1. Why summarize data?
A dataset may contain many individual values. A numerical summary helps us describe its central tendency: where its values tend to be located.
Three common measures are:
- Mean: The arithmetic average.
- Median: The middle value after ordering.
- Mode: The most frequently occurring value.
These measures answer different questions. They do not always give the same result.
Example
Consider:
2, 3, 3, 4, 18
- Mean: 6.
- Median: 3.
- Mode: 3.
The large value 18 raises the mean substantially.
Key idea: Calculating a measure correctly is only part of the task. We must also interpret what it represents.
2. Basic data vocabulary
Observation
An observation is one recorded value.
For example, if five students obtain scores:
6, 8, 8, 9, 10
there are five observations.
The score 8 occurs twice, so it contributes two observations.
Dataset size
The number of observations is called the dataset size.
We use n for this number.
For the scores above:
n = 5
Frequency
The frequency of a value is the number of times it appears.
The frequency of 8 is 2.
Distinct values versus observations
The dataset:
6, 8, 8, 9, 10
contains:
- Five observations.
- Four distinct values.
Important: The mean uses the number of observations, not the number of distinct values.
3. The arithmetic mean
The arithmetic mean is the total of all observations divided by their number.
Mean = sum of observations ÷ number of observations
The mean is often written as x̄, read as “x-bar.”
For values x₁, x₂, …, xₙ:
x̄ = (x₁ + x₂ + … + xₙ)/n
Procedure
- Add every observation, including repetitions.
- Count the observations.
- Divide the total by the count.
- Include units where appropriate.
Worked example 1: A simple mean
Find the mean of:
6, 8, 8, 9, 14
Sum:
6 + 8 + 8 + 9 + 14 = 45
Number of observations:
n = 5
Therefore:
Mean = 45/5 = 9
Verification
The mean lies between the smallest and largest values:
6 ≤ 9 ≤ 14
For any nonempty finite dataset of real numbers, the mean cannot be smaller than its minimum or larger than its maximum.
4. Mean with negative values and decimals
The mean can be negative, zero, fractional, or a value not appearing in the dataset.
Worked example 2: Negative values
Daily temperature changes are:
−3, −1, 2, 6
Their sum is:
−3 − 1 + 2 + 6 = 4
Therefore:
Mean change = 4/4 = 1 degree
Worked example 3: Decimal values
Find the mean of:
1.2, 1.5, 1.8, 2.3
Sum:
1.2 + 1.5 + 1.8 + 2.3 = 6.8
Therefore:
Mean = 6.8/4 = 1.7
The mean 1.7 is not one of the original observations. This is allowed.
5. The mean as a balance point
The deviations from the mean add to zero.
For the data:
2, 4, 6
the mean is 4.
Deviations from the mean:
−2, 0, 2
Their sum is:
−2 + 0 + 2 = 0
This explains why the mean is sometimes described as a balance point.
Useful rearrangement
From:
Mean = total/count
we obtain:
Total = count × mean
This relationship is useful for missing-value and combined-group problems.
6. Finding a missing value from the mean
Worked example 4
The mean of:
5, 7, 8, x, 12
is 9.
Find x.
Step 1: Find the required total
There are five observations.
Required total = 5 × 9 = 45
Step 2: Add the known values
5 + 7 + 8 + 12 = 32
Step 3: Subtract
x = 45 − 32 = 13
Verification
(5 + 7 + 8 + 13 + 12)/5 = 45/5 = 9
Answer: x = 13.
Updating a mean
If a dataset has n observations and mean m, its total is nm.
If one new observation a is added:
New mean = (nm + a)/(n + 1)
Worked example 5
Four scores have mean 7. A fifth score of 12 is added.
Original total:
4 × 7 = 28
New total:
28 + 12 = 40
New count:
5
New mean = 40/5 = 8
Because 12 is greater than the old mean, adding it raises the mean.
7. The median
The median is the middle value after arranging the observations in ascending order.
Essential first step
Always order the data before finding the median.
7.1 Odd number of observations
If n is odd, the median is the observation in position:
(n + 1)/2
Positions are counted from 1.
Worked example 6
Find the median of:
9, 3, 7, 5, 12
Order the values:
3, 5, 7, 9, 12
There are five observations.
Middle position:
(5 + 1)/2 = 3
Median = 7
7.2 Even number of observations
If n is even, find the observations in positions:
n/2 and n/2 + 1
The median is their arithmetic mean.
Worked example 7
Find the median of:
8, 2, 11, 4, 6, 9
Order the values:
2, 4, 6, 8, 9, 11
There are six observations.
