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SL 4.2 · Organize and display discrete and continuous data

Learn to organize and display discrete and continuous data through clear examples and targeted practice.

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Statistics and Probability

Choosing tables and graphs that preserve what the data mean

Data are easier to understand when they are organized and displayed in a way that matches how they were collected. A graph can reveal a pattern quickly, but a poorly chosen graph can hide information or suggest something misleading. This lesson focuses on the choices behind frequency tables, bar charts, histograms, and cumulative frequency graphs. You will connect the original values to a table, then to a graph, and interpret what the display says in its real context.

What you will learn

1. Begin with the type of data

A data set is a collection of observations. Before drawing a graph, identify what each observation measures and its units. Discrete data take separate, countable values, often whole-number counts such as the number of books borrowed. Continuous data are measurements that can take any value in a range, such as a time or a mass. Recorded measurements may be rounded, but the quantity itself is still continuous.
A frequency table organizes observations by value or interval. Frequency means the number of observations in a category. For discrete data, list each value and its frequency. For a large data set of continuous measurements, group values into class intervals. Intervals should not overlap, and their boundaries should cover the data without gaps. State the units and explain the interval convention, especially at shared boundaries.
For discrete data, a bar chart is usually suitable: each category has its own bar, and gaps separate categories. The bar height represents frequency. For continuous data, a histogram is suitable: adjacent rectangles represent intervals, so there are no gaps between them. The horizontal axis shows measurement intervals, not separate categories.

2. Make frequency displays meaningful

When all histogram classes have the same width, the bar heights can be the frequencies. If widths differ, frequency alone cannot be used as the height: a wider interval would cover more horizontal distance and could look more important simply because it is wider. Instead use frequency density, which adjusts frequency for class width. The area of each rectangle then represents frequency.
For a class interval, calculate its width by subtracting its lower boundary from its upper boundary. Divide the class frequency by this width to get frequency density. The vertical axis must be labelled “frequency density” (and its units, if relevant). Check each rectangle by multiplying its height by its width: the result should equal the class frequency.
A cumulative frequency table records the running total up to each class boundary. It is useful for a cumulative frequency graph, sometimes called an ogive. Plot cumulative frequency against the upper class boundary, beginning at the lower boundary with cumulative frequency zero, then join the points with a smooth increasing curve or a sequence of straight segments. Read-offs from this graph are estimates because grouped data do not reveal the exact position of every observation within a class.
frequency density=frequencyclass width\text{frequency density}=\frac{\text{frequency}}{\text{class width}}

3. Interpret and check your display

A display is a model of the recorded data, not a replacement for them. A frequency table gives exact counts by value or class. A graph makes the overall pattern easier to see. A histogram can show where measurements are concentrated and whether the distribution is spread across a range. A cumulative frequency graph shows how many observations lie at or below a chosen boundary.
Graphing technology can help create a histogram or cumulative frequency graph and check arithmetic, particularly for many observations. Enter the raw values or class information carefully, choose suitable class boundaries, and verify axis labels and units. For a histogram with unequal class widths, confirm that the calculator uses frequency density; some tools otherwise draw heights that do not represent the required frequencies. Technology can make a display, but it cannot decide whether the selected intervals or graph type suit the question.
When interpreting a graph, refer to the context and the scale. Avoid claiming an exact individual value from a grouped graph. For example, a cumulative frequency estimate may locate a median within an interval, but it relies on treating the observations in that interval as spread in a reasonably even way. State that the value is an estimate and use the requested accuracy.

Frequency information for the discrete data

Books borrowed0123` `4
Number of students2342` `1

Worked example

A discrete frequency table and bar chart

A school records the number of library books borrowed by each of 12 students in one week: 0, 2, 1, 3, 2, 1, 0, 4, 2, 1, 3, 2. Organize the results and describe a suitable display.
  1. List the possible observed values
    The observations are counts, so they are discrete. The smallest recorded count is 0 and the largest is 4. Include each integer value in this range, even if its frequency is zero.
  2. Count each value
    Tallying the list gives the frequency for each number of books. Check that the frequencies add to the number of students.
    2+3+4+2+1=122+3+4+2+1=12
  3. Choose and describe the graph
    Draw a bar chart with number of books on the horizontal axis and number of students on the vertical axis. Separate the bars because the values are distinct count categories. The tallest bar is at 2 books, so that was the most common count.
Answer: The frequencies for 0, 1, 2, 3, and 4 books are respectively 2, 3, 4, 2, and 1. A separated-bar chart is suitable.
Check: The frequency total is 12, matching the number of students.

