A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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 L-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, coefficient of variation, 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
Compute and interpret the coefficient of variation (CV) for comparing relative spread across valid ratio scales
Determine quartiles (Q1, Q2, Q3), the interquartile range (IQR), and the five-number summary
Detect potential outliers using the fence rule
Select the most appropriate measure of spread for a given context based on the presence of outliers and distribution shape
Section 2: Prerequisites
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Calculating measures of spread requires total fluency with the measures of center and distribution shapes you mastered in L-DS-3.
From L-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 L-DS-3: The Median. The middle value when data is sorted. It is , the anchor used to split the data when finding quartiles and the IQR.
From L-DS-3: Skewness. Read skewness from the distribution’s longer tail. The mean is often pulled toward that tail, so its position relative to the median is a clue — not proof by itself.
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: .
Retrieval Warm-up — from L-DS-2 and L-DS-3
A researcher posts the histogram of weekly running distances (km) for 200 recreational runners. The histogram has bars touching and a peak at the 30–40 km class. On either side of that peak, the bar heights form an approximately mirror-symmetric pattern, the tails have similar lengths, and there are no isolated extreme bars.
Which statement correctly interprets this histogram?
A professor records the exam scores for a class of 30 students and finds: and median .
Which conclusion is most defensible?
Success Factor:
Where L-DS-4 diverges from L-DS-3: In L-DS-3, we summarized an entire distribution into a single “centre” value. In L-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
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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 (Coefficient of Variation) compares relative spread across valid ratio scales. C5 (Quartiles & IQR) gives you a resistant alternative. C6 (Outlier Detection) puts IQR to work. After C5, you will apply the absolute and resistant measures 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.
Step 1 of 3 — range only
Figure 1. Both datasets have range = 60. Step through the three stages: (1) range only — the orange brackets look identical; (2) reveal the values — the bracket freezes while Dataset A’s tight cluster and Dataset B’s even spread emerge; (3) overlay SD and IQR — the resistant measures differ sharply (Dataset A IQR = 6, Dataset B IQR = 45), showing why we need spread measures the range can’t provide.
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 L-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 ? Suppose the observations are an independent random sample from a population with finite variance . If you compute squared deviations from the sample mean and divide by , that statistic is biased downward in expectation:
A particular sample can be more or less spread out than the population; the claim is about the long-run average over repeated samples. Dividing by instead removes this downward bias. The smaller the sample, the larger the correction; 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. The denominator is where that lost degree of freedom appears in the formula; under the random-sample assumptions above, 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.
Figure 2. Drag any dot left or right. Each coloured square has side length equal to the observation’s absolute deviation from the selected centre, so its area is the squared deviation. In population mode, is the mean area of the squares. In sample mode, is their total area divided by , so the dashed summary square is a corrected square, not a literal average of the coloured squares. Its side equals or , respectively.
Each draw takes n independent values with replacement. A particular sample can be narrower
or wider than the population. Across many draws, the ÷n variance is low on average; use the
n−1 option to update both the sample bracket and the ratio cloud. Arrowheads in the >1.6
lane are real ratios beyond the main axis, not values equal to 1.6.
Figure 3. Start in “Why ÷n runs small”. Each draw is an independent sample with replacement; its spread bracket may be narrower or wider than the population’s. Across repeated draws, however, the divisor- variance ratio averages toward , below 1. Select “Use n−1 for bracket and cloud” to display the sample SD and the unbiased variance ratios, whose average tends toward 1. The expert view shows both running averages against their theoretical targets. These guarantees require independent random sampling from a population with finite variance.
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.
Figure 4. Each point sits inside one of three labelled zones: green (±1s) — within one standard deviation of the mean; gold (±1s–2s) — between one and two standard deviations; red (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 red dot back in to see it turn gold, then green. If a band continues beyond the fixed 0–100 ruler, an arrow and numeric label preserve its true endpoint. Note: the exact percentages inside each band depend on the distribution’s shape — with only a handful of points they will jump around. You will formalize the fixed 68–95 percentages for bell-shaped data when you reach the Normal model.
