Session 7 — Dispersion

Data Description | Part II

Session 6 pinned down “The Wandering Fork”’s central tendency — mean €628.34, median €643.63, mode €666.67 — three closely-clustered landmarks that describe the “typical” day. 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”.

Learning objectives

Compute the range and the interquartile range.
Compute the sample variance and sample standard deviation, using both the definitional and the shortcut (computational) formula.
Compute and interpret the coefficient of variation.
Use the interactive calculator below to see how editing a sample changes its dispersion indicators live.

Theory

Definition — range and interquartile range

The range is \(\max - \min\): simple, but sensitive to a single extreme value. The interquartile range \(IQR = Q_3-Q_1\) (Session 5) measures the spread of only the central 50% of the data, making it more robust to extreme values.

Formula — 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).

Formula — 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).

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 from session-06.qmd’s central tendency companion (below \(35\%\)), a moderate level of relative variability: noticeably variable, but not wildly erratic day to day.

Using Excel

Concept	Excel function / steps
Range	`=MAX(range)-MIN(range)`
Sample variance	`=VAR.S(range)`
Sample standard deviation	`=STDEV.S(range)`
Coefficient of variation	`=STDEV.S(range)/AVERAGE(range)`
Interquartile range	`=QUARTILE.EXC(range,3)-QUARTILE.EXC(range,1)`

Proof / derivation

Derivation — the variance shortcut (computational) formula (click to expand)

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.

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.

Interactive demo — central tendency & dispersion calculator

Edit the six values 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.

#| standalone: true
#| components: [viewer]
#| viewerHeight: 560

from shiny import App, render, ui
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
import numpy as np

BG = '#1C2E22'
FG = '#D2CCC0'

app_ui = ui.page_fluid(
    ui.tags.style(f"""
        body {{ background-color: {BG}; color: {FG}; padding: 12px; font-family: sans-serif; margin:0; }}
        .form-label {{ color: {FG} !important; }}
        .form-range {{ accent-color: steelblue; width: 100%; }}
        .info-box {{
            background: #152119; border-radius: 6px; padding: 10px 14px;
            margin: 8px 0; border-left: 3px solid steelblue;
            font-family: monospace; font-size: 0.9em;
        }}
    """),
    ui.h5("🧮 Central tendency & dispersion calculator", style=f"color:{FG}"),
    ui.p(
        "Move the sliders to change a 6-day revenue sample; the mean, "
        "variance, standard deviation, and CV recompute live.",
        style=f"color:{FG}; font-size:0.88em; margin-bottom:6px;",
    ),
    ui.input_slider("d1", "Day 1 (€):", min=300, max=950, value=460.48, step=1),
    ui.input_slider("d2", "Day 2 (€):", min=300, max=950, value=500.02, step=1),
    ui.input_slider("d3", "Day 3 (€):", min=300, max=950, value=418.65, step=1),
    ui.input_slider("d4", "Day 4 (€):", min=300, max=950, value=785.79, step=1),
    ui.input_slider("d5", "Day 5 (€):", min=300, max=950, value=719.66, step=1),
    ui.input_slider("d6", "Day 6 (€):", min=300, max=950, value=886.56, step=1),
    ui.output_ui("info"),
    ui.output_plot("plot", height="280px"),
)

def server(input, output, session):
    def sample():
        return np.array([input.d1(), input.d2(), input.d3(), input.d4(), input.d5(), input.d6()], dtype=float)

    @output
    @render.ui
    def info():
        x = sample()
        mean = x.mean()
        median = float(np.median(x))
        var = x.var(ddof=1)
        std = x.std(ddof=1)
        cv = std / mean * 100 if mean else 0
        return ui.HTML(f"""
            <div class='info-box'>
                Mean = <strong style='color:darkorange'>{mean:.2f}</strong> &nbsp;|&nbsp;
                Median = <strong style='color:darkorange'>{median:.2f}</strong><br>
                Variance (s²) = {var:.2f} &nbsp;|&nbsp;
                Std. dev. (s) = {std:.2f} &nbsp;|&nbsp;
                CV = {cv:.2f}%
            </div>
        """)

    @output
    @render.plot
    def plot():
        x = sample()
        mean = x.mean()

        fig, ax = plt.subplots(figsize=(7, 3.2))
        fig.patch.set_facecolor(BG)
        ax.set_facecolor(BG)
        for spine in ax.spines.values():
            spine.set_edgecolor(FG)
        ax.tick_params(colors=FG)

        days = np.arange(1, 7)
        ax.scatter(days, x, color='steelblue', zorder=5, s=70)
        ax.axhline(mean, color='darkorange', lw=1.8, linestyle='--', label=f'Mean = {mean:.2f}')
        for d, v in zip(days, x):
            ax.plot([d, d], [mean, v], color=FG, lw=1, alpha=0.5)
        ax.set_xticks(days)
        ax.set_xlabel("Day", color=FG)
        ax.set_ylabel("Revenue (€)", color=FG)
        ax.set_title("Deviations from the mean", color=FG)
        ax.legend(fontsize=8.5, facecolor=BG, edgecolor='none', labelcolor=FG)
        ax.grid(alpha=0.15, color=FG)
        plt.tight_layout()
        return fig

app = App(app_ui, server)

Exercises

Exercise 1 — Shortcut formula (click for solution)

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: \(s^2 = \frac{1}{n-1}\left(\sum x_i^2 - n\bar{x}^2\right) = \frac{1}{4}(2140 - 5\times400) = \frac{1}{4}(2140-2000) = 35\).

Exercise 2 — Comparing variability across units (click for solution)

“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 (click for solution)

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.