1. Why do we need measures of dispersion?
Measures of center do not describe how far observations are spread out.
Consider:
Dataset A: 4, 5, 6
Dataset B: 0, 5, 10
Both datasets have:
- Mean = 5.
- Median = 5.
However, Dataset B is more spread out.
Dispersion describes the spread or variability of data.
This lesson introduces:
- Range: Distance between the largest and smallest observations.
- Variance: Average squared deviation from the mean.
- Standard deviation: Square root of the variance.
A useful description of numerical data often needs both:
- A measure of center.
- A measure of dispersion.
2. Range
The range is:
Range = maximum value − minimum value
Worked example 1
Find the range of:
8, 3, 11, 5, 7
Maximum = 11.
Minimum = 3.
Range = 11 − 3 = 8
Range with negative values
For:
−6, −2, 1, 4
Range = 4 − (−6) = 10
What range tells us
Range describes the full distance from the smallest observation to the largest.
However, it uses only those two extreme values.
For example:
A: 0, 5, 5, 5, 10
B: 0, 0, 5, 10, 10
Both have range 10, but their internal distributions differ.
Key lesson: Equal ranges do not imply equal overall dispersion.
3. Deviations from the mean
A deviation measures how far an observation lies above or below the mean.
Deviation = observation − mean
For an observation x and dataset mean x̄:
Deviation = x − x̄
Example
For:
2, 4, 6
the mean is:
x̄ = 4
The deviations are:
- 2 − 4 = −2.
- 4 − 4 = 0.
- 6 − 4 = 2.
Why not average the signed deviations?
Their sum is:
−2 + 0 + 2 = 0
In fact, deviations from the arithmetic mean always sum to zero.
Positive and negative deviations cancel, even when the observations are widely spread.
Therefore, their ordinary average cannot measure dispersion.
The solution: Square the deviations
Squaring makes every deviation nonnegative:
(−2)² = 4
0² = 0
2² = 4
The squared deviations no longer cancel.
4. Variance
Variance is the average of the squared deviations from the mean.
For n observations:
Variance = sum of squared deviations ÷ n
We write the variance as σ², read as “sigma squared.”
σ² = [(x₁ − x̄)² + (x₂ − x̄)² + … + (xₙ − x̄)²]/n
Denominator convention
Throughout this lesson:
Divide by n, the number of observations.
Some statistical estimation problems use n − 1. That is a different convention and must not be substituted into this lesson’s calculations.
The questions below concern the descriptive variance of the listed data.
Calculation procedure
- Find the mean.
- Subtract the mean from each observation.
- Square every deviation.
- Add the squared deviations.
- Divide by the number of observations.
5. Worked example 2: Calculating variance step by step
Find the variance of:
2, 4, 4, 6
Step 1: Find the mean
x̄ = (2 + 4 + 4 + 6)/4 = 16/4 = 4
Step 2: Calculate deviations and squares
| Observation x | Deviation x − 4 | Squared deviation |
|---|---|---|
| 2 | −2 | 4 |
| 4 | 0 | 0 |
| 4 | 0 | 0 |
| 6 | 2 | 4 |
| Total | 0 | 8 |
Step 3: Divide by n
There are four observations.
σ² = 8/4 = 2
Variance = 2
Verification
The squared deviations are nonnegative, so the variance must be nonnegative.
The result is positive because the observations are not all equal.
6. Standard deviation
Standard deviation is the nonnegative square root of variance.
We write it as σ, read as “sigma.”
σ = √(σ²)
For the previous example:
Variance = 2
Therefore:
Standard deviation = √2 ≈ 1.414
Why take a square root?
Squaring deviations changes their units.
If observations are measured in centimeters:
- Deviations are in centimeters.
- Squared deviations are in square centimeters.
- Variance is in square centimeters.
- Standard deviation returns to centimeters.
Interpretation
Standard deviation measures spread around the mean using all observations.
It is the root-mean-square deviation, not the ordinary average of the absolute distances from the mean.
A smaller standard deviation indicates observations that are more tightly clustered around their mean.
A larger standard deviation indicates greater spread.
It does not mean that every observation is exactly one standard deviation from the mean.
