Tales by Dots and Lines · Lesson 6 of 13
Mean and Median from Frequencies and Dot Plots
“Use repeated observations efficiently and interpret the information a display actually contains.”
• Read frequencies from tables and stacked dots. • Calculate a mean using value × frequency products. • Find median positions using running frequencies. • Interpret minimum, maximum, and limits of what a dataset reveals.
A class records family sizes, and many students give the same answer. Writing the number 4 eleven separate times is possible, but a frequency table records the same information more compactly: family size 4, frequency 11. The table changes the presentation, not the underlying data.
To calculate correctly, imagine expanding each row back into repeated observations. This will show why the denominator is the total frequency and why the numerator includes multiplication. We will also find a median without writing a long ordered list.
Mean and Median with Frequencies
The family-size table below contains 36 responses. A frequency of 3 beside family size 7 means three students reported families of size 7. It does not mean there are three family members in those families. Read the value column and the frequency column as different kinds of information.
The number of times a value occurs in a dataset.
| Family size | Frequency | Size × frequency |
|---|---|---|
| 3 | 3 | 9 |
| 4 | 11 | 44 |
| 5 | 9 | 45 |
| 6 | 7 | 42 |
| 7 | 3 | 21 |
| 8 | 1 | 8 |
| 9 | 1 | 9 |
| 10 | 1 | 10 |
Problem
Find the mean for the displayed family-size responses.
- 1.Add frequencies: 3 + 11 + 9 + 7 + 3 + 1 + 1 + 1 = 36.
- 2.Add the products: 9 + 44 + 45 + 42 + 21 + 8 + 9 + 10 = 188.
- 3.Divide total by count: 188 ÷ 36 = 5.222… .
- 4.The mean is approximately 5.22 people per reported family. This does not suggest that any one family contains a fraction of a person.
Averaging the eight distinct family sizes gives 6.5, but it treats common and rare sizes as equally frequent. The dataset has 36 observations, not 8.
Locating the Median with Running Counts
A running frequency, also called cumulative frequency, tells you how many observations you have counted up to a particular value. There are 3 observations equal to 3. Including the eleven observations equal to 4 takes us to 14. Including the nine equal to 5 takes us to 23.
Therefore the 4s occupy positions 4 through 14 and the 5s occupy positions 15 through 23. When a median position falls inside one of these ranges, the value at that position is immediately known. The cumulative count is a position boundary; it is not the data value itself.
| Family size | Running frequency | Positions |
|---|---|---|
| 3 | 3 | 1–3 |
| 4 | 14 | 4–14 |
| 5 | 23 | 15–23 |
| 6 | 30 | 24–30 |
| 7 | 33 | 31–33 |
| 8 | 34 | 34 |
| 9 | 35 | 35 |
| 10 | 36 | 36 |
Problem
Find the median of the 36 family-size observations.
- 1.Because 36 is even, locate the 18th and 19th observations.
- 2.Both positions lie in the range 15–23.
- 3.Every observation in that range equals 5.
- 4.The median is (5 + 5) ÷ 2 = 5 people. We do not average the position numbers 18 and 19.
Reading a Dot Plot as a Frequency Table
In a dot plot, each dot represents one observation. Dots stacked above the same number show its frequency. An empty position may mean frequency zero; it does not mean a missing observation if the scale and dataset are complete.
Suppose the plot records how many times ten students used a cycle during one week. We can calculate an average number of uses, but the graph does not record the dates or the lengths of individual rides. If a weekly count exceeds seven, at least one day must contain more than one use. Counts of seven or fewer do not settle that question.
Problem
Find the mean and median in the displayed cycle-use plot. Did everyone cycle at least once?
- 1.Read the ordered data: 0, 1, 1, 2, 2, 2, 3, 3, 4, 6. There are 10 observations.
- 2.The total is 24 uses, so the mean is 24 ÷ 10 = 2.4 uses per student.
- 3.The fifth and sixth observations are both 2, so the median is 2.
- 4.Everyone did not cycle: one student has value 0. The plot cannot tell us which particular days the rides occurred.
A Complete Description of Dart Trials
Different summaries answer different questions. Minimum and maximum describe the observed endpoints. The mean uses every trial count. The median describes the middle participant. Together these summaries provide more information than any one of them alone.
In the following table, a trial count records how many throws a participant needed to hit the centre. Frequencies of zero at 2 and 3 must remain visible, but they contribute nothing to the sum. The minimum observed value is the smallest value with nonzero frequency.
| Throws needed | Participants |
|---|---|
| 1 | 1 |
| 2 | 0 |
| 3 | 0 |
| 4 | 1 |
| 5 | 4 |
| 6 | 9 |
| 7 | 12 |
| 8 | 15 |
| 9 | 10 |
| 10 | 10 |
Problem
Find the count, minimum, maximum, mean, and median of the dart-trial data.
- 1.The total frequency is 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62.
- 2.The minimum observed trial count is 1 and the maximum is 10.
- 3.The weighted total is 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473. The mean is 473 ÷ 62 ≈ 7.63 throws.
- 4.There are 27 participants up to 7 throws and 42 up to 8 throws. Both middle positions, 31 and 32, are therefore 8.
- 5.The median is 8 throws. The mean and median are close, but they answer different questions.
Translate Between Displays
Use the values 1, 1, 2, 4, 4, 4 to practise moving between forms.
- Make a frequency table and include zero-frequency values if you show a complete integer scale.
- Draw a dot plot with one dot per observation.
- Calculate the mean from both the list and the table.
- Locate the two middle positions using running frequencies.
The total is 16 and the count is 6, so the mean is 8/3, approximately 2.67. The middle values are 2 and 4, giving median 3. Both displays should produce the same answers.
Quiz
A value 6 has frequency 4. How much does it contribute to the total?
Values 2 and 5 have frequencies 3 and 1. What is the mean?
For 20 observations, which positions determine the median?
In the family-size table, what is the 19th value?
A student has 8 cycle uses in 7 days. What must be true?
It represents four copies of 6, with total 6 × 4 = 24.
Practice Problems
- Values 1, 2, 3, 4 have frequencies 2, 3, 4, 1. Find the count and mean.
- Find the median for the frequency table in the previous question.
- A table lists 4 with frequency 0, 5 with frequency 2, and 6 with frequency 3. Find the minimum observed value and mean.
- All students in the displayed cycle dataset cycle one extra time next week. Find the new mean and median.
- A student has 6 cycle uses in a week. Can we conclude that they never rode twice in one day?
- Explain why the dart mean is not the mean of the distinct numbers 1 through 10.
Step 1: The count is 2 + 3 + 4 + 1 = 10. Step 2: The weighted total is 2 + 6 + 12 + 4 = 24. Step 3: The mean is 2.4.
Key Takeaways
• Frequency counts repetitions; use it in both the sum and the observation count. • Running frequencies locate median positions without expanding a long list. • Read one dot as one observation unless a legend says otherwise. • A display can support some conclusions while leaving other details unknown.