The middle values are in positions 3 and 4:
6 and 8
Therefore:
Median = (6 + 8)/2 = 7
The median does not have to be an observed value.
8. The mode
The mode is the value with the greatest frequency.
Worked example 8: One mode
For:
2, 4, 4, 4, 7, 7, 9
the value 4 occurs three times, more often than any other value.
Mode = 4
More than one mode
For:
1, 2, 2, 3, 4, 4, 5
both 2 and 4 occur twice, and no other value occurs as often.
Modes: 2 and 4
This dataset is called bimodal.
No mode
For:
1, 3, 5, 7
all values occur once. In the convention used in this course, the dataset has no mode.
More generally, when all distinct values occur equally often, this course reports no mode. Some references use different conventions, so assessment questions should make the intended interpretation clear.
Mode versus frequency
For:
3, 3, 3, 5, 6
- Mode: 3.
- Frequency of the mode: 3.
These happen to be the same number here, but they represent different quantities.
For:
8, 8, 8, 2, 5
- Mode: 8.
- Frequency of the mode: 3.
9. Mean, median, and mode compared
| Measure | How to find it | Key feature |
|---|---|---|
| Mean | Add all observations and divide by their number. | Uses every value and is sensitive to extreme values. |
| Median | Order the data and find the middle value or average the two middle values. | Depends on central positions and is less sensitive to extreme values. |
| Mode | Identify the most frequent value. | Describes the most common observation; may not be unique. |
Worked example 9: Calculate all three
Find the mean, median, and mode of:
4, 7, 5, 7, 10, 3
Order the data:
3, 4, 5, 7, 7, 10
Mean:
(3 + 4 + 5 + 7 + 7 + 10)/6 = 36/6 = 6
Median:
(5 + 7)/2 = 6
Mode:
7
Answer: Mean = 6; median = 6; mode = 7.
The mean and median agree here, but the mode differs.
10. Working with frequency tables
A frequency table records each distinct value and how many times it occurs.
Example dataset
| Value x | Frequency f | Product x × f | Cumulative frequency |
|---|---|---|---|
| 1 | 2 | 2 | 2 |
| 2 | 3 | 6 | 5 |
| 3 | 4 | 12 | 9 |
| 4 | 1 | 4 | 10 |
| Total | 10 | 24 | — |
The expanded dataset is:
1, 1, 2, 2, 2, 3, 3, 3, 3, 4
Worked example 10: Mean from frequencies
Each value contributes value × frequency to the total.
Mean = sum of (value × frequency) ÷ sum of frequencies
Therefore:
Mean = 24/10 = 2.4
Do not calculate (1 + 2 + 3 + 4)/4. That ignores the different frequencies.
Median from frequencies
There are ten observations, so use positions 5 and 6.
Cumulative frequency tells us:
- Positions 1–2 contain 1.
- Positions 3–5 contain 2.
- Positions 6–9 contain 3.
- Position 10 contains 4.
The middle values are 2 and 3.
Median = (2 + 3)/2 = 2.5
Mode from frequencies
The largest frequency is 4, belonging to the value 3.
Mode = 3
Important: The mode is the data value 3, not its frequency 4.
11. Combined means
Suppose two groups have different sizes.
To find the combined mean, combine their totals and counts.
Combined mean = (first group total + second group total)/(combined count)
If group sizes are n₁ and n₂, with means m₁ and m₂:
Combined mean = (n₁m₁ + n₂m₂)/(n₁ + n₂)
Worked example 11
Group A contains six students with mean score 70.
Group B contains four students with mean score 85.
Find the combined mean.
Group A total:
6 × 70 = 420
Group B total:
4 × 85 = 340
Combined total:
420 + 340 = 760
Combined count:
6 + 4 = 10
Combined mean = 760/10 = 76
Why averaging the means is wrong here
The calculation:
(70 + 85)/2 = 77.5
gives equal weight to the two groups, even though their sizes differ.
A simple average of the two group means is guaranteed to give the combined mean when the groups have equal sizes.
Limitation: Group medians or modes alone do not generally determine the combined median or mode. The underlying data or sufficient frequency information is needed.
12. Extreme values and choosing a measure
An extreme value is unusually far from most other observations. Formal rules for identifying outliers are not required here.
Worked example 12
Five monthly payments, in hundreds of currency units, are:
20, 22, 23, 25, 110
Mean:
(20 + 22 + 23 + 25 + 110)/5 = 200/5 = 40
Median:
23
There is no mode under the course convention.
