Two players on the same basketball team finished the season with identical scoring averages: 18.4 points per game. On paper, they look interchangeable. But ask their coach and you will hear two completely different players.
Player A scored between 15 and 22 points in nearly every game — a metronome of consistency. Player B once scored 45 points in a single game and followed it up with 6 points the next night, then 12, then 38, then 8. Same average. Wildly different performance.
The coach knows something the box-score average hides: consistency matters. A player you can count on every night is more valuable than one who oscillates between unstoppable and invisible — even if their season averages are identical. The missing piece of information is called spread, or variability: how far, on average, do individual performances stray from the centre?
In DS-3, you mastered the mean, median, and mode — measures that answer the question “What is a typical value?” In this lesson, you will answer the complementary question: “How typical is typical?” Two datasets can share the same centre and yet tell entirely different stories. The numbers that capture those differences — range, variance, standard deviation, quartiles, and the interquartile range — are the subject of this lesson.
After this lesson, you will be able to:
Compute the range of a dataset and explain why it is sensitive to extreme values
Calculate both population variance and sample variance , including Bessel’s correction ()
Compute and interpret the standard deviation in the original units of the data
Determine quartiles (Q1, Q2, Q3), the interquartile range (IQR), and the five-number summary
Detect potential outliers using the fence rule
Compute and interpret the coefficient of variation (CV) for comparing spread across different units or scales
Select the most appropriate measure of spread for a given context based on the presence of outliers and distribution shape
Section 2: Prerequisites
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Calculating measures of spread requires total fluency with the measures of center and distribution shapes you mastered in DS-3.
From DS-3: The Mean (). The average of all values. You will subtract this from every data point to compute the variance and standard deviation.
From DS-3: The Median. The middle value when data is sorted. This is the “center” used for the IQR and range.
From DS-3: Skewness. Right-skewed (tail to the right, mean > median) or left-skewed (tail to the left, mean < median).
Order of Operations (PEMDAS): You will be working with formulas like . Remember: subtract first (parentheses), then square (exponents), then sum (addition), and finally divide.
Retrieval Checkpoint
A dataset of 5 exam scores is: . What is the mean () of this dataset?
Retrieval Warm-up — from DS-2 and DS-3
A researcher posts the histogram of weekly running distances (km) for 200 recreational runners. The histogram has bars touching, a peak at the 30–40 km class, and tails of equal length on both sides. Which statement correctly interprets this histogram?
A professor records the exam scores for a class of 30 students and finds: and median . Which of the following is the most defensible conclusion?
Success Factor:
Where DS-4 diverges from DS-3: In DS-3, we summarized an entire distribution into a single “centre” value. In DS-4, we measure how far the data points spread out around that centre. The mean and median tell you where the data sit; the variance, standard deviation, and IQR tell you how tightly they cluster there. Both pieces of information are required to understand a dataset — one without the other is an incomplete description.
Section 3: Core Concepts
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Navigation tip: Six concepts build the complete vocabulary of spread. C1 (Range) is the simplest but least reliable. C2 (Variance) introduces the mathematical engine. C3 (Standard Deviation) makes it interpretable. C4 (Quartiles & IQR) gives you a resistant alternative. C5 (Outlier Detection) puts IQR to work. C6 (Coefficient of Variation) lets you compare spread across different units. After C4, you will apply everything with the Spread Explorer visualization.
C1 — Range
The range is the distance between the smallest and largest values. It is the simplest possible measure of spread — and the most fragile.
Intuitively: if you only know the minimum and maximum, you know the full width of the data but nothing about how the values are arranged inside that width. Two datasets with the same range can look completely different.
Range
where is the largest value and is the smallest value in the dataset.
Properties:
Uses only two values (the extremes) — ignores everything in between
Has the same units as the original data
Is highly sensitive to outliers: one extreme value can make the range arbitrarily large
Is always non-negative; zero means all values are identical
Worked mini-example:
Dataset A: → Range =
Now add one outlier — a value of 95:
Dataset B: → Range =
The majority of values still sit between 4 and 11, but the range now claims the data span 91 units. One extreme observation completely dominates the range. This is the fundamental trade-off: the range is easy to compute but catastrophically unreliable when outliers are present.
“The range tells you how spread out the typical values are.” The range tells you the distance between the two most extreme values — and nothing about the values in between. Two datasets can have identical ranges of 50 with entirely different internal structures: one might have all values clustered within 5 units of each other with two distant outliers, while another might have values uniformly spread across the full 50-unit span. The range cannot distinguish these cases.
Toggle to reveal what the range hides.
Both datasets have range = 60. Toggle between “range only” and “all values” to see how the range conceals radically different internal structures. Dataset A clusters five values within 6 units; Dataset B spreads them uniformly. The orange bracket looks identical for both.
C2 — Variance: Population vs. Sample
If the range is too crude (only two values), we need a measure that uses every data point. The natural idea: measure how far each value is from the centre, then average those distances.
The immediate problem: deviations are both positive and negative, and they always sum to exactly zero (the algebraic centre property from DS-3). Averaging them gives zero every time — useless.
The fix: square each deviation before averaging. Squaring makes every term non-negative, so they can’t cancel. The result is called the variance.
Variance
Population variance (when you have data for the entire population):
where is the population mean and is the population size. We write and read “sigma squared.”
Sample variance (when you have a sample and want to estimate the population variance):
where is the sample mean and is the sample size. We write and read “s squared.”
The denominator (instead of ) is called Bessel’s correction.
Why ? Here is the intuition: when you draw a sample, it rarely captures the most extreme values in the population. The typical sample is slightly less spread out than the full population it came from. If you divided by , your variance estimate would systematically understate the true population variance. Dividing by — a slightly smaller number — inflates the estimate just enough to correct for this built-in underestimation. The smaller the sample, the bigger the correction needed; the gap between and is largest when is small.
