A library assistant wants to describe an ordinary day’s borrowing. Another person wants to plan for the total number of books handled over a week. They may use the same records but choose different summaries. Calculating an average is only half the task; the other half is explaining what that number tells the reader.
Here we compare mean, median and mode using invented borrowing records and a separate frequency table. Each observation has a clear meaning. The examples use exact values or counts, not class intervals. They show how an unusually large value can change the mean and why that does not automatically make the value a mistake.
Keep the observation and the summary separate
The arithmetic mean is the sum of all observations divided by the number of observations. It shares the total equally. The median is the middle value after sorting; with an even number of observations it is the mean of the two middle values. The mode is a value occurring with the greatest frequency. There may be more than one mode, or no useful unique mode.
Before doing arithmetic, finish this sentence: “One observation represents …”. It might be one day’s borrowing count, one learner’s waiting time or one packet’s mass. A frequency tells you how many observations have a particular value. It is not an additional observation to add to the measurement total.
Worked example 1: summarise a seven-day record
An invented library record gives daily numbers of books borrowed as 8, 10, 9, 10, 12, 11, 10. One observation is one day’s count. Calculate all three averages and describe an ordinary day.
The total is 8 + 10 + 9 + 10 + 12 + 11 + 10 = 70 books. There are seven days, so the mean is 70 ÷ 7 = 10 books per day. This does not mean every day had exactly ten borrowings. It means that seven days at ten would produce the same total.
Sort the observations: 8, 9, 10, 10, 10, 11, 12. With seven values, the middle position is (7 + 1) ÷ 2 = 4. The fourth value is 10, so the median is 10 books. The position number 4 is not the median count. Finally, 10 appears three times, more often than any other count, so the mode is 10 books.
All three summaries agree in this particular dataset. A reasonable sentence is: “The typical daily count was around ten; observed counts ran from eight to twelve.” That describes these seven records. It does not establish what will happen next month or in every library.
Worked example 2: include an unusually busy day
Now consider a separate eight-day record containing those seven counts and a special-event day with 38 books borrowed. The extra day is known to be a genuine observation. Compare the summaries with the original week.
The new total is 70 + 38 = 108, and there are now eight observations. Mean = 108 ÷ 8 = 13.5 books per day. Sorted values are 8, 9, 10, 10, 10, 11, 12, 38. The fourth and fifth values are both 10, so median = (10 + 10) ÷ 2 = 10. The mode remains 10.
The mean rose by 3.5 books per day while the median and mode stayed unchanged. The high value contributes its full amount to the sum, but it does not move either middle value in this example. The fractional mean is legitimate even though no individual day can have half a book borrowed.
If the question asks about an ordinary non-event day, the median of 10 is a more representative centre here than 13.5. If it asks about the actual total work across all eight days, the mean remains useful: 13.5 × 8 recovers 108. Neither average is universally best. State the purpose and mention the event rather than quietly deleting it.
Calling 38 an unusually large value is a description relative to these records. We are not applying a formal outlier threshold in this lesson. In real records, check the source: was it an event, a recording error, or a count covering several days? Correct an error only when evidence supports the correction. An inconvenient genuine observation is still data.
Worked example 3: use a discrete-frequency table
A fictional reading club records the number of books each of 13 participants completed in a fixed month. Calculate the mean, median and mode from the frequency table.
| Books, x | Participants, f | Books contributed, xf | Ordered positions |
|---|---|---|---|
| 0 | 2 | 0 | 1–2 |
| 1 | 3 | 3 | 3–5 |
| 2 | 5 | 10 | 6–10 |
| 3 | 2 | 6 | 11–12 |
| 5 | 1 | 5 | 13 |
| Total | 13 | 24 | 13 observations |
The weighted total is Σxf = 24 books, while Σf = 13 participants. Therefore mean = 24 ÷ 13 = 1.85 books per participant, to two decimal places. There are five rows of values, but five is not the number of participants.
The median is the seventh ordered observation because (13 + 1) ÷ 2 = 7. Positions six through ten contain the value 2, so median = 2 books. The highest frequency is 5, attached to x = 2. Thus mode = 2 books, not 5. Frequency and observed value must not swap roles.
A wrong calculation gives 24 ÷ 5 = 4.8. Its denominator counts the distinct values rather than the people. You can check the correct mean by expanding the data: 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 5. This expanded list still totals 24 and contains 13 observations. The zero-book participants belong in the denominator even though they add zero to the numerator.
Explain the choice in a complete sentence
“Use the median because of the outlier” is a start, but support it with the effect you calculated. A stronger answer says that adding 38 raises the mean from 10 to 13.5 while the median remains 10, so the median better describes a routine day in these records. For a common size or category, the mode can answer a different question: which value occurs most often?
Do not claim a median always stays fixed when data are added. Adding or removing an observation can change the middle positions. Nor does an equal mean imply identical datasets. Counts 6, 10, 14 and 9, 10, 11 both have mean 10, but their spreads differ. A centre should be read alongside the actual distribution when that difference matters.
Three practice tasks with solved answers
Practice 1: two middle values
Find the three averages of 7, 4, 7, 9, 5, 10.
Answer: Sorted: 4, 5, 7, 7, 9, 10. Sum = 42, so mean = 42 ÷ 6 = 7. Median = (7 + 7) ÷ 2 = 7; mode = 7. There are two middle positions, even though their values happen to agree.
Practice 2: two modes
Values 1, 2 and 4 have frequencies 3, 2 and 3. Find the averages and explain the modal result.
Answer: Total frequency = 8; weighted sum = 3 + 4 + 12 = 19. Mean = 2.375; middle positions four and five both contain 2, so median = 2. Values 1 and 4 each appear three times, making the distribution bimodal.
Practice 3: replace an extreme observation
Waiting times are 5, 6, 6, 7, 26 minutes. Compare mean and median. Then suppose source checking proves 26 was a typing error for 6.
Answer: Initially mean = 50 ÷ 5 = 10 minutes, median = 6 minutes. After the verified correction, sum = 30, mean = 6 minutes and median remains 6 minutes. The correction is justified by evidence, not by a preference for a smaller mean.
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Choose the chart as well as the average
An average summarises the values, while a chart shows how those values are arranged. Continue with three original chart-choice cases: columns compare categories, a line follows ordered monthly totals, and an XY scatter chart preserves paired numerical measurements and their spacing. Its repaired-chart examples show how a wrong axis or data mapping can distort the evidence. Keep the units in both the chart labels and the average you report.
Frequently asked questions
Must the mean be an observed value?
No. It is an equal-share summary of the total and may be fractional even when every observation is a whole-number count.
Why sort before finding the median?
The median concerns position in numerical order. The middle value in the original recording order may have no statistical significance.
What happens with an even number of observations?
After sorting, take the arithmetic mean of the values in positions n/2 and n/2 + 1. Do not average the smallest and largest values.
Can a dataset have two modes?
Yes. If two values share the greatest frequency, both are modes. Do not choose one just because it appears earlier in the list.
What if every value appears once?
There is no uniquely most frequent value. In the usual school convention, report no mode rather than naming an arbitrary value.
Should I remove an outlier before calculating?
Only with a stated, defensible reason. Investigate possible errors and report genuine unusual values; removing them silently changes the question.
Does frequency go in the mean denominator?
Yes: add all frequencies to obtain the observation count. The numerator is the sum of value times frequency, not the sum of frequencies alone.
Background references
The original examples and diagrams above were independently calculated. These public educational references support the underlying methods; they do not establish a current Kenyan grade allocation.
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