Interpretation
The mean is 40, although four of the five values are 25 or less.
The large value 110 pulls the mean upward.
The median 23 better describes the middle payment in this dataset.
Does this mean the mean is wrong?
No.
The mean correctly describes the total divided equally among all five observations.
The choice depends on the question:
- For equal sharing of a total: mean.
- For the middle observation or a center less affected by extremes: median.
- For the most common recorded value: mode.
No single measure is best for every purpose.
13. How changes to data affect the measures
Adding the same constant
If every observation increases by c:
- Mean increases by c.
- Median increases by c.
- Each mode increases by c.
Multiplying by a positive constant
If every observation is multiplied by a positive number k:
- Mean is multiplied by k.
- Median is multiplied by k.
- Each mode is multiplied by k.
The frequencies remain unchanged.
Worked example 13
A dataset has:
- Mean 6.
- Median 5.
- Unique mode 4.
Every observation is changed using:
New value = 2 × original value + 3
Then:
- New mean: 2 × 6 + 3 = 15.
- New median: 2 × 5 + 3 = 13.
- New mode: 2 × 4 + 3 = 11.
Changing only one observation
If one observation increases by d while the dataset size n stays unchanged:
New mean = old mean + d/n
The median and mode do not have an equivalent universal update rule. Their changes depend on positions and frequencies.
14. Common mistakes and verification
Mistake 1: Finding the median before ordering
The middle entry in the original list need not be the median.
Mistake 2: Dividing by the number of distinct values
Repeated observations still count separately.
Mistake 3: Using only one middle value when n is even
Average the two central observations.
Mistake 4: Reporting the largest value as the mode
The mode is the most frequent value, not necessarily the largest.
Mistake 5: Reporting the frequency as the mode
If 7 occurs five times, the mode is 7, not 5.
Mistake 6: Ignoring frequencies when calculating a mean
Use value × frequency for each row.
Mistake 7: Averaging unequal-sized group means directly
Combine totals and counts.
Mistake 8: Assuming equal means imply identical datasets
Different datasets can have the same mean but very different patterns.
For example:
4, 5, 6 and 0, 5, 10
both have mean 5 and median 5, but their values are distributed differently.
Verification checklist
- Have I counted every observation?
- Have I included repeated values?
- Is the data ordered for the median?
- Have I used the correct middle position or positions?
- Is the mode a data value with maximum frequency?
- Does the mean lie between the minimum and maximum?
- Do the frequencies add to the stated dataset size?
- Have I weighted group means by group sizes?
- Are units and rounding appropriate?
15. Independent practice
Show the calculations and order the data where needed.
Practice 1
Find the mean, median, and mode:
3, 5, 5, 7, 10
Practice 2
Find the mean, median, and mode:
12, 4, 7, 4, 9, 6
Practice 3
Find the mean and median:
−4, −1, 0, 3, 7
Practice 4
The mean of:
6, 8, x, 10
is 9.
Find x.
Practice 5
Five observations have mean 12. A new observation of 18 is added.
Find the new mean.
Practice 6
Find all modes of:
2, 3, 3, 5, 6, 6, 8
Practice 7
Find the mean, median, and mode from the frequency table.
| Value | Frequency |
|---|---|
| 2 | 1 |
| 4 | 3 |
| 6 | 2 |
| 8 | 2 |
Practice 8
Group A has eight observations with mean 10.
Group B has two observations with mean 20.
Find the combined mean.
Practice 9
Compare the mean and median of:
5, 6, 7, 8, 9
with those of:
5, 6, 7, 8, 49
Which measure changes more?
Practice 10
A dataset has mean 8, median 7, and unique mode 6.
Every value is multiplied by 3 and then reduced by 2.
Find the new mean, median, and mode.
16. Practice solutions
Solution 1
The data is already ordered.
Sum:
3 + 5 + 5 + 7 + 10 = 30
Mean = 30/5 = 6
The third observation is 5.
Median = 5
The most frequent value is 5.
Mode = 5
Solution 2
Order the data:
4, 4, 6, 7, 9, 12
Sum:
42
Mean = 42/6 = 7
Middle values:
6 and 7
Median = 6.5
Mode = 4
Solution 3
Sum:
−4 − 1 + 0 + 3 + 7 = 5
Mean = 5/5 = 1
The third observation is 0.
