5️⃣ Position and boxplot

The business question. “The Wandering Fork” has 120 days of daily revenue data per franchise. The owner wants to know: what is a “typical” day, how spread out are the days, and are there any wildly atypical days? This module teaches you to locate specific points within a distribution — the median, quartiles, and percentiles — and to build the boxplot, the single most information-dense chart in descriptive statistics. We start with two trucks small enough to compute by hand, and only then write the formulas.

1 Theory

1.1 Same median, different weeks

Two Wandering Fork trucks recorded their revenue over the same six days. The numbers are a simplified extract chosen so that you can compute everything by hand.

Day	Truck A “Regular week” (€)	Truck B “Festival week” (€)
1	420	420
2	540	1,100
3	300	300
4	500	500
5	380	380
6	460	460

The two weeks differ on day 2 only. Is Truck B’s €1,100 simply a very good day, or so unusual that a boxplot should flag it? And what would you say about the “typical” day of each truck? Look at the figure before computing anything (start with step 1).

miniA = [420, 540, 300, 500, 380, 460]
miniB = [420, 1100, 300, 500, 380, 460]

miniStats = xs => {
  const s = [...xs].sort((a, b) => a - b);
  const n = s.length;
  const q = p => {
    const r = p * (n + 1), k = Math.floor(r), f = r - k;
    return f === 0 ? s[k - 1] : s[k - 1] + f * (s[k] - s[k - 1]);
  };
  const q1 = q(0.25), me = q(0.5), q3 = q(0.75);
  const iqr = q3 - q1, lo = q1 - 1.5 * iqr, hi = q3 + 1.5 * iqr;
  const inside = s.filter(v => v >= lo && v <= hi);
  return {s, n, q1, me, q3, iqr, lo, hi,
    wl: Math.min(q1, ...inside), wh: Math.max(q3, ...inside),
    out: s.filter(v => v < lo || v > hi)};
}

miniFmt = v => Math.round(v).toLocaleString("en-US")

viewof miniStep = {
  const labels = ["1 · Sorted days", "2 · Quartiles", "3 · Fences and box"];
  const form = html`<form class="ojs-step-picker">Show: ${labels.map((l, i) =>
    html`<label> <input type="radio" name="miniStep" value="${i + 1}"
      ${i === 0 ? "checked" : ""}> ${l} </label>`)}</form>`;
  form.value = 1;
  form.oninput = () => {
    form.value = +form.querySelector("input:checked").value;
  };
  return form;
}

{
  const W = 660, H = 350, BLUE = "steelblue", ORANGE = "darkorange";
  const RED = "#d9534f";
  const x = v => 90 + (v - 200) * (W - 120) / 1000;
  const trucks = [["Truck A", miniStats(miniA), 90],
                  ["Truck B", miniStats(miniB), 220]];
  let g = "";
  trucks.forEach(([name, st, yc]) => {
    g += `<text x="10" y="${yc + 5}" fill="currentColor"
      font-weight="bold">${name}</text>`;
    g += `<line x1="${x(200)}" x2="${x(1200)}" y1="${yc}" y2="${yc}"
      stroke="currentColor" stroke-opacity="0.2"/>`;
    if (miniStep === 3) {
      g += `<line x1="${x(st.wl)}" x2="${x(st.q1)}" y1="${yc}" y2="${yc}"
        stroke="currentColor" stroke-width="1.5"/>
        <line x1="${x(st.q3)}" x2="${x(st.wh)}" y1="${yc}" y2="${yc}"
        stroke="currentColor" stroke-width="1.5"/>
        <rect x="${x(st.q1)}" y="${yc - 25}" width="${x(st.q3) - x(st.q1)}"
        height="50" fill="${BLUE}" fill-opacity="0.35" stroke="currentColor"/>
        <line x1="${x(st.me)}" x2="${x(st.me)}" y1="${yc - 25}"
        y2="${yc + 25}" stroke="${ORANGE}" stroke-width="3"/>`;
      [[st.lo, "lower fence"], [st.hi, "upper fence"]].forEach(([f, t]) => {
        if (f >= 200 && f <= 1200) {
          g += `<line x1="${x(f)}" x2="${x(f)}" y1="${yc - 40}"
            y2="${yc + 40}" stroke="${RED}" stroke-dasharray="5 3"/>
            <text x="${x(f)}" y="${yc - 46}" text-anchor="middle"
            fill="${RED}" font-size="11">${t} ${miniFmt(f)}</text>`;
        }
      });
    }
    if (miniStep === 2) {
      [[st.q1, "Q1", -34], [st.me, "Me", 48], [st.q3, "Q3", -34]]
        .forEach(([v, t, dy]) => {
          g += `<line x1="${x(v)}" x2="${x(v)}" y1="${yc - 28}"
            y2="${yc + 28}" stroke="${ORANGE}" stroke-width="2"/>
            <text x="${x(v)}" y="${yc + dy}" text-anchor="middle"
            fill="${ORANGE}" font-size="12">${t} = ${miniFmt(v)}</text>`;
        });
    }
    st.s.forEach((v, i) => {
      const isOut = miniStep === 3 && st.out.includes(v);
      g += `<circle cx="${x(v)}" cy="${yc}" r="6"
        fill="${isOut ? "none" : BLUE}" stroke="${isOut ? RED : BLUE}"
        stroke-width="2"/>`;
      if (miniStep === 1) {
        g += `<text x="${x(v)}" y="${yc - 14}" text-anchor="middle"
          fill="currentColor" font-size="11">${i + 1}</text>`;
      }
    });
  });
  for (let t = 200; t <= 1200; t += 100) {
    g += `<line x1="${x(t)}" x2="${x(t)}" y1="${H - 50}" y2="${H - 44}"
      stroke="currentColor"/>
      <text x="${x(t)}" y="${H - 28}" text-anchor="middle"
      fill="currentColor" font-size="12">€${t}</text>`;
  }
  g += `<line x1="${x(200)}" x2="${x(1200)}" y1="${H - 50}" y2="${H - 50}"
    stroke="currentColor"/>`;
  return html`<svg viewBox="0 0 ${W} ${H}" width="100%" role="img"
    aria-label="Daily revenue of Truck A and Truck B on a number line">
    ${g}</svg>`;
}

