HS4 - Histograms and frequency density

Frequency density

In a histogram, bar area represents frequency. Use

Class width is upper boundary minus lower boundary. The vertical height is frequency per unit of the measured variable, rather than frequency itself.

Worked example 1: construct equal-width bars

Question. Draw a histogram for these waiting times, using frequency density on the vertical axis.

Plan. Calculate each width and density, then draw touching rectangles over the actual intervals.

All widths are five minutes, so the bar heights are

Use a uniform horizontal time scale and a vertical density scale starting at zero. The left equal-width diagram shows the resulting bars.

Answer. Heights $2,4,3$ people per minute over intervals zero–five, five–ten and ten–fifteen minutes.

Check. The areas are $5(2)=10$, $5(4)=20$, $5(3)=15$, matching every frequency and total $45$.

Worked example 2: read frequencies from equal-width bars

Question. A histogram of seedling growth has classes $0<L\le4$, $4<L\le8$, $8<L\le12$ centimetres. Their density heights are $3,1.5,2.5$ plants per centimetre. Find each frequency and the total.

Plan. Multiply every height by its four-centimetre width.

Answer. Frequencies $12,6,10$, total $28$ plants.

Check. Dividing those counts by four recovers the three displayed heights. A density such as $1.5$ need not be a whole number, even though the complete class contains a whole number of plants.

Watch out

Do not swap the ratio: density is frequency divided by width, not width divided by frequency. Check by multiplying the proposed height back by the width.

Unequal class intervals

For unequal classes, calculate density separately for each actual class width. Horizontal widths must use one consistent numerical scale. Bar area, rather than height, compares class frequencies. A taller bar shows a greater concentration per unit, which need not mean a larger total count.

Worked example 3: continuous data with unequal widths

Question. Construct a histogram for these journey times. Which class contains most people?

Plan. Divide frequency by the correct width, then draw each rectangle from its own lower to upper boundary.

For example, the narrow ten–fifteen-minute class has height $25/5=5$; the wider fifteen–thirty-minute class has height $30/15=2$. Use heights $2,5,2,1$, with the widths shown by the horizontal scale.

Answer. The $15<t\le30$ class contains most people: thirty, despite having a shorter bar than the ten–fifteen-minute class.

Check. Areas $10(2),5(5),15(2),20(1)$ give $20,25,30,20$, total $95$. The tallest bar has area twenty-five, not the largest area.

Worked example 4: integer marks and class boundaries

Question. Marks are whole numbers. Draw a histogram for classes ten–nineteen, twenty–twenty-nine and thirty–forty-nine, with frequencies $20,30,40$.

Plan. Replace the integer class limits by adjoining halfway boundaries, then calculate widths and densities.

The halfway numbers are plotting boundaries, not possible recorded marks. The final bar is twice as wide as the first two.

Answer. Touching bars with heights $2,3,2$ over the stated boundaries.

Check. The first group contains ten possible integers, so its width is ten, not $19-10=9$. The last bar's area is $20(2)=40$, correctly double the first bar's area despite their equal heights.