truckA = [590, 610, 630, 630, 640]
truckB = [370, 470, 670, 720, 870]
function stats(xs) {
const n = xs.length;
const mean = xs.reduce((a, b) => a + b, 0) / n;
const dev = xs.map(x => x - mean);
const sq = dev.map(d => d * d);
const sumDev = dev.reduce((a, b) => a + b, 0);
const ss = sq.reduce((a, b) => a + b, 0);
const s2 = ss / (n - 1);
return {n, mean, dev, sq, sumDev, ss, s2, s: Math.sqrt(s2)};
}
fmt = v => Math.round(v).toLocaleString("en-US")
signed = v => (v > 0 ? "+" : v < 0 ? "−" : "") + fmt(Math.abs(v))
viewof step = {
const labels = ["1 · Dots", "2 · Deviations", "3 · Squares"];
const form = html`<form class="ojs-step-picker">Show: ${labels.map((l, i) =>
html`<label> <input type="radio" name="step" value="${i + 1}"
${i === 0 ? "checked" : ""}> ${l} </label>`)}</form>`;
form.value = 1;
form.oninput = () => {
form.value = +form.querySelector("input:checked").value;
};
return form;
}The business question. “The Wandering Fork” has a mean daily revenue of €628.34 (Harbour, 30 days) — but two businesses can share the exact same mean and still be very different to run: one reliably steady, the other wildly erratic. Dispersion indicators measure exactly that variability, and they are what turns “the average day” into “the average day, ± how much”. We start with two trucks small enough to compute by hand, and only then write the formulas.
1 Theory
1.1 Same mean, different days
Two Wandering Fork trucks recorded their revenue over the same five days. The numbers are a simplified extract chosen so that you can compute everything by hand.
| Day | Truck A “Steady” (€) | Truck B “Erratic” (€) |
|---|---|---|
| 1 | 590 | 370 |
| 2 | 610 | 470 |
| 3 | 630 | 670 |
| 4 | 630 | 720 |
| 5 | 640 | 870 |
| Mean | 620 | 620 |
Both trucks earn €620 on an average day. Which one would you rather manage — and what would you tell the bank about each one’s revenue? Look at the figure before computing anything (start with step 1).
1.2 Range and interquartile range
Try it.
For each truck, find the best day, the worst day and the gap between them.
| Truck | Max (€) | Min (€) | Max − Min (€) |
|---|---|---|---|
| A “Steady” | ? | ? | ? |
| B “Erratic” | ? | ? | ? |
Check your answer
- Truck A: \(640 - 590 = 50\)
- Truck B: \(870 - 370 = 500\)
- Truck B’s best and worst days are 10 times further apart than Truck A’s, even though both average €620.
See it.
In step 1 of the figure, Truck B’s dots stretch across the almost the whole axis; Truck A’s huddle around the mean line.
In plain words: the range uses only the two most extreme values; the IQR uses the middle 50%, so a single outlier cannot distort it.
NoteDefinition — range and interquartile range
The range is \(\max - \min\): simple, but sensitive to a single extreme value.
The interquartile range \(IQR = Q_3-Q_1\) (Module 5) measures the spread of only the central 50% of the data, making it more robust to extreme values.
Recall. You computed \(Q_1\) and \(Q_3\) in Module 5; the IQR is simply the gap between them, read off the boxplot as the length of the box.
1.3 Sample variance and standard deviation
The range looks at two days only. To use every day, measure how far each day is from the mean.
Try it. Complete the deviation table for Truck A (\(n = 5\), \(\bar{x} = 620\)), working column by column.
| Day \(i\) | \(x_i\) (€) | \(x_i-\bar{x}\) (€) | \((x_i-\bar{x})^2\) (€²) |
|---|---|---|---|
| 1 | 590 | ? | ? |
| 2 | 610 | ? | ? |
| 3 | 630 | ? | ? |
| 4 | 630 | ? | ? |
| 5 | 640 | ? | ? |
| Total | 3,100 | ? | ? |
- Fill the deviation column \(x_i-\bar{x}\), then add it up. What do you notice?
- Square each deviation and add up the squares.
- Divide that total by \(n-1 = 4\).
- Take the square root of the result.
Your turn: build the same table for Truck B.
Check your answer
Truck A
| Day \(i\) | \(x_i\) | \(x_i-\bar{x}\) | \((x_i-\bar{x})^2\) |
|---|---|---|---|
| 1 | 590 | −30 | 900 |
| 2 | 610 | −10 | 100 |
| 3 | 630 | 10 | 100 |
| 4 | 630 | 10 | 100 |
| 5 | 640 | 20 | 400 |
| Total | 3,100 | 0 | 1,600 |
- \(1{,}600 / 4 = 400\) €²
- \(\sqrt{400} = 20\) €.