C4 — Coefficient of Variation (CV)
Range, variance, and standard deviation remain tied to the scale of the original data. This makes raw standard deviations impossible to compare across datasets measured in different units or operating at very different magnitudes. Is a standard deviation of 10 cm “more spread” than a standard deviation of 8 kg? The question is meaningless without a common relative scale.
The coefficient of variation answers this limitation of standard deviation by expressing SD 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) and a mean meaningfully away from zero. 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 as relative variability measures because both variables have meaningful zeroes. 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 relative variability across ratio-scale variables with different units (height vs. weight)
Comparing relative variability across the same ratio-scale variable measured at different magnitudes
Quality control: CV compares relative variability across product scales; any acceptable threshold must come from the specific process specification, not a universal cutoff
“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
Drag a mean slider toward zero — the coloured 0→x̄ segment shrinks while the green ±s bracket
keeps its full size, even if its lower endpoint crosses below zero. The bracket shows x̄−s to
x̄+s as a scale comparison; it is not a bound on the data. As the mean approaches zero,
spread becomes huge relative to the centre and CV explodes.
Figure 5. Both datasets are drawn on one shared ruler that extends below zero so the full to bracket remains visible. The coloured segment is the distance from 0 to the mean (); the green bracket is one standard deviation (). CV is simply how large that green spread bracket is relative to the coloured 0→ segment. Drag the mean of Dataset A toward 5 — the coloured segment collapses while the bracket keeps its size, so CV skyrockets even though is unchanged. The highlighted value is the relatively more variable dataset.
C5 — 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: summarize the middle 50% of the data. The spread of that middle half is the interquartile range (IQR), which gives extreme observations much less influence than the range or standard deviation.
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 L-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 always the value at position and Q3 is always at position .” Quartiles do not have one universal computational convention. This course uses the exclusive median-of-halves convention: when is odd, omit the overall median before finding the medians of the lower and upper halves. Other conventions — including Tukey hinges and software interpolation rules — can produce slightly different quartile values for the same dataset. State the convention when exact quartile values matter.
“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 different resistance properties. The IQR is resistant: changing an extreme value often leaves Q1 and Q3 unchanged, provided the change does not alter which observations determine those quartiles. It is not absolutely immune — especially in a small dataset, an extreme observation can participate in a half-sample median and change the IQR. Variance and standard deviation are sensitive to every observation because squaring deviations gives extremes disproportionate weight. The range is the most fragile: changing one extreme changes it directly. Choosing the IQR for skewed data or data with outliers, and standard deviation for roughly symmetric data without outliers, produces a more faithful description of spread.
Press Next to begin.
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Figure 6. 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 30 with 95. The upper fence is 47.5, but the whisker ends at 25 — the largest observed value at or below that fence — while 95 is plotted separately as an outlier.
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.
Figure 7. 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. If an SD endpoint lies beyond the fixed 0–100 ruler, an arrow and numeric label preserve the true off-scale value.
C6 — 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 not flagged by this rule.
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.
Conventional value: 1.5× (about 0.7% of a normal population lies beyond its theoretical
fences)
Figure 8. The seven gray values form a fixed reference dataset; its quartiles and IQR determine the fences. Drag the separate, held-out dot to see whether that candidate observation would be flagged by those fixed reference fences. The candidate is not added to the reference data, so moving it does not recompute Q1, Q3, or IQR. Use the multiplier slider to explore how different thresholds change the flagging rule — notice that 1.5× is a convention, not a physical law.
Section 4: Worked Examples
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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—and does not reveal—about these recorded sales and the broader 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.
Among these 10 recorded sales, prices run from $285K to $315K, so the observed span is $30K. Most of the values are clustered within that band, and no single sale is far separated from the others.