{
  const line = (name, st) => miniStep === 1
    ? `${name}: sorted days ${st.s.map(miniFmt).join(", ")}; n = ${st.n}, so the middle is at position (n + 1)/2 = ${(st.n + 1) / 2}.`
    : miniStep === 2
    ? `${name}: ranks ${[0.25, 0.5, 0.75].map(p => p * (st.n + 1)).join(", ")} give Q1 = ${miniFmt(st.q1)}, Me = ${miniFmt(st.me)}, Q3 = ${miniFmt(st.q3)}.`
    : `${name}: IQR = ${miniFmt(st.iqr)}; fences ${miniFmt(st.lo)} and ${miniFmt(st.hi)}; outliers: ${st.out.length ? st.out.map(miniFmt).join(", ") : "none"}.`;
  return html`<p><em>${line("Truck A", miniStats(miniA))}<br>${line("Truck B", miniStats(miniB))}</em></p>`;
}

1.2 Median

Try it. Sort Truck A’s six days from smallest to largest, then find the middle of the list.

1. At which position (or positions) of the sorted list is the middle?
2. What is the median of Truck A?
3. Now drop day 6 (460): the median of the remaining five days?

Check your answer

1. Sorted: 300, 380, 420, 460, 500, 540. With \(n = 6\) the middle lies between positions 3 and 4, that is at position \((6+1)/2 = 3.5\).
2. The average of the 3rd and 4th values: \((420+460)/2 = 440\).
3. Sorted: 300, 380, 420, 500, 540. With \(n = 5\) the middle is position \((5+1)/2 = 3\), so the median is 420.

Truck B’s median is also 440: its extra-large day 2 is simply the largest value and does not move the middle.

See it. Step 1 of the figure sorts each truck’s days and numbers them; the median sits halfway between the dots ranked 3 and 4.

In plain words: sort the days, then take the one in the middle (or the average of the two in the middle).

NoteDefinition — median

The median \(Me\) is the value that splits an ordered series into two equal halves: at least 50% of observations are \(\leq Me\) and at least 50% are \(\geq Me\). For \(n\) ordered values, if \(n\) is odd the median is the value at position \((n+1)/2\); if \(n\) is even, it is the average of the two values at positions \(n/2\) and \(n/2+1\).

1.3 Quartiles (exclusive method)

The median cuts the days in two halves. Quartiles cut them in four quarters.

Try it. Use Truck A’s sorted days (300, 380, 420, 460, 500, 540, \(n = 6\)). For each quartile, compute its rank \(p\,(n+1)\), say between which two sorted values it falls, and interpolate.

Quartile	\(p\)	Rank \(p\,(n+1)\)	Between values	Value (€)
\(Q_1\)	0.25	?	?	?
\(Q_2 = Me\)	0.5	?	?	?
\(Q_3\)	0.75	?	?	?

Your turn: compute the three quartiles of Truck B. What do you notice?

Check your answer

Truck A

Quartile	\(p\)	Rank \(p\,(n+1)\)	Between values	Value (€)
\(Q_1\)	0.25	1.75	1st (300) and 2nd (380)	\(300 + 0.75\times 80 = 360\)
\(Q_2 = Me\)	0.5	3.5	3rd (420) and 4th (460)	\(420 + 0.5\times 40 = 440\)
\(Q_3\)	0.75	5.25	5th (500) and 6th (540)	\(500 + 0.25\times 40 = 510\)

Truck B (sorted: 300, 380, 420, 460, 500, 1,100): the ranks are the same, so \(Q_1 = 360\) and \(Me = 440\) exactly as for Truck A, and \(Q_3 = 500 + 0.25\times(1100-500) = 650\).

The €560 jump of day 2 (540 → 1,100) moved the maximum a lot, but \(Q_1\) and the median did not move at all, and \(Q_3\) moved by 140.