Truck B
| Day \(i\) | \(x_i\) | \(x_i-\bar{x}\) | \((x_i-\bar{x})^2\) |
|---|---|---|---|
| 1 | 370 | −250 | 62,500 |
| 2 | 470 | −150 | 22,500 |
| 3 | 670 | 50 | 2,500 |
| 4 | 720 | 100 | 10,000 |
| 5 | 870 | 250 | 62,500 |
| Total | 3,100 | 0 | 160,000 |
- \(160{,}000 / 4 = 40{,}000\) €²
- \(\sqrt{40{,}000} = 200\) €.
- On a typical day Truck B is about €200 away from its mean; Truck A only about €20.
Why is the deviation total always 0? The mean is the balance point of the data: the days above it exactly offset the days below it. So the plain average of the deviations is always 0 and tells you nothing. Squaring removes the signs — every distance counts as positive — and gives big misses more weight than small ones.
Why divide by 4 and not 5? Suppose you know Truck A’s first four deviations: −30, −10, 10, 10. Because the total must be 0, the fifth deviation has to be +20. Only 4 of the 5 deviations are free to vary, so we average over \(n-1 = 4\).
See it. Step 2 of the figure draws each deviation as a bar from the mean (sum = 0). Step 3 draws each squared deviation as a square: the variance is the size of the “average square”, and \(s\) is its side.
In plain words: measure how far each observation is from the mean, square those distances (so they are all positive), average them (with \(n-1\)), and take the square root to get back to the original unit.
TipFormula: sample variance and standard deviation
For a sample of \(n\) values \(x_1,\dots,x_n\) with mean \(\bar{x}\), the sample variance is \[ s^2 = \frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^2, \] i.e. the average squared distance of each observation from the mean (with \(n-1\), not \(n\), correcting for the fact that \(\bar{x}\) is itself estimated from the same sample).
The sample standard deviation \(s=\sqrt{s^2}\) returns to the original unit (€, not €²).
Notation decoder: sample variance and standard deviation
| Symbol | Read aloud | In our example (Truck A) |
|---|---|---|
| \(n\) | “n” | 5 days |
| \(x_i\) | “x sub i” | the value on day \(i\), e.g. \(x_1 = 590\) |
| \(\bar{x}\) | “x-bar” | the mean, 620 |
| \(x_i-\bar{x}\) | “x sub i minus x-bar” | the deviation column: −30, −10, 10, 10, 20 |
| \((x_i-\bar{x})^2\) | “… squared” | the squares column: 900, 100, 100, 100, 400 |
| \(\sum_{i=1}^{n}\) | “add up, for i from 1 to n” | the Total row: 1,600 |
| \(\frac{1}{n-1}\) | “divided by n minus 1” | ÷ 4 |
| \(s^2\) | “s squared” | 400 €² |
| \(s=\sqrt{s^2}\) | “s” | 20 € |
Whole formula as a sentence: “the sample variance is the sum, over all \(n\) days, of each day’s squared distance from the mean, divided by \(n-1\); the standard deviation is its square root.”
If you are interested to understand why we need to divide by \(n-1\) rather \(n\), see Bessel’s correction.
1.4 Coefficient of variation
Try it. Truck B’s standard deviation is €200. Is that a lot?
- Compute \(s / \bar{x} \times 100\%\) for Truck A and Truck B.
- A coffee cart in the same fleet has a mean of €62 and \(s = €20\) — exactly Truck A’s \(s\). Is it as steady as Truck A?
Check your answer
- Truck A: \(20/620 \times 100\% = 3.23\%\). Truck B: \(200/620 \times 100\% = 32.26\%\).
- Coffee cart: \(20/62 \times 100\% = 32.26\%\) — the same relative variability as the erratic Truck B. A €20 swing is tiny next to €620 but huge next to €62. The size of \(s\) only means something relative to the mean.
In plain words: divide the standard deviation by the mean and express as a percentage. A \(CV\) of 20% means the typical deviation from the mean is about 20% of the mean itself.
TipFormula: coefficient of variation
\[ CV = \frac{s}{\bar{x}} \times 100\%. \]
Because it is unit-free, \(CV\) is the only dispersion indicator that lets you compare the relative variability of two variables measured on different scales or units (e.g., revenue in € vs. customer count).