Median = 0
Solution 4
Required total:
4 × 9 = 36
Known total:
6 + 8 + 10 = 24
x = 36 − 24 = 12
Solution 5
Original total:
5 × 12 = 60
New total:
60 + 18 = 78
New count:
6
New mean = 78/6 = 13
Solution 6
Both 3 and 6 occur twice. Every other value occurs once.
Modes: 3 and 6
Solution 7
Total frequency:
1 + 3 + 2 + 2 = 8
Weighted total:
2 × 1 + 4 × 3 + 6 × 2 + 8 × 2 = 42
Mean = 42/8 = 5.25
Expanded data:
2, 4, 4, 4, 6, 6, 8, 8
Middle values are the fourth and fifth observations:
4 and 6
Median = 5
The highest frequency belongs to 4.
Mode = 4
Solution 8
Group A total:
8 × 10 = 80
Group B total:
2 × 20 = 40
Combined mean = (80 + 40)/(8 + 2) = 120/10 = 12
Solution 9
First dataset:
Mean = 35/5 = 7
Median = 7
Second dataset:
Mean = 75/5 = 15
Median = 7
The mean increases by 8. The median does not change.
The mean changes more.
Solution 10
Apply the transformation to each measure:
New mean = 3 × 8 − 2 = 22
New median = 3 × 7 − 2 = 19
New mode = 3 × 6 − 2 = 16
17. Lesson self-check
This is a formative check within 04.4. It does not replace a separate scheduled statistics assessment.
Check 1
Find the mean of:
2, 4, 4, 6, 9
A. 4
B. 5
C. 6
D. 25
Check 2
Find the median of:
10, 2, 8, 4, 6, 12
A. 6
B. 8
C. 9
D. 7
Check 3
Find the mode of:
1, 2, 2, 3, 3, 3, 4
A. 3
B. 2
C. 4
D. 7
Check 4
The mean of:
5, 7, x, 11
is 8.
Find x.
A. 7
B. 8
C. 9
D. 10
Check 5
Three observations have mean 10, and seven observations have mean 20.
What is the combined mean?
A. 15
B. 17
C. 30
D. 16
Check 6
For the dataset:
4, 5, 6, 7, 100
which statement is correct?
A. The median is 100.
B. The mode is 100.
C. The mean is smaller than the median.
D. The mean is 24.4 and the median is 6.
18. Self-check answers and explanations
Check 1: B
Mean = (2 + 4 + 4 + 6 + 9)/5 = 25/5 = 5
Check 2: D
Order the data:
2, 4, 6, 8, 10, 12
The two middle values are 6 and 8.
Median = (6 + 8)/2 = 7
Check 3: A
The value 3 occurs three times, more frequently than any other value.
Mode = 3
Check 4: C
Required total:
4 × 8 = 32
Known total:
5 + 7 + 11 = 23
x = 32 − 23 = 9
Check 5: B
Combined total:
3 × 10 + 7 × 20 = 30 + 140 = 170
Combined count:
10
Combined mean = 170/10 = 17
Check 6: D
Sum:
4 + 5 + 6 + 7 + 100 = 122
Mean = 122/5 = 24.4
The middle observation is 6.
Median = 6
The large value 100 raises the mean substantially.
19. Revision guidance
- Difficulty with means: Check the total and observation count separately.
- Difficulty with medians: Order the data and mark the central position or positions.
- Difficulty with modes: Create a frequency count.
- Difficulty with missing values: Use total = count × mean.
- Difficulty with frequency tables: Multiply each value by its frequency.
- Difficulty with combined means: Recover group totals before combining.
- Difficulty with interpretation: Explain what question each measure answers.
For every incorrect or guessed answer:
- Identify the measure requested.
- Write the correct method.
- Show the calculation.
- Check the result against the ordered data or frequency table.
20. Lesson review
You should now be able to:
- Distinguish observations from distinct values.
- Calculate the arithmetic mean.
- Find medians for odd and even dataset sizes.
- Identify one mode, multiple modes, or no mode under the stated convention.
- Calculate all three measures from a frequency table.
- Recover missing values from a known mean.
- Update a mean when an observation is added.
- Combine group means using their sizes.
- Explain the effect of an extreme value.
- Select a measure suited to the question.
- Describe how adding or positively scaling every value changes the measures.
Essential rules
Mean = total of observations ÷ number of observations
Total = number of observations × mean
Median = middle value after ordering, or the mean of the two middle values
Mode = most frequently occurring value or values
Frequency-table mean = sum of (value × frequency) ÷ total frequency
Combined mean = combined total ÷ combined count
Completion target: Calculate and interpret all three measures correctly, then independently correct every missed practice or self-check question.