7. Basic properties of dispersion
Property 1: Nonnegative values
Range ≥ 0
Variance ≥ 0
Standard deviation ≥ 0
A negative variance is a calculation error.
Property 2: Zero dispersion
Variance is zero exactly when all observations are equal.
For:
7, 7, 7, 7
- Mean = 7.
- Every deviation = 0.
- Variance = 0.
- Standard deviation = 0.
- Range = 0.
Property 3: Data order does not matter
Rearranging observations does not change the mean, range, variance, or standard deviation.
Property 4: Standard deviation and variance rank spread consistently
If one dataset has a larger variance than another, it also has a larger standard deviation.
This follows because the square-root function increases for nonnegative inputs.
Property 5: Comparisons need context
Compare dispersion meaningfully when datasets measure the same quantity using the same units.
A larger standard deviation does not automatically mean worse performance. Whether greater variability is desirable depends on the context.
8. Worked example 3: Same mean, different spread
Compare:
Dataset A: 4, 5, 6
Dataset B: 0, 5, 10
Both have mean 5.
Dataset A
Squared deviations:
1, 0, 1
Variance = 2/3
Standard deviation = √(2/3) ≈ 0.816
Range = 2
Dataset B
Squared deviations:
25, 0, 25
Variance = 50/3
Standard deviation = √(50/3) ≈ 4.082
Range = 10
Conclusion
Dataset B is more dispersed according to all three measures.
The same mean does not imply the same variability.
9. A second formula for variance
Variance can also be calculated using:
Variance = mean of the squared observations − square of the mean
In symbols:
σ² = (x₁² + x₂² + … + xₙ²)/n − x̄²
Understand the difference
These are not the same:
- Mean of squares: Add x₁², x₂², and so on, then divide by n.
- Square of the mean: Find x̄ first, then square it.
Why the formula works
Expanding a squared deviation gives:
(x − x̄)² = x² − 2xx̄ + x̄²
When averaged over the dataset:
- The average of x² is the mean of squares.
- The average of x is x̄.
Therefore:
Variance = mean of squares − 2x̄² + x̄²
So:
Variance = mean of squares − x̄²
Worked example 4
Find the variance and standard deviation of:
1, 2, 3, 4
Mean:
x̄ = 10/4 = 2.5
Mean of squares:
(1² + 2² + 3² + 4²)/4 = 30/4 = 7.5
Variance:
σ² = 7.5 − 2.5² = 7.5 − 6.25 = 1.25
Standard deviation:
σ = √1.25 = √5/2 ≈ 1.118
Calculation caution
Keep exact fractions or sufficient decimal precision during intermediate calculations.
Premature rounding can distort the final variance.
For small datasets, the deviation table is often easiest to check.
10. Frequency tables
When a value occurs several times, its squared deviation must be counted that many times.
Formulas
n = total frequency
Mean = sum of (frequency × value) ÷ total frequency
Variance = sum of [frequency × squared deviation] ÷ total frequency
Alternatively:
Variance = sum of (frequency × value²) ÷ total frequency − mean²
Worked example 5
Find the mean, variance, and standard deviation for:
| Value x | Frequency f | f × x | f × (x − 4)² |
|---|---|---|---|
| 2 | 1 | 2 | 4 |
| 4 | 2 | 8 | 0 |
| 6 | 1 | 6 | 4 |
| Total | 4 | 16 | 8 |
Step 1: Find the mean
Mean = 16/4 = 4
Step 2: Find the variance
Variance = 8/4 = 2
Step 3: Find the standard deviation
Standard deviation = √2 ≈ 1.414
Important
There are four observations, not three.
The expanded data is:
2, 4, 4, 6
Divide by the total frequency, not the number of rows.
11. Extreme values and limitations of range
Variance and standard deviation are sensitive to extreme observations because deviations are squared.
Worked example 6
Compare:
A: 0, 5, 5, 5, 10
B: 0, 0, 5, 10, 10
Both have:
Mean = 5
Range = 10
Dataset A
Squared deviations:
25, 0, 0, 0, 25
Variance = 50/5 = 10
Standard deviation = √10
Dataset B
Squared deviations:
25, 25, 0, 25, 25
Variance = 100/5 = 20
Standard deviation = √20 = 2√5
Interpretation
Dataset B has more observations far from the mean.