The formal name for what you “lose” is a degree of freedom. When you compute from your sample and then use it to compute deviations, the deviations are no longer all independent: once you know of them, the last is forced (it must make the sum zero). You have observations but only genuinely free pieces of information about spread. Both framings — “sample underestimates population spread” and “one degree of freedom is spent estimating the mean” — are two angles on the same underlying fact. The denominator is where that lost degree of freedom appears in the formula, and dividing by it makes an unbiased estimator of .
Worked mini-example:
Sample data: . First, .
Deviations: , , , .
Squared deviations: , , , .
Sum of squared deviations: .
If we had mistakenly divided by : . Bessel’s correction gives a larger (and unbiased) estimate.
Computational formula (often easier for hand calculation):
For the same data: , , .
Same result — no need to compute deviations individually.
“The sample variance formula uses , so it must mean we subtract 1 from the sample size in all calculations.” Bessel’s correction applies only to the denominator of the variance formula — and by extension to the standard deviation’s variance component. It does not mean you use when computing the mean ( still divides by ), when counting observations, or when computing the range. The is specific to variance estimation and the degrees-of-freedom logic that justifies it.
Drag any dot left or right. Each coloured square has area = (deviation)².
Drag any dot left or right. Each coloured square has side length = |xᵢ − x̄| and area = (xᵢ − x̄)². The variance is the average area of all squares. Toggle between population (÷N) and sample (÷N−1) to see how Bessel’s correction slightly inflates the estimate.
The chart auto-extends as you add draws — lines never vanish.
After ~50 draws with n=4: orange settles on its dashed target (below σ²); green tracks the true σ² line.
Click “Draw 25” to build up runs quickly. Two dashed targets are shown: the green target (true σ²) where ÷(n−1) converges, and the orange target (σ²×(n−1)/n) where ÷n converges — always below σ². With n=4 the bias is 25% and clearly visible; try n=10 to see how a larger sample shrinks but never eliminates the gap. The chart auto-extends so lines never disappear.
C3 — Standard Deviation
The variance has a units problem: if your data are measured in dollars, the variance is in dollars squared. If your data are in centimetres, the variance is in cm squared. These squared units are difficult to interpret. The fix is straightforward: take the square root.
Standard Deviation
Population standard deviation:
Sample standard deviation:
The standard deviation has the same units as the original data. It is the most widely reported measure of spread in statistical practice.
Continuing the mini-example:, so .
Interpretation: the typical observation deviates from the mean of 5 by about 2.58 units. This is directly interpretable because 2.58 is in the same units as the original data (unlike 6.67 “squared units”).
Connecting variance and standard deviation: Every time you see a variance, you can recover the standard deviation by taking the square root. Every time you see a standard deviation, you can recover the variance by squaring it. They are two representations of the same information — the variance in squared units, the standard deviation in original units.
“The variance and standard deviation are interchangeable — just use whichever is convenient.” While they encode the same information, they live on different scales. A variance of 25 means the standard deviation is 5, but a variance of 0.25 means the standard deviation is 0.5. When variance is less than 1, the standard deviation is actually larger than the variance. This trips up intuition: a “small” variance of 0.04 corresponds to a standard deviation of 0.2 — five times larger. Always check which quantity you are holding before comparing.
Each point sits inside one of three zones: green (±1s) — within one standard deviation of the mean; gold (±1s–2s) — between one and two standard deviations; orange (beyond ±2s) — unusually far from the centre. Drag any point outward and watch the green band widen, the SD value climb, and the proportion inside ±1s drop. Drag the orange dot back in to see it turn gold, then green.
C4 — Quartiles and the Interquartile Range (IQR)
The range is unreliable because it uses only the extremes. The variance and standard deviation use every value — but that means they, too, are sensitive to outliers (an extreme value contributes a massive squared deviation). We need a spread measure that is resistant — like the median is for centre.
The solution: focus on the middle 50% of the data. Ignore the lowest 25% and the highest 25%. The spread of the middle half is the interquartile range (IQR).
Quartiles and IQR
After sorting the data in ascending order:
Q1 (first quartile): the median of the lower half of the data (25th percentile)
Q2 (second quartile): the median of the full dataset (50th percentile)
Q3 (third quartile): the median of the upper half of the data (75th percentile)
The five-number summary is:
The IQR is resistant to outliers: changing the most extreme value (without crossing Q1 or Q3) does not change the IQR at all.
How to find quartiles (standard method):
Sort the data.
Find Q2 (the median of the entire dataset).
Find Q1 (the median of all values below Q2).
Find Q3 (the median of all values above Q2).
Worked mini-example: ()
Sorted already. Q2: position → Q2 = 15.
Lower half (values below Q2): (, even). Median of lower half: → Q1 = 7.5.
Upper half (values above Q2): (, even). Median of upper half: → Q3 = 23.5.
Five-number summary:.
The IQR is the robust counterpart to standard deviation. When your data are symmetric with no outliers, use the standard deviation (it uses more information). When your data are skewed or have outliers, use the IQR (it is resistant). This parallels the mean/median decision rule from DS-3 exactly.
“The median of the lower half is always a data value.” Not necessarily. In the example above, Q1 = 7.5 is not in the dataset. Quartiles — like the median — are positional summaries that may fall between data values. This is especially common when the halves have an even number of elements.
“Q1 is the value at position and Q3 is at position .” This interpolation method is one of several conventions for computing quartiles. The “median of halves” method described above is the most intuitive and is the standard approach in introductory statistics (Tukey’s method). Different software packages use different conventions, which can produce slightly different quartile values for the same dataset. For this course, use the median-of-halves method.
“IQR, variance, and range all measure spread — they’re just different computations of the same underlying thing.” They measure fundamentally different aspects of spread and have completely different resistance properties. The IQR is resistant to outliers — it ignores the top and bottom 25% of the data entirely, so extreme values have zero influence on it. The variance and standard deviation are sensitive to every observation, including outliers, because squaring deviations gives extreme values disproportionate weight. The range is the most extreme — a single outlier can make it arbitrarily large. Choosing the IQR for skewed data (parallel to choosing the median for centre) and the standard deviation for symmetric data (parallel to choosing the mean) is not a matter of taste — it is essential for an accurate description of spread.