Notation decoder: coefficient of variation
| Symbol | Read aloud | In our example |
|---|---|---|
| \(CV\) | “coefficient of variation” | 3.23% (A), 32.26% (B) |
| \(s\) | “s” | 20 € (A), 200 € (B) |
| \(\bar{x}\) | “x-bar” | 620 € for both |
| \(\times 100\%\) | “as a percentage” | 0.0323 → 3.23% |
Whole formula as a sentence: “the coefficient of variation is the standard deviation expressed as a percentage of the mean.” The € cancel out, which is why \(CV\) has no unit.
1.5 Shortcut (computational) formula for variance
Try it. Compute Truck A’s variance a second way, without any deviation.
- Square each value and add the squares: \(590^2 + 610^2 + 630^2 + 630^2 + 640^2\).
- Compute \(n\bar{x}^2 = 5 \times 620^2\).
- Subtract, then divide by \(n-1 = 4\).
- Redo step 2 with a rounded mean of 621. What happens?
Check your answer
- \(348{,}100 + 372{,}100 + 396{,}900 + 396{,}900 + 409{,}600 = 1{,}923{,}600\).
- \(5 \times 384{,}400 = 1{,}922{,}000\).
- \((1{,}923{,}600 - 1{,}922{,}000)/4 = 1{,}600/4 = 400\) — the same \(s^2\) as the deviation table.
- \(5 \times 621^2 = 1{,}928{,}205\), so the result becomes \((1{,}923{,}600 - 1{,}928{,}205)/4 = -1{,}151.25\): a negative variance, which is impossible. The shortcut subtracts two large, nearly equal numbers, so a tiny rounding of \(\bar{x}\) ruins it. Always keep the mean unrounded.
In plain words: square every value and add them up, subtract \(n\) times the squared mean, and divide by \(n-1\). Same answer, no deviation column.
TipFormula: variance shortcut
\[ s^2 = \frac{1}{n-1}\left(\sum_{i=1}^{n} x_i^2 - n\bar{x}^2\right). \]
This needs only \(\sum x_i^2\) and \(\bar{x}\), one pass over the data instead of first computing every deviation \((x_i-\bar{x})\) and then squaring it. In Excel, use SUMSQ(range) for \(\sum x_i^2\).
Notation decoder: variance shortcut
| Symbol | Read aloud | In our example (Truck A) |
|---|---|---|
| \(\sum_{i=1}^{n} x_i^2\) | “sum of x sub i squared” | square first, then add: 1,923,600 |
| \(\left(\sum_{i=1}^{n} x_i\right)^2\) | “square of the sum” — not the same! | add first, then square: \(3{,}100^2 = 9{,}610{,}000\) |
| \(n\bar{x}^2\) | “n times x-bar squared” | \(5 \times 620^2 = 1{,}922{,}000\) |
| \(\frac{1}{n-1}\) | “divided by n minus 1” | ÷ 4 |
Whole formula as a sentence: “the variance is the sum of the squared values minus \(n\) times the squared mean, divided by \(n-1\).”
2 Worked example
Using “The Wandering Fork”’s daily_revenue_eur (\(n=30\), \(\bar{x}=628.34\)), verified in data-model.md:
| Statistic | Value |
|---|---|
| Range | 478.97 |
| Sample variance \(s^2\) | 18,793.33 |
| Sample std. dev. \(s\) | 137.09 |
| Coefficient of variation \(CV\) | 21.82% |
Step 1 — range. \(886.56 - 407.59 = 478.97\): over the 30-day period, the best day out-earned the worst by almost €480.
Step 2 — variance and standard deviation. Squaring each day’s deviation from \(\bar{x}=628.34\), summing, and dividing by \(n-1=29\) gives \(s^2 = 18{,}793.33\) €²; taking the square root gives \(s = 137.09\) €. On a typical day, revenue is about €137 away from the €628.34 average, in either direction.
Step 3 — coefficient of variation. \(CV = 137.09/628.34 \times 100\% = 21.82\%\) — using the guide below, a moderate level of relative variability: noticeably variable, but not wildly erratic day to day.
Step 4 — shortcut formula check. \(\sum x_i^2 = 12{,}389{,}403.98\), \(n\bar{x}^2 = 30 \times 628.3417^2 = 11{,}844{,}397.50\), difference \(= 545{,}006.48\), divided by \(29 = 18{,}793.33\) — identical to the definitional result. Keep the mean unrounded here: with \(\bar{x}=628.34\) the result drifts to about \(18{,}795\), because the shortcut subtracts two very large, nearly equal numbers.