That evidence describes only these 10 sales. Without knowing how the homes and time period were selected, whether the properties are comparable, or how this span compares with the broader market, the range does not establish that the whole neighbourhood is homogeneous or predict what a future buyer will pay.
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 that these 10 observed values are all within a narrow band, so the range describes their span clearly. I will not generalize from that span to the whole neighbourhood without evidence that the sales represent the broader market. If I saw a value far from the cluster, I would also flag the range as potentially misleading and supplement it with the 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 calculating, use the range from 4.8 to 5.5 (span = 0.7) and a mean near 5.1 to predict whether the standard deviation will be closer to 0.2 or to 0.5.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
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 exact sample variance is mm², which rounds to the requested mm². The sample standard deviation is mm. The typical bearing diameter deviates from the mean of 5.1 mm by about 0.26 mm. These six measurements are tightly clustered; establishing process stability would require repeated monitoring over time against the process specification.
Answer to prediction: The standard deviation is 0.26 mm. The two largest absolute deviations from the mean are 0.4 mm and 0.3 mm, while the other four are 0.2 mm or less, so a typical distance from the mean should be a few tenths of a millimetre. The computed value of 0.26 mm fits that reasoning. The range is 0.7 mm, but range and standard deviation answer different questions: the range measures the full span from the smallest to largest value, whereas standard deviation summarizes distances from the mean.
Example 3 — Five-Number Summary and IQR
Scenario: A dietician records daily sodium intake (mg) for 11 patients:
Find the five-number summary, the IQR, and determine if there are any outliers.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Solution
Step 1: Sort the data.
(odd).
Step 2: Identify the minimum and maximum.
,.
Step 3: Find Q2 (median).
Position: . The 6th value in the sorted list is 2150. → Q2 = 2150 mg.
(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. Q2 is 200 mg above Q1 and 250 mg below Q3, so the middle half is fairly balanced but not exactly symmetric. A full judgment about shape would use the complete distribution rather than quartile spacing alone.
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.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
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: .
For even , split the sorted data into two equal halves:
Lower half: .
Upper half: .
The analyst’s lower-half description omitted 26, but Q1 happened to remain 24:
Q3 also remains 29.5, so the analyst’s IQR and fences were already correct:
The value 55 exceeds 37.75, so it is flagged as a potential outlier. In this dataset, correcting Q2 and the even- split does not change Q1, Q3, the IQR, the fences, or the outlier flag.
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 upper fence of 37.75). 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
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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)
The range is the simplest spread measure — — and the most fragile, because those two extreme values are the only ones it sees. Compute the range for each dataset below; type your answer, and a wrong-method value returns the specific misread it encodes (reporting the maximum alone, the minimum, or misreading an extreme).
Watch what one extreme value does: add a single far-off observation and the range can jump several-fold, even though almost every other data point is unchanged. That fragility — one extreme value dominating the whole measure — is exactly why range gives way to variance, standard deviation, and IQR for the rest of this lesson.
Problem 2 — Sample Variance and Standard Deviation (C2 + C3)
A new dataset each time. Study the worked mean, sum of squared deviations, and sample variance, then take over the final step: convert the variance to the sample standard deviation . The final answer is yours to compute and check.
Problem 3 — Five-Number Summary and IQR Method (C5)
A fresh dataset each time. Work through the quartiles one step at a time — the median , then and as medians of the halves, then the IQR. Answer each step, or take the solo path and pull the steps up only if you need them.
Problem 4 — Coefficient of Variation (C4)
Compute the requested CV from a fresh cross-unit comparison. Type the number rather than selecting from numeric options; a wrong-method value returns feedback for forgetting , using the other variable’s CV, or reporting the raw standard deviation. Then use the displayed CVs to decide which variable is more variable relative to its mean.
Section 6: Independent Practice
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Work through these problems without guided steps. Attempt each problem first, then expand the solution whenever you are ready to compare your reasoning. Concepts are interleaved across all six measures of spread.