See it. Step 2 of the figure marks the three quartiles on each truck’s sorted days; the rank tells you where to read them.

In plain words: to find \(Q_1\), compute \(0.25 \times (n+1)\); if the result is 7.75, take the value 75% of the way between the 7th and 8th smallest observations. This is exactly the position formula from OpenStax §2.2, and it matches Excel’s QUARTILE.EXC.

TipFormula — quartiles (exclusive method)

The three quartiles \(Q_1, Q_2 (=Me), Q_3\) split an ordered series into four equal quarters. Following the exclusive convention used throughout this course (Excel’s QUARTILE.EXC), the position of \(Q_p\) (\(p=0.25, 0.5, 0.75\)) in the ordered data is \[ \text{rank}(Q_p) = p\,(n+1), \] interpolating linearly between the two surrounding observations when the rank is not a whole number.

Notation decoder: quartiles (exclusive method)

Symbol	Read aloud	In our example (Truck A)
\(n\)	“n”	6 days
\(p\)	“p”	0.25, 0.5 or 0.75
\(\text{rank}(Q_p)\)	“the rank of Q sub p”	1.75, 3.5, 5.25
\(p\,(n+1)\)	“p times n plus 1”	\(0.25 \times 7 = 1.75\)
\(Q_1, Q_2, Q_3\)	“Q one, Q two, Q three”	360, 440, 510 €
\(Q_2 (=Me)\)	“Q two, which is the median”	440 €

Whole formula as a sentence: “the quartile at fraction \(p\) sits at rank \(p\) times \(n+1\) in the sorted data; when the rank is not a whole number, take the value that fraction of the way between the two neighbouring observations.”

1.4 Interquartile range and outlier fences

Try it. For each truck, use the quartiles you just found.

	\(Q_1\)	\(Q_3\)	\(Q_3 - Q_1\)	\(Q_1 - 1.5\,(Q_3-Q_1)\)	\(Q_3 + 1.5\,(Q_3-Q_1)\)
Truck A	360	510	?	?	?
Truck B	360	650	?	?	?

Is any day of either truck outside its interval of fences?

Check your answer

	\(Q_1\)	\(Q_3\)	\(Q_3 - Q_1\)	Lower fence	Upper fence
Truck A	360	510	150	\(360 - 225 = 135\)	\(510 + 225 = 735\)
Truck B	360	650	290	\(360 - 435 = -75\)	\(650 + 435 = 1{,}085\)

• Truck A: all days lie between 300 and 540, inside \([135, 735]\): no outlier.
• Truck B: day 2 (1,100) is above 1,085: a high outlier. The other five days are inside \([-75, 1085]\).

Note that the outlier itself pushed the upper fence up (735 → 1,085); even so, it still sticks out.

See it. Step 3 of the figure draws the two fences as dashed lines: Truck B’s €1,100 lies beyond the upper fence, Truck A’s days all lie inside. (Truck B’s lower fence, −75, is off the axis.)

In plain words: the IQR measures the spread of the central 50% of the data. The fences extend 1.5 IQRs beyond the box — anything further out is an outlier.

TipFormula — interquartile range and outlier fences

\[ IQR = Q_3 - Q_1, \qquad \text{lower fence} = Q_1 - 1.5\,IQR, \qquad \text{upper fence} = Q_3 + 1.5\,IQR. \] Any observation outside \([\text{lower fence}, \text{upper fence}]\) is flagged as a statistical outlier on the boxplot.

Notation decoder: interquartile range and outlier fences

Symbol	Read aloud	In our example
\(IQR\)	“inter-quartile range”	150 € (A), 290 € (B)
\(Q_3 - Q_1\)	“Q three minus Q one”	\(510-360\) (A), \(650-360\) (B)
\(1.5\,IQR\)	“one and a half IQRs”	225 € (A), 435 € (B)
lower fence	“the lower fence”	135 € (A), −75 € (B)
upper fence	“the upper fence”	735 € (A), 1,085 € (B)
\([\text{lower fence}, \text{upper fence}]\)	“the interval from lower fence to upper fence”	\([135, 735]\) (A), \([-75, 1085]\) (B)

Whole formula as a sentence: “the IQR is the distance from the first to the third quartile; the fences lie one and a half IQRs below \(Q_1\) and above \(Q_3\), and any value outside them is an outlier.”

1.5 Boxplot

Try it. Complete the five-number summary of each truck, then decide which days are plotted as separate dots.

	\(\min\)	\(Q_1\)	\(Me\)	\(Q_3\)	\(\max\)	Outliers
Truck A	?	?	?	?	?	?
Truck B	?	?	?	?	?	?

Check your answer

	\(\min\)	\(Q_1\)	\(Me\)	\(Q_3\)	\(\max\)	Outliers
Truck A	300	360	440	510	540	none
Truck B	300	360	440	650	1,100	1,100

Truck A’s whiskers reach its minimum (300) and maximum (540). Truck B’s box runs from 360 to 650 and its €1,100 day is drawn as an individual dot. With only six days, the largest day inside the fences (500) is already inside the box, so Truck B has no visible upper whisker; with more days a whisker normally extends beyond the box.