2.1 Interpreting \(CV\)
| \(CV\) range | Interpretation |
|---|---|
| \(CV < 15\%\) | Low relative variability — data is tightly clustered around the mean |
| \(15\% \le CV \le 35\%\) | Moderate relative variability — noticeable but not extreme |
| \(CV > 35\%\) | High relative variability — data is widely spread relative to the mean |
“The Wandering Fork”’s \(CV = 21.82\%\) falls in the moderate band.
3 Proof / derivation
CautionDerivation: the variance shortcut (computational) formula
Starting from the definitional formula and expanding the square:
\[ \sum_{i=1}^{n}(x_i-\bar{x})^2 = \sum_{i=1}^n x_i^2 - 2\bar{x}\sum_{i=1}^n x_i + n\bar{x}^2. \]
Since \(\sum_{i=1}^n x_i = n\bar{x}\), the middle term becomes \(-2\bar{x}(n\bar{x}) = -2n\bar{x}^2\), so
\[ \sum_{i=1}^{n}(x_i-\bar{x})^2 = \sum_{i=1}^n x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 = \sum_{i=1}^n x_i^2 - n\bar{x}^2. \]
Dividing by \(n-1\) gives the shortcut formula
\[ s^2 = \frac{1}{n-1}\left(\sum_{i=1}^{n} x_i^2 - n\bar{x}^2\right), \]
which needs only \(\sum x_i^2\) and \(\bar{x}\) — one pass over the data instead of first computing every deviation \((x_i-\bar{x})\) and then squaring it. Both formulas give the identical result, \(s^2=18{,}793.33\), for “The Wandering Fork”; the shortcut form is simply faster to compute by hand or with SUMPRODUCT/SUMSQ in Excel.
4 Visual intuition
Variance is literally the average squared distance from the mean: picture each of the 30 days as a point on a number line at its revenue value, and \(\bar{x}=628.34\) as a fixed pivot. Variance answers “on average, how far (squared) do the points sit from the pivot?” — a wide, flat spread of points gives a large variance; a tight cluster around the pivot gives a small one.
Step 3 of the two-truck figure makes this literal: each square’s area is one squared deviation, the variance is their total divided by \(n-1\) — the “average square” — and the standard deviation \(s\) is the side of that average square. Truck B’s average square is 100 times larger in area than Truck A’s, but only 10 times wider.
5 Interactive demo — central tendency & dispersion calculator
Move the six sliders below and watch the mean, variance, standard deviation, and coefficient of variation update live. Starting values are a 6-day sample of “The Wandering Fork”’s daily_revenue_eur.
6 Exercises
All exercises below are done by hand; the Excel ones are in the homework.
Exercise 1 (Shortcut formula) A sample of \(n=5\) values has \(\sum x_i = 100\) and \(\sum x_i^2 = 2140\). Find the sample variance.
Solution
\(\bar{x}=100/5=20\).
Shortcut formula:
\[ s^2 = \frac{1}{n-1}\left(\sum x_i^2 - n\bar{x}^2\right) \] \[ s^2= \frac{1}{4}(2140 - 5\times400) = \frac{1}{4}(2140-2000) = 35 \]
Exercise 2 (Comparing variability across units) “The Wandering Fork”’s items_sold has mean \(74.40\) and sample standard deviation \(16.22\) (from data-model.md). Which variable — daily_revenue_eur or items_sold — has higher relative variability?
Solution
- \(CV_{\text{revenue}} = 137.09/628.34 = 21.82\%\);
- \(CV_{\text{items}} = 16.22/74.40 = 21.80\%\).
- The two are almost identical (about 21.8%) even though the raw units (€ vs. items sold) are completely different, exactly the kind of comparison only \(CV\), not \(s\) alone, can make fairly.
Exercise 3 (Effect of an outlier on the range) If one exceptionally bad day of \(\text{daily\_revenue\_eur}=150\) replaced the current minimum (407.59), how would the range change, and would the standard deviation be affected in the same direction?
Solution
- New range \(= 886.56 - 150 = 736.56\) (up from 478.97)
- The range, using only two values, reacts strongly to a single extreme point. The standard deviation would also increase (an extra-large deviation from the mean adds to \(\sum(x_i-\bar{x})^2\)), but by a much smaller relative amount, since it accounts for all 30 days rather than just the two most extreme.
Exercise 4 (Formula to words)
Read \(\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2\) aloud as one sentence. Which indicator is it?
