Difficulty: Introductory | Prerequisites: Measures of central tendency (Part 1), summation notation (Σ), square roots
Big picture: Knowing the centre of a dataset is only half the story. Two datasets can share the same mean but look completely different if one is tightly clustered and the other is widely scattered. Measures of variability tell you how spread out the data is, and they are essential for almost every inferential procedure in statistics. This section builds directly on the summation table from Part 1, so make sure you are comfortable with ΣX, ΣX², and (X − X̄) before continuing.
Variability describes how much the individual observations in a dataset differ from the centre. The main measures covered here are variance (average squared deviation from the mean), standard deviation (the square root of variance, back in the original units), coefficient of variation (standard deviation as a percentage of the mean), and quartiles (values that split the sorted data into four equal parts).
Sample variance (s²)
The average of the squared deviations from the mean, with one adjustment: you divide by (n − 1) instead of n. The formula is s² = [ΣX² − (ΣX)²/n] / (n − 1). Think of it as a measure of how far, on average, each observation sits from the mean, but in squared units.
Sample standard deviation (s)
The square root of the variance: s = √s². This brings the measure of spread back into the same units as the original data. In simple terms, if the data are in parts per million, the standard deviation is also in parts per million.
Coefficient of variation (CV)
The standard deviation expressed as a percentage of the mean: CV = (s / X̄) × 100%. This is a unitless measure of relative spread, which makes it useful for comparing variability between datasets measured in different units or on different scales.
Quartiles (Q₁ and Q₃)
Values that divide the sorted data into four equal parts. Q₁ (the first quartile, 25th percentile) separates the lowest 25% of data. Q₃ (the third quartile, 75th percentile) separates the lowest 75%. Together with the median (Q₂), they describe the shape and spread of the distribution. In simple terms, Q₁ is the "median of the lower half" and Q₃ is the "median of the upper half."
Interquartile range (IQR)
Q₃ − Q₁. It captures the range of the middle 50% of the data and is resistant to outliers.
Mean deviation (of a single observation)
The signed difference between an observation and the mean: (X − X̄). A negative value means the observation falls below the mean; a positive value means it falls above.
The computational formula (easier for hand calculation) is:
s² = [ΣX² − (ΣX)² / n] / (n − 1)
Using the Quiz 1 data (X values: 4, 2, 4, 8, 3, 6, 8, 2, 5, 8):
ΣX = 50
ΣX² = 302
n = 10
Plugging in:
s² = [302 − (50)² / 10] / (10 − 1)
s² = [302 − 2500/10] / 9
s² = [302 − 250] / 9
s² = 52 / 9
s² = 5.78 (rounded to two decimal places)
Why divide by (n − 1) instead of n? This is called Bessel's correction. A sample tends to underestimate the true population variability, and dividing by (n − 1) corrects for that bias. The denominator (n − 1) is also called the degrees of freedom.
s = √s²
s = √5.78
s = 2.40
The standard deviation tells you that the typical observation sits roughly 2.4 parts per million away from the mean of 5.
CV = (s / X̄) × 100%
CV = (2.40 / 5) × 100%
CV = 48%
A CV of 48% is fairly high, indicating substantial variability relative to the mean. The CV is especially useful when comparing datasets with different units or different magnitudes, because it normalises the spread.
The method used in this course:
Step 1: Sort the data in ascending order.
Sorted: 2, 2, 3, 4, 4, 5, 6, 8, 8, 8
Step 2: Compute the position of Q₁ and Q₃.
Q₁ position = 0.25 × n = 0.25 × 10 = 2.5
Q₃ position = 0.75 × n = 0.75 × 10 = 7.5
Step 3: Round up to the next whole number to find the data position.
Q₁: round 2.5 up to position 3 → the 3rd value in the sorted list = 3
Q₃: round 7.5 up to position 8 → the 8th value in the sorted list = 8
So Q₁ = 3 and Q₃ = 8. The IQR = 8 − 3 = 5.
Note: different textbooks and software use slightly different quartile methods. This course uses the "round up" approach. Make sure you use the method your professor specifies.
The quiz asked: what is the sum of the mean deviations of 3 and 5?
Mean deviation of X = 3: (3 − 5) = −2
Mean deviation of X = 5: (5 − 5) = 0
Sum = −2 + 0 = −2
This is a targeted calculation, not the total Σ(X − X̄). The total across all observations is always 0, but the sum of deviations for a subset of observations is generally not zero.