Press Next to begin.
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Step through the six-stage box plot construction: sorted data → Q2 (median) → Q1 and Q3 → draw the box → add whiskers → outlier demo. Step 6 replaces the value 30 with 95, showing the whisker stop at the fence and the outlier plotted as a separate circle.
Spread Explorer — See All Four Measures at Once
The visualization below lets you experience how range, IQR, variance, and standard deviation respond when data points move. Drag any dot horizontally. Watch what happens to each measure. Toggle between population () and sample () formulas to see Bessel’s correction in action.
Interactive: Drag data points to see how range, IQR, variance, and standard deviation update in real time. The orange band shows the range; the blue box spans Q1 to Q3 (IQR); the green marker indicates the standard deviation; the purple bar represents the variance. Toggle between population and sample formulas using the control above.
C5 — Outlier Detection via Fences
Detecting outliers is not just a matter of eyeballing the data — we need a systematic rule. The most common criterion uses the IQR and is called the fence method.
Outlier Detection — Fence Method
Lower fence:
Upper fence:
Any value below the lower fence is a potential outlier.
Any value above the upper fence is a potential outlier.
Values between the fences are considered within the “expected” range.
The multiplier is a convention — it corresponds roughly to standard deviations in a normal distribution, meaning about 0.7% of observations in a normal population would fall outside the fences by chance alone.
All values in are between and . No outliers detected.
Now modify the dataset: replace 30 with 95. The new data: .
Recompute: Q2 stays 15. Lower half still gives Q1 = 7.5. Upper half is now → median of upper half: → Q3 = 23.5.
IQR still 16. Fences: still and .
The value 95 exceeds the upper fence of 47.5 → 95 is flagged as a potential outlier.
“Potential” is important. The fence rule flags values that are unusually far from the bulk of the data. It does not automatically mean the value is an error or should be discarded. An outlier might be a data-entry mistake, a measurement error, or a genuinely unusual but valid observation. Always investigate before deciding to remove or retain.
“If the lower fence is negative but my data cannot be negative (e.g., counts, prices, times), I should set the lower fence to 0.” The fence is a mathematical rule applied uniformly. A negative lower fence simply means no values are low enough to be flagged as outliers on the low end — which makes sense for bounded data. Do not adjust the fence value; interpret what it tells you about the data.
Drag the dot to test different values.
Typical value: 1.5× (flags ≈0.7% of normal data)
Drag the dot to reposition the new observation. It turns red the moment it crosses a fence. Use the multiplier slider to explore how different thresholds (0.5×, 1.0×, 1.5×, 2.0×, 3.0× IQR) change what counts as an outlier — notice that 1.5× is a convention, not a physical law.
C6 — Coefficient of Variation (CV)
All the spread measures so far — range, variance, standard deviation, IQR — have units tied to the original data. This makes it impossible to compare spread across datasets measured in different units. Is a standard deviation of 10 cm “more spread” than a standard deviation of 8 kg? The question is meaningless without a common scale.
The coefficient of variation solves this by expressing the standard deviation as a percentage of the mean:
Coefficient of Variation
where is the sample standard deviation and is the sample mean.
CV is unitless — it is a ratio, so units cancel.
CV measures spread relative to the centre rather than in absolute terms.
CV is meaningful only for ratio-scale data (data with a true zero, where ratios make sense). It is not appropriate for interval-scale data (like temperature in Celsius) because the zero is arbitrary.
Worked mini-example: Two biology students measure different traits of the same plant species.
Student A measures leaf lengths: cm, cm.
Student B measures leaf masses: g, g.
Even though the standard deviations are in completely different units (cm vs. g), the CVs are comparable. Leaf mass (CV = 36.5%) is relatively more variable than leaf length (CV = 22.6%) — by a factor of about .
CV is especially useful for:
Comparing variability across variables with different units (height vs. weight)
Comparing variability across the same variable measured on different scales (income in Canada vs. income in India)
Quality control: a CV below 10% often indicates a stable manufacturing process
“A larger CV always means the data are more spread out.” The CV combines information about both spread () and centre (). A dataset with a huge standard deviation can have a modest CV if its mean is also huge. Conversely, a dataset with a tiny standard deviation can have a large CV if its mean is near zero. Always check both components before interpreting the CV.
CVA =—
CVB =—
Dataset A
Dataset B
Try dragging x̄ close to zero — watch CV explode. A small mean makes any SD look enormous relative to centre.
Use the sliders to adjust the mean and SD for each dataset independently. Try dragging the mean of Dataset A toward 5 — watch CV skyrocket even with the same SD. Then increase the SD while holding the mean constant and notice the proportional response. The highlighted value is the relatively more variable dataset.
Section 4: Worked Examples
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Four examples with progressively less scaffolding. Example 1 is fully narrated so you can see the complete thought process. By Example 4, you are doing the thinking and catching errors.
Example 1 — Computing the Range and Recognizing Its Limitations
Scenario: A real-estate agent records the sale prices (in thousands of dollars) of 10 homes in a neighbourhood:
285, 310, 295, 305, 290, 315, 300, 308, 292, 302
Find the range and discuss what it reveals about the housing market.
Step 1: Identify the minimum and maximum.
Sorting makes this easiest — but for range, you only need to find the two extremes.
Scanning the list: the smallest value is 285, the largest is 315.
Step 2: Compute the range.
The sale prices span $30,000 from lowest to highest.
Step 3: Interpret in context.
All 10 homes sold between $285K and $315K — a narrow span of $30K. The range suggests a homogeneous neighbourhood where homes are similarly valued. A buyer looking here can expect prices in this tight band.
Sanity check: Are the data clustered? Most values sit between 290 and 315 — the minimum of 285 is not an extreme outlier, just the lowest-priced home in a consistent market. The range of 30K is a meaningful description here because there are no extreme outliers.