For \(x_1=4\), \(x_2=6\), \(x_3=8\), compute \(\sum_{i=1}^{3} x_i\) and \(\sum_{i=1}^{3} x_i^2\). Is \(\left(\sum x_i\right)^2\) equal to \(\sum x_i^2\)?
Solution
- “Take each observation’s distance from the mean, square it, add up the \(n\) squares, and divide by \(n-1\).” It is the sample variance \(s^2\).
- \(\sum x_i = 4+6+8 = 18\); \(\sum x_i^2 = 16+36+64 = 116\). But \(\left(\sum x_i\right)^2 = 18^2 = 324 \neq 116\): “square then add” is not “add then square” — the most common slip in the shortcut formula.
Exercise 5 (Words to formula) Write in symbols:
- “add up the squares of all \(n\) observations”;
- “the standard deviation as a percentage of the mean”;
- “the distance between the 3rd observation and the mean”.
Solution
- \(\sum_{i=1}^{n} x_i^2\).
- \(CV = \dfrac{s}{\bar{x}} \times 100\%\).
- \(x_3-\bar{x}\) (negative if the 3rd observation is below the mean).
7 Common mistakes
WarningWatch out
- Using
VAR.P/STDEV.Pwhen you needVAR.S/STDEV.S. The course standard is the sample formulas (denominator \(n-1\)). The population variants (denominator \(n\)) are for when you have the entire population — rare in business. - Forgetting the square root. Variance is in squared units (€²); standard deviation is in the original unit (€). Always report the one that matches your interpretation.
- Computing \(CV\) with the median instead of the mean. \(CV = s/\bar{x}\), not \(s/Me\). The mean is the balance point that pairs with variance.
- Using the range as your only dispersion measure. The range ignores everything between min and max; always report \(s\) or \(IQR\) alongside it.
- Comparing standard deviations across different units. €137 vs 16 items — meaningless without \(CV\). Always use \(CV\) for cross-variable or cross-dataset comparisons.
8 Exit check
Five minutes, no calculator needed except for question 4.
- In one sentence, what does the standard deviation \(s\) measure?
- Why does \(\sum(x_i-\bar{x})\) always equal 0?
- Read \(s^2=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2\) aloud as a sentence.
- Compute \(s^2\) for the values 2, 4, 4, 6, 9.
- You want to compare the spread of revenue (in €) with the spread of items sold. Which indicator do you use, and why?
Answers
- The typical distance between an observation and the mean, in the original unit (e.g., “on a typical day, revenue is about €137 away from the average”).
- The mean is the balance point: deviations above it exactly cancel deviations below it.
- “The sum, over all \(n\) observations, of each squared distance from the mean, divided by \(n-1\).”
- \(\bar{x} = 25/5 = 5\); deviations −3, −1, −1, 1, 4; squares 9, 1, 1, 1, 16; total 28; \(s^2 = 28/4 = 7\).
- The coefficient of variation \(CV = s/\bar{x}\times100\%\): it has no unit, so € and items can be compared.
9 Homework — Excel lab
Reproduce, in Excel, every indicator you computed by hand in class, with VAR.S, STDEV.S, SUMSQ and a \(CV\) formula, then apply them to the full dataset.
Estimated time: about 50–60 minutes. Any spreadsheet application works, at home or in a campus computer room.
Follow-up: a 10-minute debrief at the start of Module 7. Bring your Step 0 results and any question on Homework 1–3.
9.1 Using Excel
The complete list of function names, with their French equivalents, is on the Excel functions page.
| Concept | Excel function / steps |
|---|---|
| Range | =MAX(range)-MIN(range) |
| Sample variance | =VAR.S(range) |
| Sample standard deviation | =STDEV.S(range) |
| Population variance | =VAR.P(range) |
| Population standard deviation | =STDEV.P(range) |
| Coefficient of variation | =STDEV.S(range)/AVERAGE(range) |
| Interquartile range | =QUARTILE.EXC(range,3)-QUARTILE.EXC(range,1) |
| Sum of squares (for shortcut) | =SUMSQ(range) |
Sample vs population: This is a descriptive statistics course. We use the sample formulas (
.S) by default because real business data are almost always samples. The population variants (.P) are shown for completeness.
For the full table (120 days per franchise) use the Excel Table tblFork and add the franchise as one more condition, for example =VAR.S(FILTER(tblFork[daily_revenue_eur], tblFork[franchise]="Harbour")).