See it. Step 3 of the figure is exactly this picture: a box from \(Q_1\) to \(Q_3\), an orange line at the median, whiskers, and any outlier as an open dot.

In plain words: a boxplot is the five-number summary drawn on a number line — the box holds the central half of the days, and dots flag the days that are too far away to belong to the pack.

NoteDefinition — boxplot

A boxplot draws the five-number summary (\(\min, Q_1, Me, Q_3, \max\), or the fences if there are outliers) as a box (from \(Q_1\) to \(Q_3\), split by a line at \(Me\)) with whiskers extending to the most extreme non-outlier values; outliers are marked individually.

1.6 Percentiles

Quartiles are three special cut points. A percentile is the same idea for any fraction \(p\) of the days.

Try it. Truck C recorded four Saturdays: 320, 400, 480, 640 € (\(n = 4\), so the rank is \(p \times 5\)).

1. Find the ranks of \(P_{20}\), \(P_{40}\), \(P_{60}\) and \(P_{80}\), and read their values.
2. Find \(P_{30}\) and \(P_{70}\) by interpolation.
3. What rank would \(P_{90}\) have? What does it mean?

Check your answer

1. Ranks 1, 2, 3, 4, so \(P_{20}=320\), \(P_{40}=400\), \(P_{60}=480\), \(P_{80}=640\).
2. \(P_{30}\) has rank 1.5: \(320 + 0.5\times 80 = 360\). \(P_{70}\) has rank 3.5: \(480 + 0.5\times 160 = 560\).
3. Rank \(0.9 \times 5 = 4.5\) — beyond the 4th (last) value. With four observations the exclusive method cannot place \(P_{90}\) (Excel’s PERCENTILE.EXC returns an error); the largest percentile it can give is \(P_{80}\).

See it. Each percentile is a position on the sorted line; \(P_{30}\) and \(P_{70}\) fall between two values.

Sorted value (€)	320	400	480	640
Rank	1	2	3	4
Percentile at this rank	\(P_{20}\)	\(P_{40}\)	\(P_{60}\)	\(P_{80}\)

Between ranks 1 and 2 lies \(P_{30}\) (360); between ranks 3 and 4 lies \(P_{70}\) (560).

In plain words: the \(p\)-th percentile is the value below which a fraction \(p\) of the days lie; to find it, go to rank \(p\,(n+1)\) in the sorted data, interpolating if the rank is not whole.

TipFormula — percentiles (exclusive method)

For any percentile \(P_p\) (\(0 < p < 1\)), the position in the ordered data is \(\text{rank}(P_p) = p\,(n+1)\), interpolating when the rank is not an integer. In Excel: =PERCENTILE.EXC(range, p).

Notation decoder: percentiles (exclusive method)

Symbol	Read aloud	In our example (Truck C)
\(P_p\)	“the p-th percentile”	\(P_{30} = 360\) €
\(p\)	“p”	0.3 for \(P_{30}\)
\(0 < p < 1\)	“p strictly between 0 and 1”	0.9 is allowed, but its rank is out of range for \(n=4\)
\(\text{rank}(P_p)\)	“the rank of P sub p”	1.5 for \(P_{30}\)
\(p\,(n+1)\)	“p times n plus 1”	\(0.3 \times 5 = 1.5\)
\(n\)	“n”	4 Saturdays

Whole formula as a sentence: “the percentile at fraction \(p\) sits at rank \(p\) times \(n+1\) in the sorted data, interpolating when that rank is not a whole number.”

1.7 Grouped-data interpolation

Sometimes you only have a class table, not the raw days. Truck A’s six days, grouped in classes of width 100 (as in Module 3), look like this:

Class (€)	Count	Cumulative count
[300, 400[	2	2
[400, 500[	2	4
[500, 600[	2	6

Try it. For each of \(Q_1\), \(Me\) and \(Q_3\), compute the target cumulative position \(p\cdot n\), find the class that contains it, and interpolate linearly inside that class.

	Target \(p\cdot n\)	Class	\(N_{\text{before}}\)	\(n_{\text{class}}\)	Estimate (€)
\(Q_1\)	?	?	?	?	?
\(Me\)	?	?	?	?	?
\(Q_3\)	?	?	?	?	?

Check your answer

	Target \(p\cdot n\)	Class	\(N_{\text{before}}\)	\(n_{\text{class}}\)	Estimate (€)
\(Q_1\)	1.5	[300, 400[	0	2	\(300 + \frac{1.5-0}{2}\times100 = 375\)
\(Me\)	3	[400, 500[	2	2	\(400 + \frac{3-2}{2}\times100 = 450\)
\(Q_3\)	4.5	[500, 600[	4	2	\(500 + \frac{4.5-4}{2}\times100 = 525\)

Compare with the raw-data values 360, 440 and 510: close, but not identical. The class table has forgotten where the days sit inside each class.