What would change with an outlier? Suppose one home sold for $850K (a fully renovated heritage property). The range would jump from 30 to — nearly 19 times larger. The single $850K sale would dominate the range, making it say “prices span $565K” when 9 of 10 homes are still within a $30K band. This is why range is unreliable when outliers are present.
Metacognitive narration: “I notice the values are all within a narrow band — nothing looks like an outlier — so I conclude the range is a reasonable description of spread here. If I saw a value far from the cluster, I would compute the range but also flag it as potentially misleading, and supplement with IQR or standard deviation.”
Example 2 — Computing Sample Variance and Standard Deviation
Scenario: A quality-control technician measures the diameter (mm) of 6 ball bearings from a production run:
5.1, 4.8, 5.3, 5.0, 4.9, 5.5
Compute the sample variance and sample standard deviation . Round to 2 decimal places.
Before computing: the values range from 4.8 to 5.5 (span = 0.7). The mean should be near 5.1. Do you expect the standard deviation to be closer to 0.2 or to 0.5? Why?
Step 1: Compute the sample mean.
The mean is 5.1 mm. (This is exactly one of the data values — a coincidence.)
Step 2: Compute deviations and their squares.
5.1
0.0
0.00
4.8
−0.3
0.09
5.3
+0.2
0.04
5.0
−0.1
0.01
4.9
−0.2
0.04
5.5
+0.4
0.16
Check: deviations sum to ✓.
Step 3: Sum the squared deviations.
Step 4: Divide by .
Step 5: Take the square root for .
Interpretation: The sample variance is mm², and the sample standard deviation is mm. The typical bearing diameter deviates from the mean of 5.1 mm by about 0.26 mm. This small standard deviation (relative to the mean) indicates a consistent manufacturing process.
Answer to prediction: The standard deviation is 0.26. The range (0.7) overestimates typical spread because it’s driven by the two extreme values. The standard deviation, averaging across all observations, gives a more moderate (and usually more accurate) measure of typical deviation. The prediction “closer to 0.2” was correct — the range divided by 4 (a rough heuristic for SD in small samples) gives , and the actual SD of 0.26 is in that ballpark.
Example 3 — Five-Number Summary and IQR
Scenario: A dietician records daily sodium intake (mg) for 11 patients:
(odd). Median position: . The 3rd value: 1950. → Q1 = 1950 mg.
Step 5: Find Q3 (median of the upper half).
Upper half (values above Q2): .
(odd). Median position: 3. The 3rd value: 2400. → Q3 = 2400 mg.
Step 6: Compute the IQR.
Five-number summary: (all in mg).
Step 7: Outlier check.
Lower fence: .
Upper fence: .
All 11 values fall between 1275 and 3075. No outliers detected.
Interpretation: The middle 50% of patients consume between 1950 mg and 2400 mg of sodium daily — an IQR of 450 mg. The full dataset spans from 1750 mg to 2600 mg. The distribution appears roughly symmetric (Q2 is exactly centred between Q1 and Q3: and ).
Example 4 — Outlier Detection — Find the Error
Scenario: An analyst computes quartiles for the monthly sales ($ thousands) of a retail store over the past 12 months:
22, 25, 28, 24, 23, 55, 26, 27, 30, 24, 29, 31
The analyst presents the following analysis. Find the error.
Analyst’s work:
Sort: ().
With , the median position is the 6th value: Q2 = 26.
Lower half (below Q2): . Q1 = 24 (the 3rd value).
Upper half (above Q2): . Q3 = .
IQR = .
Lower fence: .
Upper fence: .
The value 55 exceeds the upper fence of 37.75, so 55 is an outlier. It should be removed from the dataset.
Task: Identify the two errors in the analyst’s work — one computational, one interpretive.
Show Solution — Errors Identified
Error 1 — Q2 method for even : For (even), Q2 is the average of the 6th and 7th values, not the 6th value alone. The analyst treated even like odd .
Correct Q2: .
This changes the lower half from (5 values) to (6 values), and consequently changes Q1:
Correct IQR: (not 5.5). Correct fences:
The value 55 still exceeds 37.0, so it is still flagged as a potential outlier. Here the computational error did not change the outlier detection result — but for a borderline value (say, 37.5), the analyst’s wrong upper fence of 37.75 would have failed to flag it while the correct fence of 37.0 would have.
Error 2 — Automatic removal: “It should be removed from the dataset.” The fence method flags values for investigation, not deletion. The value 55 might represent a legitimate exceptional month — a holiday-season surge, a large contract, or a promotional event. Removing it without a substantive reason would discard valid information.
Correct conclusion: “The value 55 is flagged as a potential outlier (it exceeds the correct upper fence of 37.0). It should be investigated — is it a data-entry error, or a genuinely exceptional month? If the value is valid, retain it and note it in the analysis.”
Key takeaway: The fence method detects potential outliers — it does not license automatic deletion. Statistical rules flag values for human investigation. The decision to remove or retain is a substantive judgment, not a mechanical one.
Section 5: Guided Practice
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Work through each problem with immediate feedback. Every distractor names the specific error you would make by choosing it — use these rationales to catch misunderstandings before they solidify.
Problem 1 — Range and Outlier Sensitivity (C1)
An ecologist counts the number of bird species observed at 8 wetland sites during a morning survey:
12, 8, 15, 10, 9, 14, 7, 11
Question 1a: What is for this dataset?
Question 1b: What is the range?
Question 1c: Later, the ecologist discovers a 9th site with 37 bird species. What is the new range?
Notice how one new observation changed the range from 8 to 30 — nearly a 4× increase. This is the range’s fragility in action: a single extreme value dominates it completely.
Problem 2 — Sample Variance and Standard Deviation (C2)
A new scenario each time. Apply the formula and take the square root.
Five tomato plants produce fruit with the following counts: 8, 12, 9, 11, 10.
(a) What is ?
(b) What is ?
(c) What is the sample variance (to 1 decimal place)?
(d) What is the sample standard deviation (to 2 decimal places)?
Six runners log their times (seconds) for a 100 m sprint: 12.1, 11.8, 12.5, 12.0, 11.6, 12.4.