The ranges are equal, but Dataset B has greater variance and standard deviation.
Key lesson: Range describes only the extremes; variance and standard deviation account for every observation.
12. Adding a constant to every observation
Suppose every observation increases by c.
The mean also increases by c.
The deviations remain unchanged because:
(x + c) − (x̄ + c) = x − x̄
Therefore:
- Mean increases by c.
- Range stays unchanged.
- Variance stays unchanged.
- Standard deviation stays unchanged.
Worked example 7
Original data:
1, 3, 5
Add 10 to every value:
11, 13, 15
Original mean:
3
New mean:
13
In both datasets, deviations from the mean are:
−2, 0, 2
Therefore, both have:
Range = 4
Variance = 8/3
Standard deviation = √(8/3)
Adding a constant changes location, not dispersion.
13. Multiplying every observation by a constant
Suppose each observation is multiplied by k.
Each deviation is also multiplied by k.
Each squared deviation is multiplied by k².
Therefore:
New variance = k² × original variance
New standard deviation = |k| × original standard deviation
New range = |k| × original range
The absolute value is necessary because range and standard deviation cannot be negative.
General transformation
For:
y = kx + c
the measures change as follows:
| Measure | After y = kx + c |
|---|---|
| Mean | k × original mean + c |
| Range | |k| × original range |
| Variance | k² × original variance |
| Standard deviation | |k| × original standard deviation |
Worked example 8
A dataset has:
- Mean = 4.
- Range = 8.
- Variance = 9.
- Standard deviation = 3.
Every value is transformed using:
y = −2x + 5
Then:
New mean = −2 × 4 + 5 = −3
New range = 2 × 8 = 16
New variance = (−2)² × 9 = 36
New standard deviation = 2 × 3 = 6
The negative multiplier reverses the order of the observations, but dispersion remains nonnegative.
Special case: k = 0
Every transformed observation becomes c.
Therefore, range, variance, and standard deviation all become zero.
14. Interpreting consistency
Worked example 9
Two machines fill containers.
Their recorded fill volumes have:
- Machine A: mean 500 mL, standard deviation 2 mL.
- Machine B: mean 500 mL, standard deviation 7 mL.
Which machine has more consistent recorded volumes?
Machine A, because its smaller standard deviation indicates less variation around its mean.
What this does not prove
A smaller standard deviation alone does not establish accuracy relative to a target.
For example:
- Machine C could have mean 480 mL and standard deviation 1 mL.
- Its fills are consistent, but centered below a 500 mL target.
Consistency concerns spread. Accuracy relative to a target also requires considering location.
15. Common mistakes and verification
Mistake 1: Averaging signed deviations
They sum to zero. Square them before averaging for variance.
Mistake 2: Forgetting the final division
The sum of squared deviations is not the variance until it is divided by n.
Mistake 3: Using n − 1 automatically
This lesson uses n. Follow the convention stated in the problem.
Mistake 4: Confusing variance with standard deviation
Standard deviation is the square root of variance.
Mistake 5: Squaring a negative deviation incorrectly
(−3)² = 9, not −9.
Mistake 6: Ignoring frequencies
A value occurring five times contributes five squared deviations.
Mistake 7: Multiplying variance by k instead of k²
Scaling every observation by 3 multiplies variance by 9.
Mistake 8: Thinking a shift changes dispersion
Adding 10 to every observation leaves variance and standard deviation unchanged.
Mistake 9: Averaging group variances without enough information
A combined variance depends on group sizes, group variances, and differences between group means.
Do not assume it is the simple average of the group variances.
Verification checklist
- Is the mean correct?
- Have I included every observation?
- Do the signed deviations sum to zero, allowing for rounding?
- Are all squared deviations nonnegative?
- Have I divided by the total number of observations?
- Is variance nonnegative?
- Does standard deviation squared reproduce variance?
- Have I used frequencies correctly?
- Are variance units squared and standard-deviation units original?
- Have I avoided premature rounding?
16. Independent practice
Use the variance convention that divides by n.