9.2 Step 0 — check your class results
- Type Truck A’s five values (590, 610, 630, 630, 640) in
A2:A6and Truck B’s (370, 470, 670, 720, 870) inB2:B6. - Below each column, compute
=AVERAGE(A2:A6),=MAX(A2:A6)-MIN(A2:A6),=VAR.S(A2:A6),=STDEV.S(A2:A6)and=STDEV.S(A2:A6)/AVERAGE(A2:A6)(format as a percentage). Repeat for column B. - For Truck A, check the shortcut formula:
=(SUMSQ(A2:A6)-COUNT(A2:A6)*AVERAGE(A2:A6)^2)/(COUNT(A2:A6)-1). - Compare every result with your hand-computed deviation table.
Expected results
| Indicator | Truck A | Truck B |
|---|---|---|
| Mean | 620 | 620 |
| Range | 50 | 500 |
VAR.S |
400 | 40,000 |
STDEV.S |
20 | 200 |
| \(CV\) | 3.23% | 32.26% |
| Shortcut formula | 400 | 40,000 |
If your shortcut cell is off, check that it references the AVERAGE cell rather than a rounded mean you typed in.
9.3 Homework 1 (Excel) — Dispersion for Campus revenue
In wandering-fork.xlsx, filter franchise to “Campus” and compute the range, sample variance, sample standard deviation, and \(CV\) for daily_revenue_eur (120 days). How do they compare to Harbour’s?
How to. Filter franchise to “Campus”, or wrap the range in FILTER(tblFork[daily_revenue_eur], tblFork[franchise]="Campus") inside each function; use MAX−MIN, VAR.S, STDEV.S and STDEV.S/AVERAGE, then repeat with “Harbour” to compare.
Solution
Campus 120-day: range = 1383.74, \(s^2 = 94{,}511.80\), \(s = 307.43\), \(CV = 67.56\%\). Over the same 120 days, Harbour has range = 998.79, \(s = 216.01\), \(CV = 52.24\%\). Campus has higher absolute and relative dispersion than Harbour (range 1384 vs 999, \(CV\) 67.6% vs 52.2%) — consistent with the right-skewed, event-driven revenue pattern.
9.4 Homework 2 (Excel) — Shortcut formula with SUMSQ
Using the Harbour 30-day daily_revenue_eur data, verify the shortcut formula: compute =SUMSQ(range) and =AVERAGE(range), then apply the shortcut formula manually. Confirm it matches =VAR.S(range).
How to. Put =SUMSQ(range), =AVERAGE(range) and =COUNT(range) in three cells, type the shortcut formula pointing to those three cells, and put =VAR.S(range) in a fourth cell to compare.
Solution
SUMSQ = 12,389,403.98, AVERAGE = 628.3417, \(n=30\). Shortcut: \((12{,}389{,}403.98 - 30 \times 628.3417^2) / 29 =
18{,}793.33\), matching VAR.S exactly (point to the AVERAGE cell rather than typing 628.34, or rounding shifts the result).
9.5 Homework 3 (Excel) — Payroll skew: dispersion
Open payroll-skew.csv. Compute the range, sample standard deviation, and \(CV\) for the 11 salaries. Why is the \(CV\) so high?
How to. Open the CSV, select the 11 salaries, and use MAX−MIN, STDEV.S and STDEV.S/AVERAGE; format the \(CV\) cell as a percentage.
Solution
Range = 154,834, \(s = 44{,}788\), \(CV = 94.8\%\). The \(CV\) is close to 100% because the standard deviation (driven by the CEO’s €180,000 salary) is almost as large as the mean (€47,231) — a hallmark of extreme right skew where the mean is pulled far above the bulk of the data.
10 Further reading
| Source | Where | What it adds | Time | Verified |
|---|---|---|---|---|
| OpenStax IBS 2e | §2.7 Measures of the Spread of the Data (book pages 82–90) | Sample variance, standard deviation, grouped formulas, Chebyshev, Empirical Rule, CV | ~8 pages | checked against the PDF contents (T046) |
| Khan Academy | Unit Summarizing quantitative data in the Statistics and probability course; look for lessons on variance, standard deviation, and coefficient of variation | Short videos and practice on dispersion measures | ~30–45 min (estimate) | link check pending (T048) |
OpenStax IBS 2e: Alexander Holmes, Barbara Illowsky and Susan Dean, Introductory Business Statistics 2e, OpenStax, Rice University, https://openstax.org/details/books/introductory-business-statistics-2e, licensed under CC BY-NC-SA 4.0. Sections are cited, not reproduced.