(a) What is ?
(b) What is (to 2 decimal places)?
(c) What is (to 3 decimal places)?
(d) What is (to 2 decimal places)?
A café records daily pastry sales over 7 days: 24, 30, 22, 28, 26, 20, 25.
(a) What is ?
(b) What is ?
(c) What is (to 2 decimal places)?
(d) What is (to 2 decimal places)?
Eight containers hold the following liquid volumes (mL): 250, 248, 253, 251, 249, 252, 247, 250.
(a) What is ?
(b) What is ?
(c) What is ?
(d) What is ?
Four packages weigh (kg): 3.2, 3.8, 3.5, 3.1.
(a) What is ?
(b) What is ?
(c) What is ?
(d) What is (to 2 decimal places)?
Problem 3 — Five-Number Summary and IQR Method (C4, C5)
A fresh dataset each time. Sort, find the five-number summary, compute IQR and fences, and identify any outliers.
Ten students’ scores on a statistics quiz (out of 20): 14, 8, 17, 11, 15, 9, 13, 18, 12, 16.
(a) After sorting, what is the five-number summary?
(b) What is the IQR?
(c) Are there any outliers?
Seven delivery trucks travel the following distances (km): 45, 72, 38, 61, 83, 49, 67.
(a) After sorting, what is the five-number summary?
(b) What is the IQR?
(c) Are there any outliers?
Seven pharmacies report the following prescription counts per day: 142, 98, 175, 115, 160, 88, 205.
(a) After sorting, what is the five-number summary?
(b) What is the IQR?
(c) Are there any outliers?
Eight car batteries last these numbers of months: 36, 48, 30, 42, 54, 24, 60, 38.
(a) After sorting, what is the five-number summary?
(a) After sorting, what is the five-number summary?
(b) Are there any outliers?
Problem 4 — Coefficient of Variation (C6)
Each variant gives summary statistics for two variables measured in different units. Compute the CV for each variable and determine which is relatively more variable. Recall: .
A sports scientist measures two traits in a squad of 8 athletes: sprint speed ( m/s, m/s) and vertical jump height ( cm, cm).
(a) What is the CV for sprint speed (to 1 decimal place)?
(b) The CV for jump height is approximately 13.1%. Which trait is relatively more variable, and what does this comparison tell the scientist?
A nutritionist compares two nutrients across 10 protein bars: protein content ( g, g) and sodium content ( mg, mg).
(a) What is the CV for protein content (to 1 decimal place)?
(b) The CV for sodium is approximately 22.7%. Which nutrient is relatively more variable across bars?
A geologist records measurements at 9 river stations: water depth ( m, m) and flow rate ( L/s, L/s).
(a) What is the CV for flow rate (to 1 decimal place)?
(b) Water depth has CV ≈ 25.0%. Which measurement is relatively more variable?
An HR analyst reports summary statistics for 12 software engineers: years of experience ( years, years) and salary ( K, K).
(a) What is the CV for years of experience (to 1 decimal place)?
(b) Salary has CV ≈ 20.0%. Which is relatively more variable, and what practical insight does this give HR?
A pharmacist tests 8 batches of a compound: purity (, ) and tablet mass ( mg, mg).
(a) What is the CV for purity (to 2 decimal places)?
(b) Tablet mass has CV = 2.50%. Which property is relatively more variable, and what does this mean for quality control?
Section 6: Independent Practice
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Work through these problems without guided steps. Commit to an answer before revealing the solution. Concepts are interleaved across all six measures of spread.
Problem 1 — Compute Sample Variance and Standard Deviation
Problem 2 — Find the Five-Number Summary and IQR
Problem 3 — Identify Outliers Using the Fence Method
Problem 4 — Range vs. Variance: Same Range, Different Spread
Problem 5 — Choosing the Right Spread Measure for the Context
A different scenario each time. No computation — apply the decision framework from Section 3. Match the measure to the distribution shape and the question being asked.
A financial advisor compares the consistency of returns for two mutual funds over 10 years. Both fund returns are approximately symmetric and bell-shaped with no outliers. The advisor wants to report the single most informative measure of consistency.
Which spread measure should be used?
An economist studies household incomes in a region. The distribution is strongly right-skewed: most households earn $30K–$70K, but a small group of very high earners pulls the right tail past $500K.
Which measure best describes the typical spread of incomes?
A biologist compares the variability of body length (cm) and body mass (g) in a sample of 50 lizards. Both distributions are approximately symmetric. The biologist asks: “Which trait is relatively more variable?”
Which measure enables this comparison?
A hospital monitors waiting times in the emergency department (minutes). The data are strongly right-skewed: most patients wait 20–60 minutes, but a few wait 4–6 hours. Hospital administration reports the mean and standard deviation. Is this appropriate?
What should they report instead?
A quality engineer compares two production lines that manufacture the same component. Line 1 produces parts with a target length of 10 mm; Line 2 produces parts with a target length of 50 mm. Both lines produce symmetric, normally distributed measurements. The engineer wants to know which line is more consistent.
Which measure should be used?
Problem 6 — Find the Error in Spread Measure Computations
A student computes the sample standard deviation for the following weekly study hours of 5 classmates: 8, 6, 10, 4, 12.
Student’s work:
hours.
Conclusion: The average study time is 8 hours, with a variance of 8 hours² and standard deviation of 2.83 hours.
What error did the student make?
An analyst finds the five-number summary for daily customer counts at a store over 10 days: 34, 28, 41, 23, 37, 45, 30, 39, 26, 42.
Analyst’s work:
Sort: 23, 26, 28, 30, 34, 37, 39, 41, 42, 45.
Q1 = 28 (the 3rd value, approximately 25% through). Q2 = (34+37)/2 = 35.5. Q3 = 41 (the 8th value, approximately 75% through).
A lab technician measures the mass (g) of 6 chemical samples: 5.12, 5.08, 5.15, 5.11, 5.09, 5.14.
Technician’s report:
. Range = 5.15 − 5.08 = 0.07 g.