Worked example

A histogram with unequal class widths

Travel times to school, in minutes, are grouped as follows: 0–10 minutes, frequency 8; 10–30 minutes, frequency 18; 30–40 minutes, frequency 10. Find the histogram heights and explain how to draw the display.
  1. Find each class width
    Subtract the lower boundary from the upper boundary. The class widths are not all equal, so the frequencies themselves cannot be used as the bar heights.
    10, 20, 1010,\ 20,\ 10
  2. Calculate frequency densities
    Divide each frequency by its class width. These values are the histogram heights; their units are students per minute.
    810=0.8,1820=0.9,1010=1.0\frac{8}{10}=0.8,\quad \frac{18}{20}=0.9,\quad \frac{10}{10}=1.0
  3. Check the rectangle areas
    For each class, multiply its density by its width. The area should reproduce the class frequency, confirming that the chosen heights represent the data correctly.
    0.8(10)=8,0.9(20)=18,1.0(10)=100.8(10)=8,\quad 0.9(20)=18,\quad 1.0(10)=10
Answer: Draw adjacent rectangles over the three time intervals with heights 0.8, 0.9, and 1.0. Label the vertical axis frequency density in students per minute.
Check: The areas of the rectangles are 8, 18, and 10, and their sum is 36 students.

Worked example

Building and using cumulative frequency

The times, in minutes, for 40 runners are grouped as follows: 0–10, frequency 5; 10–20, frequency 13; 20–30, frequency 15; 30–40, frequency 7. Find the cumulative frequencies and estimate the time below which about 20 runners finished.
  1. Calculate running totals
    Add each class frequency to the total from the preceding classes. Plot these totals against the upper class boundaries, and include the starting point at the lower boundary with cumulative frequency zero.
    5,5+13=18,18+15=33,33+7=405,\quad 5+13=18,\quad 18+15=33,\quad 33+7=40
  2. Locate the required cumulative frequency
    About 20 runners is halfway through the 40 observations. On a cumulative frequency graph, find 20 on the vertical axis, move across to the plotted curve, then down to the time axis. This level lies in the 20–30 minute class because cumulative frequency rises from 18 to 33 there.
  3. Estimate within the class
    The increase from 18 to 33 is 15 runners across 10 minutes. Reaching 20 requires an increase of 2 runners beyond 18. Assuming the runners are spread evenly within this class gives an estimated additional time of about 1.3 minutes.
    20+20−1815×10≈21.320+\frac{20-18}{15}\times 10\approx 21.3
Answer: The cumulative frequencies at the upper boundaries 10, 20, 30, and 40 minutes are 5, 18, 33, and 40. The estimated time for about 20 runners is 21.3 minutes, to 1 decimal place.
Check: The estimate is between 20 and 30 minutes, the class in which cumulative frequency increases from 18 to 33.

Common mistakes and how to avoid them

Drawing a histogram with gaps between its bars.
Correction: Histogram intervals are adjacent parts of a continuous scale, so their rectangles touch.
Using frequency as the height for every histogram class when class widths differ.
Correction: Calculate frequency density for each class so that rectangle area represents frequency.
Treating a cumulative frequency as the frequency in that single class.
Correction: Cumulative frequency is the total in that class and all earlier classes.
Reporting a value read from grouped data as exact.
Correction: State that it is an estimate because the original positions within each interval are not known.

Lesson summary

Check your understanding

Question 1

A histogram class has width 5 and frequency 15. What is its frequency density?
  1. 3
  2. 10
  3. 15
  4. 75
Show answer and explanation
3
Frequency density is frequency divided by class width, so the result is 15 divided by 5, which is 3.

Question 2

Which display is most suitable for counts of the number of pets owned by students?
  1. A bar chart with separated bars
  2. A histogram with unequal-width intervals
  3. A cumulative frequency graph with no categories
  4. A continuous scale with touching bars
Show answer and explanation
A bar chart with separated bars
The number of pets is discrete count data. A bar chart shows the separate possible counts as distinct categories.

Question 3

Cumulative frequency is 12 at one class boundary and 19 at the next. How many observations are in that class?
  1. 7
  2. 12
  3. 19
  4. 31
Show answer and explanation
7
The class frequency is the increase in cumulative frequency: 19 minus 12 equals 7.

Key terms

Discrete data
Data that take separate, countable values, such as a number of items.
Continuous data
Measurements that can take any value within a range, such as time or length.
Frequency
The number of observations in a value or class.
Class width
The difference between the upper and lower boundaries of a class interval.
Frequency density
Frequency divided by class width; used as histogram height when class widths differ.
Cumulative frequency
The running total of frequencies up to a specified class boundary.

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Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 4.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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