 30 observations. Their drawn heights differ by 3 cm.Worked example
True densities are 6 and 3
The narrower class is taller.
Density difference 3 corresponds to 3 cm
One cm of height represents density 1.
Their heights are therefore 6 cm and 3 cm
The histogram baseline represents zero density.
09 · Difference calibration
Classes of widths 4 and 8 each contain 32 observations. Their drawn heights differ by 2 cm. Find both heights.
Hint
Densities are 8 and 4, so 4 density units correspond to 2 cm.
Worked solution
One cm represents density 2. Heights are 4 cm and 2 cm.
10 · Equal densities
Classes [0,5) and [5,15) contain 10 and 20 observations. Can they have a nonzero height difference in a correct histogram?
Hint
Both densities equal 2.
Worked solution
No. They must have equal height on the same scale, despite different counts.
07 / Check what is determined
A photograph’s apparent size is not an absolute scale.
11 · No known count
Only bar widths and relative heights are available. Can absolute frequencies be recovered?
Hint
Multiplying every count by the same constant preserves the shape.
Worked solution
No, unless a total count or another calibration is supplied. Relative frequencies can be obtained by dividing each area by the total area.
12 · Enlarged image
A histogram image is doubled in both dimensions. Can its original observations/cm² factor be used on the enlarged copy?
Hint
Every area is four times larger.
Worked solution
No. Divide the old physical-area conversion by 4, or recalibrate using a known bar. Frequencies themselves do not change.
08 / Calibrate, calculate, check
Keep height, density, area and frequency distinct.
Choose consistent units. Find the area factor or density-height conversion using known information. Calculate the requested count or dimension, then check widths, totals, physical units and whether an exact count is an integer.
Section 1 of 8 · Start with a known bar