”The range of 0.07 g indicates excellent consistency in the process — all samples are within 0.07 g of each other. No further spread analysis is needed.”
What error is present?
A student computes spread for these exam scores: 65, 70, 75, 80, 85, 90, 95.
Student’s work:
. Deviations: −15, −10, −5, 0, +5, +10, +15.
.
Standard deviation = 10 points.
What error did the student make?
An HR analyst compares salaries (in $K) in two departments. Department A: 42, 48, 45, 52, 47, 44, 50, 46. Department B: 38, 42, 55, 40, 95, 44, 41, 39.
Analyst’s conclusion: “Department A range = 52 − 42 = 10. Department B range = 95 − 38 = 57. Department B is much more spread out, indicating pay inequity.”
What error is present?
Problem 7 — Multi-Step Synthesis
A forestry researcher measures the diameter at breast height (cm) of 9 maple trees in a study plot:
28, 34, 22, 31, 45, 26, 38, 24, 29
(a) Compute the mean , the sample variance , and the sample standard deviation . Round to 2 decimal places.
(b) Find the five-number summary, the IQR, and use the fence method to identify any potential outliers.
(c) The researcher adds a 10th tree with a diameter of 68 cm — a particularly old specimen. Recompute the standard deviation and IQR. Which measure changed more, proportionally? What does this tell you about their relative resistance to outliers?
Show Solution
(a) Sort the data first (useful for both parts): .
. . cm.
Deviations from : , , , , , , , , .
Squared deviations: .
.
cm².
cm.
Note on rounding: The exact mean is cm. Using 30.78 introduces small rounding errors that accumulate through the squared-deviation steps. If your answer is between 7.20 and 7.30, your calculation is correct — the discrepancy is rounding, not a formula error. Working with the exact fraction throughout eliminates this.
(b) Five-number summary from the sorted list:
Q2 (median): position → .
Lower half (below Q2): . (even). Q1 = .
Upper half (above Q2): . (even). Q3 = .
.
Five-number summary: .
IQR = cm.
Fences: Lower = . Upper = .
All values are between 8.5 and 52.5. No outliers detected.
Conclusion: Adding one outlier (68 cm) nearly doubled the standard deviation () but barely moved the IQR (). This demonstrates that the standard deviation, like the mean, is sensitive to outliers — it squares deviations, so extreme values have disproportionate influence. The IQR is resistant: it ignores the tails entirely and focuses on the middle 50%, so extreme values outside Q1–Q3 have zero effect on it.
Mixed Review — Retrieval from Earlier Lessons
These problems draw on concepts from DS-2 and DS-3. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.
Review Problem 1 — Central Tendency in a Skewed Distribution (DS-3)
A real-estate agency reports the following data for home sale prices in a suburban neighbourhood over the past year: 8 homes sold between $280K and $350K, 3 homes sold between $350K and $450K, and 1 home sold for $1.2M (a luxury waterfront property).
(a) Without computing exact values, explain which measure of central tendency — mean or median — better represents what a typical buyer paid in this neighbourhood. Justify your choice using the concept of outlier sensitivity from DS-3.
(b) The agency’s advertisement states: “Average home in our neighbourhood costs $412,000.” A buyer later finds that 10 of the 12 sales were under $400,000. Explain what statistical feature of the data allowed this advertisement to be technically accurate yet practically misleading.
Show Solution
(a) Median is the better measure. The $1.2M luxury sale is an extreme outlier on the right tail of the distribution. The mean uses every value’s magnitude: the luxury sale contributes the same weight as any other observation, but its value is roughly 3–4 times larger than the cluster of typical sales. This pulls the mean upward, far above where most buyers actually transacted. The median, by contrast, is resistant to outliers — it simply locates the middle of the sorted list and is unaffected by exactly how extreme the luxury sale is. For 12 observations (even ), the median would be the average of the 6th and 7th sorted values, both of which sit in the $280K–$450K range, giving an accurate picture of typical market prices.
(b) The mean is sensitive to extreme values (right-skewed distribution). The $1.2M sale is far enough above the other 11 sales that it pulls the mean to $412K — above even the highest of the typical sales ($350K–$450K cluster). The advertisement uses “average” (mean) without disclosing that the distribution is right-skewed, which is the condition where the mean systematically overstates the typical value. A buyer who sees “$412K average” and assumes that describes a typical transaction will have a distorted expectation. The honest summary would report the median (likely near $320K–$340K) and note the one outlier sale.
Review Problem 2 — Histogram Shape and Measure Selection (DS-2 and DS-3)
A student creates a histogram of the number of hours per week a sample of 60 university students spend on social media. The histogram shows:
A large bar at 1–5 hours and 6–10 hours
A moderate bar at 11–15 hours
Small bars at 16–20 hours and 21–25 hours
A single small bar at 41–45 hours (three students who self-identify as “heavy users”)
(a) Describe the shape of this distribution. Name the direction of skew and identify which specific feature of the histogram supports your answer.
(b) Given this shape, state which measure of centre — mean or median — the student should report and explain exactly why the other measure would misrepresent the data.
(c) The student computes a mean of 11.2 hours and a median of 8.5 hours. Are these values consistent with the shape you described? Explain briefly.
Show Solution
(a) Right-skewed distribution. The distribution has most of its data clustered at the low end (1–15 hours per week), with the frequency declining as hours increase, and a very small bar appearing far to the right at 41–45 hours. The long tail extends to the right (toward higher values), which is the defining characteristic of right skew. The three heavy users at 41–45 hours are the extreme values in the right tail.
(b) The median should be reported. The three students at 41–45 hours are outliers that pull the mean substantially upward. Because the mean uses every value’s magnitude, the 41–45 hour values contribute disproportionate influence — each of those students contributes a deviation of roughly 30–35 hours above the typical student, while the 57 other students contribute much smaller deviations. The result is a mean that sits above where the vast majority of students actually fall.