Give exact answers where possible. Decimal approximations may be added to three decimal places.
Practice 1
Find the range, variance, and standard deviation of:
3, 3, 3, 3
Practice 2
Find the mean, variance, and standard deviation of:
2, 4, 6
Practice 3
Find the range, mean, variance, and standard deviation of:
−2, 0, 2, 4
Practice 4
Use the mean-of-squares formula to find the variance of:
1, 3, 5, 7
Practice 5
A dataset has eight observations. The sum of its squared deviations from the mean is 72.
Find its variance and standard deviation.
Practice 6
Find the mean, variance, and standard deviation from this frequency table.
| Value | Frequency |
|---|---|
| 1 | 2 |
| 3 | 4 |
| 5 | 2 |
Practice 7
Compare the variances of:
A: 8, 10, 12
B: 6, 10, 14
Which is more dispersed?
Practice 8
A dataset has mean 12, range 9, and standard deviation 4.
Every value is increased by 7.
Find the new mean, range, variance, and standard deviation.
Practice 9
A dataset has mean 5 and variance 4.
Every value is transformed using:
y = 3x − 2
Find the new mean, variance, and standard deviation.
Practice 10
A dataset has range 10 and standard deviation 6.
Every value is transformed using:
y = −x/2 + 8
Find the new range, variance, and standard deviation.
17. Practice solutions
Solution 1
All observations equal 3.
Range = 0
Variance = 0
Standard deviation = 0
Solution 2
Mean:
(2 + 4 + 6)/3 = 4
Squared deviations:
4, 0, 4
Variance = 8/3
Standard deviation = √(8/3) ≈ 1.633
Solution 3
Range:
4 − (−2) = 6
Mean:
(−2 + 0 + 2 + 4)/4 = 1
Deviations:
−3, −1, 1, 3
Squared deviations:
9, 1, 1, 9
Variance = 20/4 = 5
Standard deviation = √5 ≈ 2.236
Solution 4
Mean:
(1 + 3 + 5 + 7)/4 = 4
Mean of squares:
(1 + 9 + 25 + 49)/4 = 84/4 = 21
Variance = 21 − 4² = 5
Solution 5
Variance = 72/8 = 9
Standard deviation = √9 = 3
Solution 6
Total frequency:
2 + 4 + 2 = 8
Mean:
(1 × 2 + 3 × 4 + 5 × 2)/8 = 24/8 = 3
Weighted sum of squared deviations:
2 × (1 − 3)² + 4 × (3 − 3)² + 2 × (5 − 3)²
= 2 × 4 + 4 × 0 + 2 × 4 = 16
Variance = 16/8 = 2
Standard deviation = √2 ≈ 1.414
Solution 7
Both means are 10.
Dataset A:
Variance = (4 + 0 + 4)/3 = 8/3
Dataset B:
Variance = (16 + 0 + 16)/3 = 32/3
Dataset B is more dispersed.
Its variance is four times that of A, and its standard deviation is twice that of A.
Solution 8
Adding 7 changes only the mean.
New mean = 19
New range = 9
New variance = 4² = 16
New standard deviation = 4
Solution 9
New mean = 3 × 5 − 2 = 13
New variance = 3² × 4 = 36
New standard deviation = √36 = 6
Solution 10
The multiplier is −1/2, whose absolute value is 1/2.
New range = 10/2 = 5
New standard deviation = 6/2 = 3
New variance = 3² = 9
The added constant 8 does not affect dispersion.
18. Lesson assessment — Variance & Dispersion Quiz
Suggested time: 12–15 minutes
Questions: 8 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 7/8
Instructions: Use variance calculated by dividing by n for every question.
Question 1
Find the range of:
−5, −1, 3, 7
A. 2
B. 7
C. 12
D. 14
Question 2
Find the variance of:
2, 4, 6
A. 8/3
B. 4
C. 8
D. √(8/3)
Question 3
A dataset has variance 25.
What is its standard deviation?
A. 25
B. 625
C. −5
D. 5
Question 4
Every observation in a dataset is increased by 10.
Which statement is correct?