See it. On the cumulative-count column, the target position falls between two cumulative counts; the estimate is the same fraction of the way across that class’s width.

In plain words: find which class contains the target position, then interpolate proportionally across that class’s width.

TipFormula — quantile in grouped continuous data

For a continuous variable grouped into classes, we cannot read \(Q_p\) directly from a list of raw values — we interpolate within the class that contains the target cumulative frequency. If class \([L, L+h[\) has lower cumulative count \(N_{\text{before}}\) (from all earlier classes) and count \(n_{\text{class}}\) of its own, and we seek the value at cumulative position \(p\cdot n\), linear interpolation across the class width \(h\) gives \[ Q_p = L + \frac{p\,n - N_{\text{before}}}{n_{\text{class}}}\; h. \]

Notation decoder: quantile in grouped continuous data

Symbol	Read aloud	In our example (median of Truck A)
\(Q_p\)	“Q sub p”	\(Me\), so \(p = 0.5\): estimate 450 €
\([L, L+h[\)	“the class from L up to, not including, L plus h”	[400, 500[
\(L\)	“L”	400 € (lower bound of the class)
\(h\)	“h”	100 € (class width)
\(p\cdot n\)	“p times n”	\(0.5 \times 6 = 3\)
\(N_{\text{before}}\)	“N before”	2 days in earlier classes
\(n_{\text{class}}\)	“n of the class”	2 days in the class

Whole formula as a sentence: “start at the lower bound of the class that contains the target position, and move across the class by the fraction of its days still needed to reach the target, times the class width.”

2 Worked example

Using “The Wandering Fork”’s daily_revenue_eur (\(n=30\)), verified in data-model.md:

Statistic	Value
Minimum	407.59
\(Q_1\) (exclusive)	506.61
\(Q_2 = Me\)	643.63
\(Q_3\) (exclusive)	729.99
Maximum	886.56
\(IQR = Q_3-Q_1\)	223.39
Lower fence	171.52
Upper fence	1065.07

Step 1 — locate \(Q_1\). With \(n=30\), rank\((Q_1) = 0.25\times 31 = 7.75\): interpolate 75% of the way between the 7th and 8th smallest values of the ordered daily_revenue_eur series: \(500.02 + 0.75\times(508.80-500.02) = 506.605 \approx 506.61\).

Step 2 — locate \(Q_3\). rank\((Q_3) = 0.75\times 31 = 23.25\): interpolate between the 23rd and 24th smallest values: \(725.66 + 0.25\times(742.99-725.66) = 729.9925 \approx 729.99\).

Step 3 — fences and outliers. Keeping the unrounded quartiles, \(IQR = 729.9925-506.605 = 223.3875 \approx 223.39\); lower fence \(= 506.605 - 1.5(223.3875) \approx 171.52\); upper fence \(= 729.9925+1.5(223.3875) \approx 1065.07\). The observed minimum (407.59) and maximum (886.56) both lie inside the fences — “The Wandering Fork” has zero outliers in its daily revenue, itself a useful finding: 30 days of trading, however variable, never produced a wildly atypical day.

Step 4 — draw the boxplot. Box from 506.61 to 729.99, median line at 643.63, whiskers extending to the true min (407.59) and max (886.56) since neither is an outlier.

Step 5 — grouped-data check. Using the class table from Module 3 to estimate \(Q_1\) (target cumulative position \(0.25\times30=7.5\)): cumulative count reaches 6 after [400,500[ and 12 after [500,600[, so \(Q_1\) lies in [500,600[ with \(N_{\text{before}}=6\), \(n_{\text{class}}=6\), \(h=100\): \[ Q_1 \approx 500 + \frac{7.5-6}{6}\times100 = 525. \] This is close to, but not identical to, the raw-data value of 506.61 — the difference is the same class-grouping approximation error discussed in Module 4.

3 Proof / derivation

CautionDerivation — locating a quantile in grouped continuous data (click to expand)

For a continuous variable grouped into classes, we cannot read \(Q_p\) directly from a list of raw values — we interpolate within the class that contains the target cumulative frequency. If class \([L, L+h[\) has lower cumulative count \(N_{\text{before}}\) (from all earlier classes) and count \(n_{\text{class}}\) of its own, and we seek the value at cumulative position \(p\cdot n\), linear interpolation across the class width \(h\) gives \[ Q_p = L + \frac{p\,n - N_{\text{before}}}{n_{\text{class}}}\; h. \]

Sanity check with our class table (Module 3): to locate the median (\(p=0.5\), target position \(0.5\times30=15\)) in daily_revenue_eur’s classes, the cumulative count reaches 12 after [500,600[ and 20 after [600,700[, so the median lies in [600,700[ with \(N_{\text{before}}=12\), \(n_{\text{class}}=8\), \(h=100\): \[ Me \approx 600 + \frac{15-12}{8}\times 100 = 637.5, \] close to (though not identical to, because this formula uses grouped midpoint approximations while the true value uses the raw 30 observations) the raw-data median of \(643.63\) computed above — the small gap is exactly the information lost when data is grouped into classes.