The median is resistant: it finds the middle position of the sorted list (position 30 or 30.5 for ), which will fall in the 6–10 or 11–15 hour range — accurately capturing what a typical student does. The three outliers cannot move the median significantly, regardless of whether they’re at 41 hours or 141 hours.
(c) Yes, these values are consistent. Mean (11.2) > Median (8.5) is exactly what we expect in a right-skewed distribution — the mean is pulled to the right of the median by the heavy-user outliers. The gap of 2.7 hours between the two measures is evidence of the skewness: in a symmetric distribution, they would be approximately equal.
Section 7: Mastery Check
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No hints. No guided steps. These three items measure whether the core ideas have actually landed. Take your time with the Feynman test — explaining a concept in plain language is the truest test of understanding.
Question 1 — Feynman Test (C2)
A friend taking a different statistics course just learned the formula for sample variance and is confused: “Wait — why are we dividing by instead of ? If I have 5 numbers, shouldn’t I average the squared deviations by dividing by 5? This thing feels like someone made a mistake and statisticians just ran with it.”
Explain why we use in the sample variance formula. Your explanation must be in plain language — no formulas, no jargon without defining it first. Aim for 100–200 words.
0 / 200 words
Show a Model Answer
Model answer: “When you have the entire population, you know the true centre, so dividing by N makes sense — that’s the actual average squared deviation. But with a sample, you have to estimate the centre using the sample average, which is always pulled slightly toward your sample values. This means your sample deviations tend to be just a bit smaller than the true deviations would be. Dividing by a slightly smaller number (n−1 instead of n) inflates the variance just enough to correct for this built-in underestimation. Think of it as a statistical honesty tax: since your sample is almost certainly less variable than the population it came from, you compensate by making the variance a touch larger. The smaller the sample, the bigger the correction — with 5 values, dividing by 4 really matters; with 500 values, dividing by 499 barely changes anything.”
Question 2 — Apply (Selecting Measures)
An urban planner is studying neighbourhoods in two cities. City A has house prices that are approximately symmetric and bell-shaped, with no outliers. City B has house prices that are strongly right-skewed: most homes sell for $200K–$400K, but a small cluster of luxury properties sells for $1M–$3M, forming a long right tail.
(a) For City A (symmetric, no outliers), which spread measure is most appropriate?
(b) For City B (right-skewed with luxury outliers), which spread measure is most appropriate?
(c) Briefly explain the principle behind your choices.
Show a Model Answer
The choice of spread measure mirrors the choice of centre measure from DS-3. When data are symmetric with no outliers, the standard deviation is preferred because it uses every observation and is directly connected to the mean through the concept of “average squared deviation.”
When data are skewed or contain outliers, the standard deviation becomes unreliable — extreme values contribute disproportionately large squared deviations, inflating the SD beyond what describes the typical data. The IQR is the resistant counterpart: it focuses on the middle 50% of data and ignores the tails entirely, so extreme values have zero influence.
The principle: measure of spread should match the measure of centre in terms of resistance. Mean + SD for symmetric data; Median + IQR for skewed data or data with outliers.
Question 3 — Analyze (Variance vs. Standard Deviation)
Two students analyze the same dataset — daily temperatures (°C) for a week: .
Student 1 reports: “The mean is 21°C and the variance is 4.33°C², so the standard deviation is also 4.33°C.”
Student 2 reports: “The mean is 21°C and the variance is 4.33°C², so the standard deviation is °C.”
Which student made the error, and what is the specific misconception?
Self-Assessment
How confident are you with the material in this lesson?
Still unsureFully confident
Section 8: Boss Fight
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Boss Fight — Pick Your Path. Both paths require the same statistical knowledge. The difference is in how you apply it: the Analyst works with real data to reach a conclusion; the Architect designs a measurement strategy from scratch. Choose the path that fits your thinking style — you can attempt the other afterward for extra practice.
🔬 Path A: The Analyst — Departmental Salary Equity
A company wants to know whether pay is more equitable in Engineering or Marketing. You have the raw salary data. Compute all spread measures and advise the HR director.
🏗️ Path B: The Architect — Quality-Control Study Design
A pharmaceutical company needs to monitor pill-weight consistency. No data exists yet. Design the spread-measurement protocol, choose outlier thresholds, and justify every decision.
🔬 Path A: The Analyst
A mid-sized company’s HR director wants to assess pay equity across departments. She provides the annual salaries (in thousands of CAD) for all non-managerial employees in two departments:
Engineering Department (12 employees):
62, 58, 71, 65, 68, 60, 64, 70, 67, 63, 66, 59
Marketing Department (10 employees):
48, 52, 55, 50, 95, 53, 47, 51, 54, 49
Task A1 — Compute the mean salary for each department. What do you notice?
Show Answer
Engineering: , , (thousands, so $64,420).
Marketing: , , (thousands, so $55,400).
The means differ by about $9K. Engineering has the higher average salary.
Task A2 — Compute the range and standard deviation for each department. Round SD to 2 decimal places.
Show Answer
Engineering: Range = . Deviations from : sum of squared deviations = . Sum . . (thousands, so $4,460).
Marketing: Range = . (Notice: nearly 4× Engineering’s range.)
. Deviations: . Sum . . (thousands, so $14,150).
The standard deviation in Marketing ($14,150) is more than 3× Engineering’s ($4,460).
Task A3 — Compute the five-number summary and IQR for each department.
Task A4 — Identify outliers in each department using the fence method.
Show Answer
Engineering: Lower fence = . Upper fence = . All values in [58, 71] are within [51.25, 77.25]. No outliers.
Marketing: Lower fence = . Upper fence = .
Value 95 > 61.5 → 95 is a potential outlier. All other values (47–55) are within the fences.
Task A5 — Synthesize and advise. The HR director asks you: “Which department has more equitable pay? What numbers support your conclusion?” Write your recommendation using the statistics you computed. Address both departments and explain why you chose the measures you did.
Show a Model Response
Engineering has more equitable pay. All four spread measures agree:
Range: Engineering $13K vs. Marketing $48K — but Marketing’s range is inflated by the $95K outlier.