A. The variance increases by 10.
B. The variance and standard deviation remain unchanged.
C. The standard deviation increases by 10.
D. The variance increases by 100.
Question 5
A dataset has variance 9. Every observation is multiplied by 2.
What is the new variance?
A. 18
B. 6
C. 36
D. 81
Question 6
A dataset has values 1 and 5, each occurring twice.
What is its variance?
A. 4
B. 2
C. 8
D. 16
Question 7
Two datasets measure the same quantity in the same units and have equal means.
Dataset A has standard deviation 2. Dataset B has standard deviation 5.
Which statement is correct?
A. Dataset A must contain fewer observations.
B. Every observation in B is greater than every observation in A.
C. The datasets must have different medians.
D. Dataset A has smaller variance and smaller standard deviation.
Question 8
A dataset has standard deviation 4.
Every value is transformed using:
y = −3x + 7
What is the new standard deviation?
A. −12
B. 12
C. 19
D. 144
19. Instructor answer key
Tutor LMS placement: Put the assessment in a separate Quiz item if it will be graded. Keep the answer key and solutions separate from student-facing questions until submission.
| Question | Answer | Skill assessed |
|---|---|---|
| 1 | C | Range |
| 2 | A | Variance from raw data |
| 3 | D | Standard deviation from variance |
| 4 | B | Adding a constant |
| 5 | C | Scaling variance |
| 6 | A | Repeated observations |
| 7 | D | Comparing dispersion |
| 8 | B | Scaling standard deviation |
20. Assessment solutions
Answer 1: C
Range = 7 − (−5) = 12
Answer 2: A
Mean:
4
Squared deviations:
4, 0, 4
Variance = 8/3
Answer 3: D
Standard deviation = √25 = 5
Use the nonnegative square root.
Answer 4: B
Adding the same constant to every observation shifts the mean by that constant.
Deviations from the mean remain unchanged, so variance and standard deviation remain unchanged.
Answer 5: C
Variance is multiplied by the square of the scale factor.
New variance = 2² × 9 = 36
Answer 6: A
The expanded data is:
1, 1, 5, 5
Mean:
3
Each squared deviation equals 4.
Variance = (4 + 4 + 4 + 4)/4 = 4
Answer 7: D
Dataset A has variance:
2² = 4
Dataset B has variance:
5² = 25
Therefore, A has smaller variance and standard deviation.
The given information does not determine dataset sizes or medians.
Answer 8: B
The multiplier has absolute value 3.
New standard deviation = 3 × 4 = 12
The added constant 7 does not affect dispersion.
21. Revision guidance
- Missed Question 1: Review maximum minus minimum and subtraction of negative numbers.
- Missed Question 2: Review the mean, squared deviations, and division by n.
- Missed Question 3: Distinguish variance from its square root.
- Missed Question 4: Review why a shift leaves deviations unchanged.
- Missed Question 5: Review the squared scaling rule for variance.
- Missed Question 6: Count every repeated observation.
- Missed Question 7: Review what dispersion does and does not establish.
- Missed Question 8: Use the absolute scale factor for standard deviation.
For each incorrect or guessed answer:
- State the measure requested.
- Identify the applicable formula.
- Show the calculation.
- Check nonnegativity, units, and the relationship between variance and standard deviation.
22. Lesson review
You should now be able to:
- Explain why a mean alone does not describe a dataset.
- Calculate and interpret range.
- Calculate deviations from the mean.
- Explain why signed deviations cannot measure spread by averaging.
- Calculate variance by dividing the squared-deviation total by n.
- Calculate standard deviation.
- Use the mean-of-squares formula.
- Work with frequency tables.
- Compare datasets with equal means or equal ranges.
- Explain the sensitivity of variance to extreme values.
- Apply shift and scaling rules.
- Distinguish consistency from accuracy relative to a target.
Essential rules
Range = maximum − minimum
Variance = sum of squared deviations from the mean ÷ n
Standard deviation = √variance
Variance = mean of squares − square of mean
For y = kx + c:
New mean = k × original mean + c
New variance = k² × original variance
New standard deviation = |k| × original standard deviation
New range = |k| × original range
Completion target: Score at least 7/8 on the assessment and independently correct every missed or guessed answer.