4 Visual intuition

The boxplot is the five-number summary drawn as a picture: the box’s width is the IQR (how spread out the central 50% of days are), the line’s position inside the box shows whether the median sits closer to \(Q_1\) or to \(Q_3\) (a first hint of asymmetry, formalised in Module 7), and the whisker lengths show how far the more extreme (but non-outlier) days stretch beyond the box.

Step 3 of the two-truck figure shows it on six days: Truck A’s box is 150 € wide, Truck B’s is 290 € wide, and the festival day is an open dot far to the right of everything else.

5 Interactive demo — boxplot from quartiles

Drag each slider to set the five-number summary and watch the boxplot redraw. Starting values are “The Wandering Fork”’s verified statistics.

{
  const start = [["vmin", "Minimum", 407.59], ["q1", "Q1", 506.61],
                 ["med", "Median", 643.63], ["q3", "Q3", 729.99],
                 ["vmax", "Maximum", 886.56]];
  const root = html`<div class="ojs-demo">
    <p>Move the sliders (kept in min ≤ Q1 ≤ median ≤ Q3 ≤ max order) to
      see the boxplot and IQR/fences update live.</p>
    <form>${start.map(([name, label, v]) => html`<div><label>${label}:
      <input type="range" name="${name}" min="300" max="900" step="0.01"
        value="${v}"> <output>${v.toFixed(2)}</output></label></div>`)}</form>
    <p class="ojs-demo-info"></p>
    <div class="ojs-demo-chart"></div>
  </div>`;
  const form = root.querySelector("form");

  function redraw() {
    const inputs = [...form.querySelectorAll("input")];
    inputs.forEach(el => {
      el.parentNode.querySelector("output").textContent = el.valueAsNumber.toFixed(2);
    });
    const [vmin, q1, med, q3, vmax] = inputs.map(el => el.valueAsNumber)
      .sort((a, b) => a - b);
    const iqr = q3 - q1, lo = q1 - 1.5 * iqr, hi = q3 + 1.5 * iqr;
    const outlier = vmin < lo || vmax > hi;
    root.querySelector(".ojs-demo-info").innerHTML =
      `IQR = ${iqr.toFixed(2)} · Lower fence = ${lo.toFixed(2)} ·
       Upper fence = ${hi.toFixed(2)}<br>${outlier
         ? "⚠️ min/max beyond fences — would be plotted as outliers"
         : "✅ no outliers"}`;

    const W = 660, Ht = 220, L = 30, R = 30;
    const x = v => L + (v - 300) * (W - L - R) / 600;
    const yc = 105, bh = 50;
    let g = `<text x="${W / 2}" y="18" text-anchor="middle" fill="currentColor"
      font-size="14">Boxplot from the five-number summary</text>`;
    for (let v = 300; v <= 900; v += 100) {
      g += `<line x1="${x(v)}" x2="${x(v)}" y1="40" y2="170"
        stroke="currentColor" stroke-opacity="0.15"/>
        <text x="${x(v)}" y="190" text-anchor="middle" fill="currentColor"
        font-size="11">${v}</text>`;
    }
    g += `<line x1="${x(vmin)}" x2="${x(q1)}" y1="${yc}" y2="${yc}"
        stroke="currentColor" stroke-width="1.5"/>
      <line x1="${x(q3)}" x2="${x(vmax)}" y1="${yc}" y2="${yc}"
        stroke="currentColor" stroke-width="1.5"/>
      <line x1="${x(vmin)}" x2="${x(vmin)}" y1="${yc - 15}" y2="${yc + 15}"
        stroke="currentColor" stroke-width="1.5"/>
      <line x1="${x(vmax)}" x2="${x(vmax)}" y1="${yc - 15}" y2="${yc + 15}"
        stroke="currentColor" stroke-width="1.5"/>
      <rect x="${x(q1)}" y="${yc - bh / 2}" width="${x(q3) - x(q1)}"
        height="${bh}" fill="steelblue" stroke="currentColor"/>
      <line x1="${x(med)}" x2="${x(med)}" y1="${yc - bh / 2}"
        y2="${yc + bh / 2}" stroke="darkorange" stroke-width="3"/>
      <text x="${W / 2}" y="212" text-anchor="middle" fill="currentColor"
        font-size="12">Value</text>`;
    root.querySelector(".ojs-demo-chart").innerHTML =
      `<svg viewBox="0 0 ${W} ${Ht}" width="100%" role="img"
        aria-label="Boxplot from the five-number summary">${g}</svg>`;
  }
  form.addEventListener("input", redraw);
  redraw();
  return root;
}

6 Exercises

All exercises below are done by hand; the Excel ones are in the homework.

Exercise 1 (Median position) For an ordered series of \(n=24\) values, at which position(s) do you read the median?

Solution

\(n=24\) is even, so the median is the average of the values at positions \(n/2=12\) and \(n/2+1=13\).

