0580 · 9.7Histograms

Frequency density and histogram area

0:00Imagine two groups of students with the same number of people in each group. One group has journey times spread across ten minutes. The other covers twenty minutes. Drawing equally tall bars would make the wider bar look much more important. Histograms solve this problem by making the area of each bar represent frequency. By the end of this video, you will understand why histogram heights are called frequency densities.

0:21You will draw a histogram from grouped data and recover frequencies from its bars. We will also practise completing a histogram when some information is missing. Throughout, the key idea is that width and height work together. We need three starting ideas: frequency, class intervals and rectangle area. Frequency means the number of observations in a group. A class interval gives the range of values belonging to that group. The area of a rectangle equals its width multiplied by its height, and that relationship explains how histograms work.

0:51For example, consider the class from ten up to and including twenty minutes. The board writes this as ten is less than t, and t is less than or equal to twenty. Its class width is 20 − 10, which equals ten minutes. The inequality signs show which endpoints belong to the class. A histogram displays grouped numerical data, such as journey times, heights or masses. The horizontal axis shows the measured quantity on a numerical scale.

1:17Each rectangle stretches across its actual class interval. Adjacent classes therefore have touching bars, although an empty class can leave a gap. This differs from a bar chart, where separate bars usually represent categories. A histogram bar cannot be given an arbitrary width because its width represents a numerical interval. A class covering twenty minutes must be twice as wide as one covering ten minutes. That makes unequal class widths especially important.

1:43Look at these two classes, each containing twenty students. The first has width ten and height two. The second has width twenty and height one. Their areas are equal because 10 × 2 and 20 × 1 both give twenty. The wider bar is shorter because the same frequency is spread across a larger interval. The height describes how concentrated the observations are within the class. We call this frequency density.

2:07Frequency density equals frequency divided by class width. Dividing by the width adjusts for the size of the interval, allowing classes of different widths to be represented fairly. For journey times measured in minutes, density describes the frequency per minute of class width. It does not tell us that exactly that many students have each individual journey time. We only know the total within the interval. The rectangle represents that grouped information without pretending we know every original value.

2:34Multiplying density by width reverses the division and gives the frequency again. This is also the rectangle area calculation. So the formula is not a separate rule to memorise without meaning. Histogram height is chosen precisely so that width multiplied by height gives the number of observations in the class.

Drawing and interpreting histograms

2:50Our first exam-style example asks us to draw a histogram for one hundred students. Their journey times are grouped into four classes, shown on the board. Notice that the first two classes have width ten, while the final two have width twenty. We must calculate density before choosing the bar heights. For the first class, subtract zero from ten to get a width of ten. Divide its frequency of twenty by ten to obtain density two.

3:14The second interval has width 20 − 10, which is also ten. Its density is 30 ÷ 10, giving three. The third class has width 40 − 20, giving twenty. Its frequency density is 40 ÷ 20, which equals two. For the final class, 60 − 40 gives a width of twenty. Dividing its frequency of ten by twenty gives density zero point five. Choose a horizontal scale that reaches sixty minutes and keeps all class widths proportional.

3:40The vertical scale must accommodate the greatest density, which is three. Label that axis Frequency density, not Frequency. Then draw each rectangle from its lower class boundary to its upper boundary at the calculated height. The tallest bar covers ten up to and including twenty minutes. However, the greatest frequency belongs to the class from twenty up to and including forty minutes. Its height is only two, but its width is twenty.

4:05Its area represents forty students, compared with thirty students in the tallest bar. This distinction is central to interpreting histograms. Height tells us frequency density, while area tells us frequency. A taller bar represents a greater concentration of observations, not necessarily a greater total number. Always include the width when comparing frequencies in classes of different sizes. If every class has the same width, comparing heights also compares frequencies. That works because every height is multiplied by the same width.

4:35However, the vertical axis in these histograms is still labelled Frequency density. Equal widths simplify comparisons, but they do not change the definition. The same question often continues by asking for an estimate of how many journeys took more than thirty minutes. Thirty lies inside the class from twenty up to and including forty. We assume the forty students in that class are spread evenly across it. So we use only the part of that bar from thirty to forty.

4:59That part has width 40 − 30, which equals ten. Multiplying ten by the density of two gives an estimate of twenty students. The whole final class lies above thirty and adds its ten students. So the estimate is 20 + 10, giving thirty students. It is only an estimate because we do not know the exact times inside a class.