Problem 1 — Compute Sample Variance and Standard Deviation (C2 + C3)
Problem 2 — Find the Five-Number Summary and IQR (C5)
Problem 3 — Identify Outliers Using the Fence Method (C5 + C6)
Problem 4 — Synthesize Resistant and Non-Resistant Spread (C1 + C5 + C6)
Consider the sorted dataset
Compute the range and five-number summary.
Compute the IQR and both fences.
Decide whether any value is a potential outlier.
Remove only the flagged value for a sensitivity check, recompute the range and IQR, and explain which measure changed more and why. This is a comparison exercise, not permission to delete the value in a real analysis.
Show Solution
(a) Original range
With , Q2 is the fifth value:
The lower half is , so
The upper half is , so
The five-number summary is
(b) IQR and fences
(c) Potential outlier
The value 25 exceeds 11.25, so it is flagged for investigation.
(d) Sensitivity check
Without 25, the range is
For , Q1 is 5 and Q3 is 7, so
The range falls from 21 to 4, while the IQR falls only from 2.5 to 2. The extreme value dominates the range but has limited influence on the middle-half spread.
Problem 5 — Choose a Spread Measure from Context (C1 + C3 + C4 + C5)
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.
Scenario 1. 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?
Scenario 2. 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?
Scenario 3. 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?
Scenario 4. 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 the hospital report instead?
Scenario 5. A quality engineer compares two production lines that manufacture the same component at different sizes. Line 1 has a positive sample mean close to its 10 mm target; Line 2 has a positive sample mean close to its 50 mm target. Both lines produce symmetric, normally distributed measurements. The engineer wants to compare variability as a percentage of each line’s typical part length—not absolute variability in millimetres or conformance to engineering tolerances.
Which measure should be used to compare relative consistency across the two typical part lengths?
Problem 6 — Diagnose a Spread Calculation (C2 + C3)
A worked sample-standard-deviation solution appears below with one injected misconception — dividing by instead of , forgetting to square the deviations, or reporting the variance as the standard deviation. Locate the first step where the reasoning breaks. A fresh worked solution regenerates each time.
Mixed Review — Retrieval from Earlier Lessons
These problems draw on concepts from L-DS-2 and L-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 (L-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 L-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 because it locates the middle of the sorted list. As long as the luxury sale remains the largest observation, moving it farther upward does not change the 6th and 7th positions that determine the median. Both middle values lie within the broad $280K–$450K interval described, although the grouped information does not reveal their exact values.
(b) The mean is sensitive to extreme values (right-skewed distribution). The $1.2M sale can pull the mean upward even though 10 of the 12 sales were under $400K. The grouped information supplied does not identify each sale price, so it does not let us verify the advertised $412K mean or compute an exact median. What it does establish is the misleading feature: “average” is being used for a non-resistant mean without disclosing the isolated high sale and the right-skewed distribution. A more transparent advertisement would report the median computed from the original 12 prices and disclose the $1.2M sale separately.
Review Problem 2 — Histogram Shape and Measure Selection (L-DS-2 and L-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: for , it averages the 30th and 31st sorted observations. The supplied median of 8.5 hours is consistent with both middle observations lying in the 6–10 hour bin. Moving the three largest observations farther right would not change those middle positions, provided their ordering relative to the other observations remains unchanged.
(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
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
These questions ask you to recall and apply what you’ve learned independently.
Question 1 — Feynman Test
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. Use plain language—no formulas and no jargon without defining it.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Compare with a model answer
Model answer: A sample uses its own mean as the centre, and that mean is fitted to the same observations. This makes the squared deviations a little too small, on average, for the population we want to describe. For independent observations drawn from the same population with finite variance, dividing by instead of makes the sample variance unbiased for the population variance. The correction matters most for small samples and becomes tiny for large ones.