SD: Engineering $4.46K vs. Marketing $14.15K — a 3.2× difference, largely driven by the outlier’s massive squared deviation.
IQR: Engineering $6.5K vs. Marketing $5K — the middle 50% spread is actually slightly smaller in Marketing.
Outliers: Marketing has one (the $95K salary), Engineering has none.
The IQR tells the most nuanced story: when we strip away the one extreme Marketing salary, the pay dispersion in Marketing is actually marginally tighter than in Engineering. The range and SD gave a misleadingly dramatic picture because they were dominated by a single data point. This is exactly why the IQR exists — to provide a spread measure that is not commandeered by outliers.
Recommendation: The $95K salary in Marketing should be investigated. Is it a data-entry error? A legacy contract? A role misclassified as non-managerial? Once clarified, re-run the analysis. If the $95K is valid, report the IQR alongside the median for Marketing to give the fairest picture. Engineering’s pay structure is consistent and equitable; no red flags.
Reflection: The HR director now understands that “equity” can be measured in different ways. How would you explain to a non-technical colleague why the range and SD sometimes tell a different story than the IQR? Write 2-3 sentences in the space below.
0 / 600
🏗️ Path B: The Architect
PharmaCorp manufactures a blood-pressure medication in tablet form. Each tablet must contain 50 mg of the active ingredient. The Quality Control (QC) team is designing a new monitoring protocol. Currently, tablets are weighed individually as a proxy for active-ingredient content, but the team knows that pill weight varies even when the active-ingredient dose is perfect.
You have been brought in as a statistical consultant to design the spread-measurement component of the protocol.
Task B1 — Identify the variable and its type. The QC team will measure the mass of each sampled tablet in milligrams (mg) using a high-precision scale. Which variable type is “tablet mass in mg”?
Task B2 — Choose the primary spread measure. The QC team will sample tablets per batch. Previous data suggest the masses are approximately normally distributed (bell-shaped, symmetric) with no outliers. Which spread measure should be the primary metric in the protocol?
Task B3 — Set outlier detection thresholds. The team asks you to define what constitutes an “unusual” tablet mass so they can flag individual tablets for investigation. You recommend the fence method. What information do you need to compute the fences, and what multiplier do you use?
Task B4 — Design a comparative spread analysis. The QC team also produces a 100 mg version of the same medication. The team wants to know: is the manufacturing process equally consistent for both strengths? Both products produce data that are symmetric and approximately normal.
(a) Which measure would allow a fair comparison of consistency between the 50 mg and 100 mg tablets?
(b) Suppose the 50 mg line has mg with mg, and the 100 mg line has mg with mg. Compute the CV for each and interpret.
Show Answer
The 100 mg line has a smaller CV (2.39% vs. 3.59%), meaning relative variability is lower for the higher-strength tablets. Even though the absolute SD is larger (2.4 > 1.8 mg), the process is actually more consistent relative to the target mass. This is exactly the insight CV was designed to surface.
Reflection: You now have a complete spread-measurement framework. How would you summarize your protocol in one paragraph for the QC team lead? Include which measures you chose, why, and what they should do when the protocol flags a tablet. Write below.
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Section 9: Challenge Problems
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Optional — stretch beyond the lesson objectives. These problems require connecting concepts or applying mathematical reasoning. They are not required to complete the lesson.
Challenge 1 — Proving and Connecting It to
In DS-3, you learned that the deviations from the mean sum to zero: . This was presented as an algebraic fact. But have you actually proved it?
Claim: For any set of numbers , define . Then .
Proof:
Since is constant across the sum: .
Connection to : Because , the deviations are not independent — once you know of them, the last is forced (it must make the sum exactly zero). This is the mathematical reason we lose one degree of freedom when estimating variance from a sample. The denominator (rather than ) reflects the number of genuinely independent pieces of information in the deviations.
Verify for a concrete dataset
Take . . Deviations: . Sum = 0 ✓.
If we only know the first two deviations ( and ), the third must be to make the sum zero. We don’t have three independent pieces of information — we have two. Hence degrees of freedom.
Extension: What is the minimum number of values for which can be computed? (Hint: if , what is ?)
Show Answer
If , then and the variance formula involves division by zero — undefined. You cannot measure spread with a single observation. The minimum sample size for computing sample variance is (giving degree of freedom).
This makes intuitive sense: with one value, you have a location but no information about how spread out other values might be. Variability is inherently a property of at least two observations.
Challenge 2 — Comparing Spread Across Different Units with CV
An agricultural researcher measures two traits for the same 12 corn plants:
Plant height (cm): ,
Ear mass (g): ,
(a) Compute the CV for each trait. Which trait shows greater relative variability?
Show Answer
Ear mass (CV = 19.61%) shows greater relative variability than plant height (CV = 15.33%). Even though the absolute SD of mass (42.1 g) cannot be directly compared to the SD of height (28.4 cm), the CV tells us mass is about 28% more variable relative to its mean.
(b) The researcher repeats the study on a second farm with different growing conditions. Choose which is more variable and explain.
Farm B: Height: cm, cm. Mass: g, g.
Which trait is relatively more variable at Farm B, and how does it compare to Farm A?
Farm C: Height: cm, cm. Mass: g, g.
Which farm (A, B, or C) has the most consistent (least variable) plant height, as measured by CV?
Farm D: The researcher measures height in metres instead of cm: m, m. Compute CV and compare to Farm A’s height CV. What do you notice about the effect of changing units on CV?
Section 10: Solutions Reference
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Complete, step-by-step solutions for all problems in Sections 5–9 are available on the solutions page. Solutions include worked arithmetic, common mistakes to watch for, and interpretation guidance for every problem.
If you’re stuck: Re-read the relevant Core Concept in Section 3, then find the Worked Example that maps to that concept. Example 1 → Range, Example 2 → Variance/SD, Example 3 → Five-number summary/IQR, Example 4 → Outlier detection. The solutions page shows the reasoning behind every step, not just the final answer.