Exercise 2 (Outlier check) A single unusually bad day of \(\text{daily\_revenue\_eur} = 100\) is added to “The Wandering Fork”’s data. Using the fences computed above (171.52 / 1065.07, before recomputing the quartiles for the new \(n=31\) sample), would this new day be flagged as an outlier?

Solution

Yes: \(100 < 171.52\) (the lower fence), so it would be flagged as a low outlier — a day that badly underperformed relative to the rest of the trading period.

Exercise 3 (Grouped-data interpolation) Using the class table from Module 3, estimate \(Q_1\) for daily_revenue_eur (target cumulative position \(0.25\times30=7.5\)).

Solution

Cumulative count reaches 6 after [400,500[ and 12 after [500,600[, so \(Q_1\) lies in [500,600[ with \(N_{\text{before}}=6\), \(n_{\text{class}}=6\), \(h=100\): \[ Q_1 \approx 500 + \frac{7.5-6}{6}\times100 = 525. \] This is close to, but not identical to, the raw-data value of 506.61 — the difference is the same class-grouping approximation error discussed in this module’s proof.

Exercise 4 (Formula to words)  

1. Read \(\text{rank}(Q_p) = p\,(n+1)\) aloud as one sentence. For \(n=9\) and \(p=0.75\), which rank do you get?

2. Read \(Q_3 + 1.5\,IQR\) aloud. Which indicator is it? Compute it for the values 10, 20, 30, 40, 50 (\(n=5\): \(Q_1 = 15\), \(Q_3 = 45\)).

Solution

1. “The position of the quartile (or percentile) at fraction \(p\) is \(p\) times \(n+1\).” For \(n=9\): \(0.75\times10 = 7.5\), halfway between the 7th and 8th values.

2. “Go one and a half IQRs above the third quartile.” It is the upper fence. \(IQR = 45-15 = 30\), so the upper fence is \(45 + 1.5\times 30 = 90\).

Exercise 5 (Words to formula) Write in symbols:

1. “the gap between the third and the first quartile”;
2. “the rank of the 30th percentile among \(n\) values”;
3. “a value \(x\) is a high outlier”;
4. “the median of \(n\) values when \(n\) is odd”.

Solution

1. \(IQR = Q_3 - Q_1\).
2. \(\text{rank}(P_{0.3}) = 0.3\,(n+1)\).
3. \(x > Q_3 + 1.5\,IQR\).
4. The value at position \((n+1)/2\).

7 Common mistakes

WarningWatch out

• Using QUARTILE.INC instead of QUARTILE.EXC. The course standard is the exclusive method (position \((n+1)p\)); mixing them gives inconsistent quartiles, especially on small samples.
• Forgetting that percentiles need \(0 < p < 1\) for PERCENTILE.EXC. p=0 or p=1 will error — use MIN/MAX for the extremes.
• Confusing the fence rule with standard deviations. The 1.5×IQR rule is for boxplots; the “2 standard deviations” rule is for normal distributions — they are different tools.
• Reading the median position as \(n/2\) instead of \((n+1)/2\). For odd \(n\), \((n+1)/2\) gives the exact middle position; \(n/2\) would be off by 0.5.
• Using grouped-data interpolation when raw data is available. The grouped formula is an approximation; prefer the raw-data quantile when you have the full dataset.

8 Exit check

Five minutes, no calculator needed except for question 4.

1. In one sentence, what does the IQR measure?
2. Why does one extreme day change the range a lot, but hardly move \(Q_1\) and the median?
3. Read \(\text{rank}(Q_p) = p\,(n+1)\) aloud as a sentence.
4. Compute \(Q_1\), \(Q_3\) and the \(IQR\) (exclusive method) for the five values 10, 14, 18, 22, 50. Is 50 an outlier?
5. For \(n=19\) ordered values, at which rank do you read \(Q_1\)?

Answers

1. The spread of the central 50% of the days: the distance between the first and the third quartile.
2. The range uses only the two most extreme values, so it follows an extreme day completely. The quartiles and the median depend on ranks in the middle of the sorted list, so a single day at the end barely changes them.
3. “The position of the quartile at fraction \(p\) is \(p\) times \(n+1\) in the sorted data.”
4. \(n=5\): rank\((Q_1) = 1.5\), so \(Q_1 = (10+14)/2 = 12\); rank\((Q_3) = 4.5\), so \(Q_3 = (22+50)/2 = 36\); \(IQR = 24\). Upper fence \(= 36 + 36 = 72\), so 50 is not an outlier.
5. \(0.25\times(19+1) = 5\): exactly the 5th smallest value, no interpolation.

9 Homework — Excel lab

Reproduce, in Excel, every indicator you computed by hand in class, with MEDIAN, QUARTILE.EXC, PERCENTILE.EXC, MIN and MAX, then apply them to the full dataset and build the boxplots.

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 6. Bring your Step 0 results and any question on Homework 1–3.

9.1 Using Excel

The sheet harbour_30d of wandering-fork.xlsx has the data. The complete list of function names, with their French equivalents, is on the Excel functions page.