Finding frequencies from bars

5:18Our second exam-style example starts with a completed histogram rather than a frequency table. It shows times spent waiting for a bus. We are asked to calculate the number of observations in each class and find the total. This time, we multiply each class width by its bar height. The first bar extends from zero to five, so its width is five. Multiplying that width by density two gives a frequency of ten.

5:41The next interval has width 15 − 5, which equals ten. Its frequency is 10 × 1.5, giving fifteen. For the third class, 20 − 15 gives width five. Its frequency is 5 × 4, which equals twenty. The final interval has width 40 − 20, giving twenty. Multiplying by density zero point five produces frequency ten. Adding ten, fifteen, twenty and ten gives fifty-five observations. Suppose the question also asks how many waiting times were greater than fifteen minutes.

6:12The two relevant classes are entirely to the right of fifteen. Add their frequencies, twenty and ten, to obtain thirty. We add the represented frequencies, not the bar heights. Notice that the final bar has the same frequency as the first, even though it looks much lower. Its width is four times as large, while its height is only 1/4 as large. These changes balance in the area calculation. Checking this relationship is a useful way to catch an incorrect interpretation.

Completing a histogram

6:39Our third exam-style example asks us to complete a histogram and calculate a missing frequency. The data now concern masses of parcels, measured in kilograms. Two bars are already drawn, and the frequency of the final class is given as thirty. We must use the appropriate direction of the density relationship for each part. The missing bar covers thirty up to and including fifty kilograms. Its class width is 50 − 30, which equals twenty.

7:03Divide the given frequency of thirty by twenty to obtain density one point five. Draw the rectangle across the full interval with its top at that height. Before calculating, read the bar heights carefully. Each small step on this vertical axis is zero point two. The first bar stops three steps up, at zero point six. The middle bar reaches nine steps up, which is one point eight. For the middle class, subtract twenty from thirty to obtain width ten.

7:28Multiplying ten by its density of one point eight gives eighteen parcels. The first class has width 20 − 0, giving twenty. Its frequency is 20 × 0.6, which equals twelve. Adding twelve, eighteen and thirty gives sixty parcels altogether. A last part asks for an estimate of the number of parcels heavier than twenty-five kilograms. Twenty-five lies halfway through the class from twenty up to and including thirty. The part from twenty-five to thirty has width five and density one point eight.

7:565 × 1.8 gives an estimate of nine parcels. Adding the thirty parcels in the last class gives thirty-nine. Completing this question required division for the missing height and multiplication for the missing frequencies. Choose the operation by considering what the rectangle represents. If frequency is known, spread it across the width to find density. If density is known, multiply by width to recover the frequency.

Common mistakes and exam technique

8:20A common mistake is drawing bars whose heights equal the original frequencies. With unequal class widths, this gives wider classes too much visual importance. Another mistake is drawing every bar with the same width. Use the horizontal numerical scale to place the boundaries, rather than treating the intervals as separate category labels. Students also sometimes use an upper boundary as the class width. For the class from twenty to forty, the width is twenty, not forty.

8:45Subtract the lower boundary from the upper boundary every time. Read the vertical scale carefully too, especially when its steps are decimals rather than whole numbers. In an examination, commands such as draw, interpret and calculate can appear in different combinations. You might construct a histogram, complete missing bars or determine frequencies from a graph. Show the class width and the density or frequency calculation clearly. Each line should explain how the information leads to your answer.

9:12These methods must work without a calculator on Paper two. On Paper four, a calculator is allowed, but the mathematical reasoning is unchanged. Keep exact values where possible and do not round during intermediate calculations. If a final answer is not exact, give it to three significant figures unless the question says otherwise. For a short written explanation, state that histogram bar area represents frequency. Explain that frequency density equals frequency divided by class width.

9:39If comparing unequal classes, calculate their frequencies before claiming which contains more observations. Those statements are precise because they distinguish the height from the quantity represented by the whole rectangle. The central idea is simple: class width controls the horizontal size, and frequency density controls the height. Multiplying them gives frequency, so area represents the number of observations. Divide frequency by width when drawing, and multiply width by density when interpreting.

10:06With those connections understood, histograms become meaningful pictures rather than a collection of rules.