Sample variance: s² = [ΣX² − (ΣX)²/n] / (n − 1)
Sample standard deviation: s = √s²
Coefficient of variation: CV = (s / X̄) × 100%
Quartile positions: Q₁ position = 0.25n, Q₃ position = 0.75n (round up to find the data position)
IQR: Q₃ − Q₁
Mean deviation of an observation: (X − X̄)
In the water quality scenario from the quiz, a regulatory agency might set a threshold like "the standard deviation of suspended solids shall not exceed 3 ppm over any 10-day monitoring window." The CV allows comparison across monitoring stations that measure different pollutants at different scales. Quartiles are used to set alert thresholds: if Q₃ exceeds a safe limit, the station triggers an investigation even if the mean looks fine.
"Variance and standard deviation measure the same thing." They are related, but variance is in squared units and standard deviation is in the original units. You cannot interpret variance directly as "how far values are from the mean" because the units are wrong (e.g., ppm² has no physical meaning). Always take the square root to get the standard deviation if you want an interpretable spread measure.
"Dividing by n gives the sample variance." Dividing by n gives the population variance (σ²). For a sample, you must divide by (n − 1). Using the wrong denominator is a common exam error.
"A high standard deviation means the data is bad." High variability is a property of the data, not a quality judgement. Some phenomena are naturally more variable than others.
"The CV is always useful." The CV is meaningless when the mean is zero or near zero (you would be dividing by zero or near-zero). It also does not apply well to data measured on interval scales without a true zero point (e.g., temperature in Celsius).
⚠️ The variance formula s² = [ΣX² − (ΣX)²/n] / (n − 1) is almost guaranteed to appear on the exam. Know the difference between ΣX² (sum of each value squared) and (ΣX)² (square of the sum). Mixing these up is the most common calculation mistake.
⚠️ Always show the intermediate steps: compute ΣX and ΣX² first, substitute into the formula, then simplify. Partial credit depends on visible working.
⚠️ For quartile problems, use the method taught in class (0.25n, round up). Do not use a different method from another textbook or calculator.
⚠️ Remember that CV is expressed as a percentage. Forgetting to multiply by 100 or forgetting the % sign loses marks.
⚠️ The "sum of mean deviations" question tests whether you understand that individual deviations can be summed for a subset, even though the total across all observations is always zero.
True or false: ΣX² is the same as (ΣX)².
Fill in the blank: Sample variance divides by ______ instead of n, to correct for bias.
True or false: The standard deviation is always smaller than the variance.
Fill in the blank: If s = 6 and X̄ = 20, the coefficient of variation is ______%.
True or false: The IQR is resistant to outliers.
Answers: 1. False (ΣX² = 302, but (ΣX)² = 2500 in the quiz data). 2. (n − 1). 3. False (when s² < 1, the standard deviation is larger, e.g., s² = 0.25 gives s = 0.5). 4. 30%. 5. True.
Q: Using ΣX = 50, ΣX² = 302, and n = 10, compute the sample variance.
A: s² = [302 − (50)²/10] / (10 − 1) = [302 − 250] / 9 = 52/9 = 5.78.
Q: What is the sample standard deviation?
A: s = √5.78 = 2.40.
Q: Compute the coefficient of variation.
A: CV = (2.40 / 5) × 100% = 48%.
Q: Find Q₁ and Q₃ for the sorted data {2, 2, 3, 4, 4, 5, 6, 8, 8, 8}.
A: Q₁ position = 0.25(10) = 2.5, round up to 3rd value = 3. Q₃ position = 0.75(10) = 7.5, round up to 8th value = 8. So Q₁ = 3, Q₃ = 8.
Q: What is the sum of the mean deviations of X = 3 and X = 5, given X̄ = 5?
A: (3 − 5) + (5 − 5) = −2 + 0 = −2.
Q: Explain why the sample variance divides by (n − 1) rather than n.
A: Dividing by n would systematically underestimate the true population variance because the sample mean is calculated from the same data. Dividing by (n − 1) corrects this bias. The quantity (n − 1) is called the degrees of freedom.
Variance and standard deviation reappear throughout STAT 211. They form the basis of the normal distribution (the "bell curve" is defined by its mean and standard deviation), and they feed directly into confidence intervals and hypothesis tests. Quartiles connect to boxplots (a graphical tool for displaying spread and identifying outliers) and to non-parametric methods. The coefficient of variation comes up in experimental design when comparing measurement precision across different instruments or scales.
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