Concept	Excel function / steps
Median	=MEDIAN(range)
Quartiles (exclusive)	=QUARTILE.EXC(range, 1) / (range, 2) / (range, 3)
Arbitrary percentile	=PERCENTILE.EXC(range, p), e.g. p=0.9 for \(P_{90}\)
IQR	=QUARTILE.EXC(range,3)-QUARTILE.EXC(range,1)
Boxplot	Select the data → Insert → Statistical Chart → Box and Whisker

Why QUARTILE.EXC? The course follows the \((n+1)\times p\) position formula (OpenStax §2.2), which matches Excel’s exclusive method. QUARTILE.INC (inclusive) gives different values on small samples — do not mix them.

For the full table (120 days per franchise) use the Excel Table tblFork and add the franchise as one more condition, for example =QUARTILE.EXC(IF(tblFork[franchise]="Harbour",tblFork[daily_revenue_eur]),1) (entered as an array formula in older Excel, or use FILTER in newer versions).

9.2 Step 0 — check your class results

1. Type Truck A’s six values (420, 540, 300, 500, 380, 460) in A2:A7, Truck B’s (420, 1100, 300, 500, 380, 460) in B2:B7 and Truck C’s four Saturdays (320, 400, 480, 640) in C2:C5.
2. Below column A, compute =MEDIAN(A2:A7), =QUARTILE.EXC(A2:A7,1), =QUARTILE.EXC(A2:A7,3), the IQR and the two fences (with =Q1-1.5*IQR and =Q3+1.5*IQR pointing to your cells). Repeat for column B.
3. For Truck C, compute =PERCENTILE.EXC(C2:C5,0.3), =PERCENTILE.EXC(C2:C5,0.7) and =PERCENTILE.EXC(C2:C5,0.9).
4. Compare every result with your hand computation.

Expected results

Indicator	Truck A	Truck B
MEDIAN	440	440
QUARTILE.EXC(…,1)	360	360
QUARTILE.EXC(…,3)	510	650
IQR	150	290
Lower fence	135	−75
Upper fence	735	1,085

Truck C	\(P_{30}\)	\(P_{70}\)	\(P_{90}\)
PERCENTILE.EXC	360	560	#NUM!

#NUM! is expected for \(P_{90}\): with \(n=4\) its rank (4.5) is out of range.

9.3 Homework 1 (Excel) — Build the boxplot for Campus

In wandering-fork.xlsx, filter franchise to “Campus” and build a boxplot for daily_revenue_eur (120 days). What are the five-number summary and fences? Are there outliers?

How to. Filter franchise to “Campus”, select the revenue column, use Insert → Statistical Chart → Box and Whisker, and read the five numbers with MIN, QUARTILE.EXC and MAX.

Solution

Campus 120-day: min = 23.85, Q1 = 236.07, median = 390.39, Q3 = 598.60, max = 1407.59. IQR = 362.54. Lower fence = -307.74, upper fence = 1142.40. Three days exceed the upper fence (1335.48, 1362.93 and 1407.59), so there are three high outliers and no low one — consistent with Campus’s right-skewed revenue (skewness = 1.00).

9.4 Homework 2 (Excel) — Percentile \(P_{90}\) for Harbour revenue

Compute \(P_{90}\) for Harbour’s 120-day daily_revenue_eur using PERCENTILE.EXC. What does this value mean?

How to. Wrap the revenue column in FILTER for “Harbour” inside PERCENTILE.EXC, with p=0.9.

Solution

A filter on the sheet does not change what PERCENTILE.EXC sees, so restrict the range with FILTER:

=PERCENTILE.EXC(FILTER(tblFork[daily_revenue_eur],
                       tblFork[franchise]="Harbour"), 0.9)

This gives approximately 699.63. It means 90% of Harbour trading days earned €699.63 or less; only the top 10% of days exceeded this amount.

9.5 Homework 3 (Excel) — Compare franchises with side-by-side boxplots

Create side-by-side boxplots for Harbour and Campus daily_revenue_eur (120 days each). What visual differences do you see?

How to. Put the two franchises’ revenues in two adjacent columns and select both before inserting the Box and Whisker chart; compare the boxes, medians and dots.

Solution

The two medians are almost level (~391 vs ~390), but Harbour’s box is narrower (IQR ~299 vs ~363) and Harbour has a single high outlier (€1,050.00, day 80, above its upper fence of 1016.45). Campus’s box is wider, with a long upper whisker and three high outliers (€1,335.48 to €1,407.59) — visually confirming the right skew and higher variability seen in the numbers.

10 Further reading

Source	Where	What it adds	Time	Verified
OpenStax IBS 2e	§2.2 Measures of the Location of the Data (book pages 65–74)	Quartiles, percentiles, IQR, boxplot	~9 pages	checked against the PDF contents (T046)
Khan Academy	Unit Summarizing quantitative data in the Statistics and probability course; look for lessons on quartiles, percentiles, and boxplots	Short videos and practice on position indicators and boxplots	~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.