Question 2 — Novel Application: Choose and Defend the Evidence
A city is deciding where to place a mobile housing-support office. It has summary data for weekly rent in two neighbourhoods:
Neighbourhood A: approximately symmetric, no potential outliers, mean $465, median $463, 3851.
Neighbourhood B: strongly right-skewed, several luxury rentals above $1,500, mean $612, median $438, 29574.
Prepare a short recommendation that:
selects the centre-and-spread pair that best describes a typical renter in each neighbourhood;
explains why the other pair is less trustworthy for Neighbourhood B; and
identifies which neighbourhood has the wider typical middle spread, using the supplied numbers.
Show a Model Answer
For Neighbourhood A, report the mean ($465) and standard deviation ($38) because the distribution is symmetric with no outliers; using every observation is informative here.
For Neighbourhood B, report the median ($438) and IQR ($74). The luxury rentals pull the mean upward and create very large squared deviations, so the mean and SD exaggerate what a typical renter experiences.
Comparing the resistant spreads, Neighbourhood B has the wider middle 50% because its IQR is $74 rather than $51. The evidence supports placing the office where other need indicators also point; spread alone does not measure housing hardship or determine the decision.
A worked sample-standard-deviation solution contains one authentic error. Locate the first step where the reasoning breaks; the feedback identifies the misconception and repairs it. Use “Try another” to practice rotating error types.
Self-Assessment
Still unsureFully confident
Section 8: Boss Fight
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
Boss Fight — Pick Your Path. Both paths ask you to choose spread measures, compute comparable evidence, investigate an unusual value, and limit your conclusion to what the data can support. Work solo first, then expand the complete model for your chosen path whenever you are ready.
🔬 Path A: The Analyst
Reconcile conflicting salary-spread summaries and write a careful HR evidence brief.
🏗️ Path B: The Architect
Design a total-tablet-mass monitoring rule without overstating what mass can prove.
🔬 Path A: The Analyst
An HR team is comparing non-managerial salaries in two departments. Salaries are in thousands of dollars. The team supplies these summaries:
Department
Mean
Median
Q1
Q3
Min
Max
Engineering
12
64.42
64.5
4.21
61.0
67.5
58
71
Marketing
10
55.40
51.5
14.15
49.0
54.0
47
95
Task 1 — Build a fair comparison. Compute the range, IQR, and CV for each department. Decide which results describe the middle of each department well and which are being driven by an extreme value.
Task 2 — Investigate the disagreement. Use the rule to check both departments for potential outliers. Explain why Marketing can have a much larger SD but a smaller IQR than Engineering.
Task 3 — Write a limited HR brief. State what the data support about salary dispersion and what they do not establish about pay equity. Name at least two additional variables the HR team would need before making an equity claim.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show complete Path A solution
Task 1 — Comparable spread evidence
Engineering range:
Engineering IQR:
Engineering CV:
Marketing range:
Marketing IQR:
Marketing CV:
Marketing’s range, SD, and CV are much larger, but its IQR is slightly smaller. That pattern suggests one extreme salary is dominating the non-resistant measures while the middle half remains tightly grouped.
Task 2 — Fence check
Engineering lower fence:
Engineering upper fence:
Every Engineering salary lies from 58 to 71, so none is flagged.
Marketing lower fence:
Marketing upper fence:
The value 95 exceeds 61.5, so it is a potential outlier. It should be investigated, not automatically removed. Because SD squares deviations, that single large deviation strongly increases Marketing’s SD. In this dataset, moving the 95 farther upward would leave Q1 = 49 and Q3 = 54 unchanged, so the IQR remains 5; this illustrates resistance, not absolute immunity to every change in an extreme observation.
Task 3 — Limited conclusion
This snapshot supports a claim about observed dispersion: Engineering salaries are more tightly clustered by SD and CV, while Marketing’s middle 50% is slightly tighter by IQR and one high salary drives the larger non-resistant spread. It does not establish which department is more equitable. An equity analysis would also need comparable job level, role, tenure, hours, performance criteria, and relevant protected-group information, handled under the organization’s privacy and legal review. The next step is to validate the $95K record and analyze like-for-like roles before drawing an equity conclusion.
Reflection: Which changed your interpretation more—the disagreement between SD and IQR, or the limit on what spread can establish? What would you verify first?
🏗️ Path B: The Architect
A manufacturer makes two tablet strengths. The figures below describe total tablet mass, including inactive ingredients; they do not measure active-ingredient content. The team wants a spread-monitoring rule for total mass while a separate validated assay monitors the active ingredient.
Product
Mean mass
Q1
Q3
Observed min–max
S
20
250.4 mg
3.1 mg
248.6 mg
252.0 mg
244.1–256.8 mg
L
20
501.2 mg
5.0 mg
498.5 mg
503.0 mg
487.0–508.6 mg
Task 1 — Compare consistency across scale. Choose and compute a measure that lets the team compare relative total-mass variability between the 250 mg and 500 mg products. State which product is more consistent relative to its mean.
Task 2 — Build an investigation rule. Compute the fences for both products, identify any flagged observation, and explain what a flag should trigger.
Task 3 — Write the protocol boundary. Recommend which spread summaries the dashboard should show, how the team should respond to a flag, and one claim the mass data must never be used to make.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show complete Path B solution
Task 1 — Relative variability
Product S CV:
Product L CV:
Product L has the larger absolute SD but the smaller CV, so its total mass is slightly more consistent relative to its mean.
Task 2 — Fence rule
Product S IQR:
Product S fences:
Its observed values from 244.1 to 256.8 mg stay inside the fences.
Product L IQR:
Product L fences:
The 487.0 mg observation is below 491.75 mg, so it is flagged for investigation. A flag should trigger record verification, equipment and process checks, and the organization’s approved QC procedure—not automatic deletion.
Task 3 — Protocol boundary
Show mean and SD for routine total-mass consistency, CV for comparing the two scales, and median/IQR plus fence flags so one unusual tablet cannot hide inside an average. Investigate flagged records and document the disposition before recomputing summaries.
The dashboard must never claim that low total-mass variability proves active-ingredient content uniformity. Total mass is a different measured variable; active-ingredient content requires its own validated assay and acceptance process.
Reflection: Which safeguard matters more in this protocol—relative comparison, robust outlier screening, or the measurement boundary? What failure could occur if it were omitted?
Section 9: Challenge Problems
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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 L-DS-3, you learned that the deviations from the mean sum to zero: . Now prove the identity and connect it to the denominator in sample variance.
Claim: For any set of numbers , define . Then .
Your task: Expand the summation, substitute the definition of , and show that the result is zero. Then explain why knowing any deviations forces the final deviation.
Pause and write. Record your answer and reasoning on paper or in your notes before continuing. Even if you are unsure, committing your thinking helps you see what you understood and what you need to revisit.
Show Proof and Connection
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 Relative Spread Across Ratio-Scale Variables
For ratio-scale variables with meaningful zeroes and means away from zero, the coefficient of variation removes units and lets you compare relative variability across different scales. Each problem below draws two such traits with their own means and standard deviations — compute the requested CV. A wrong-method value returns the specific mistake it encodes.
Challenge 3 — CV Under Multiplicative Unit Conversion
A researcher measures corn-plant height twice: once in centimetres( cm, cm) and once in metres( m, m) — the very same plants, only the unit changed.
Compute the CV in metres. What does changing the unit do to it?
Section 10: Solutions Reference
▾
A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so
you know where to look back.
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.
Quick-Reference Formulas
Range:
Sample Variance:
Population Variance:
Standard Deviation:
Interquartile Range:
Outlier Fences:
Coefficient of Variation:
Distribution
Centre
Spread
Symmetric, no outliers
Mean
Standard Deviation
Skewed or has outliers
Median
IQR
Comparing relative spread across ratio-scale variables with meaningful zeroes and means away from zero