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Relative Variation, Normality, and a Sales Statistics Case

Key Takeaways

  • CV equals standard deviation divided by mean.

  • Normal empirical percentages require an appropriate distributional model.

  • Investigate an outlier rather than deleting it solely to tighten the result.

Last updated: October 2026

Relative Variation, Normality, and a Sales Statistics Case

The Normal Distribution and the Empirical Rule

The Normal Distribution (Gaussian bell curve) is a symmetrical, continuous probability distribution where data clusters symmetrically around the mean.

For a normally distributed dataset, the Empirical Rule (68-95-99.7 Rule) establishes fixed mathematical boundaries:

  • 68.27% of all observations lie within ±1\pm 1 standard deviation of the mean (xˉ±1s\bar{x} \pm 1s).
  • 95.45% of all observations lie within ±2\pm 2 standard deviations of the mean (xˉ±2s\bar{x} \pm 2s).
  • 99.73% of all observations lie within ±3\pm 3 standard deviations of the mean (xˉ±3s\bar{x} \pm 3s).

Appraisal Outlier Detection

Appraisers use the empirical rule to identify and scrutinize market outliers:

  • Any closed sale falling beyond ±2\pm 2 standard deviations from the sample mean represents an atypical transaction (occurring in less than 5% of cases under normality).
  • Transactions exceeding ±3\pm 3 standard deviations are statistical outliers that warrant rigorous investigation. Appraisers must verify whether such sales involved non-arm's-length motivations, undisclosed seller concessions, unrecorded physical defects, substantial personal property, or erroneous recording data. If non-arm's-length conditions exist, the sale must be discarded or adjusted.

Coefficient of Variation (CV)

While standard deviation quantifies absolute dispersion, it cannot evaluate relative dispersion across different price levels. Comparing a $25,000 standard deviation in a $200,000 starter-home market to a $25,000 standard deviation in a $1,500,000 luxury-home market is misleading.

The Coefficient of Variation (CV) solves this by expressing standard deviation as a percentage of the arithmetic mean:

CV=sxˉ×100%\text{CV} = \frac{s}{\bar{x}} \times 100\%

Market TierMean Sale Price (xˉ\bar{x})Standard Deviation (ss)Calculated CVAppraisal Interpretation
Starter Homes (Subdivision A)$250,000$30,00030,000250,000×100%=12.0%\frac{30,000}{250,000} \times 100\% = \mathbf{12.0\%}Higher relative price dispersion in this sample; investigate differences in features and property mix.
Luxury Homes (Subdivision B)$1,200,000$36,00036,0001,200,000×100%=3.0%\frac{36,000}{1,200,000} \times 100\% = \mathbf{3.0\%}Lower relative price dispersion in this sample; price variation alone does not establish physical homogeneity.

Tip

In appraisal practice, a lower Coefficient of Variation indicates higher comparability and pricing predictability. When selecting between two competing sets of comparable sales, a lower CV describes narrower relative dispersion but does not by itself establish a more representative sample or stronger causal adjustment evidence.


Sample Size, Degrees of Freedom, and Paired Sales Verification

Degrees of Freedom (df=n−1df = n - 1)

When calculating sample variance (s2s^2) and sample standard deviation (ss), statisticians divide the sum of squared deviations by n−1n - 1 rather than nn.

  • Why? The sample mean (xˉ\bar{x}) is already calculated from the sample itself, imposing one mathematical restriction: the deviations (xi−xˉ)(x_i - \bar{x}) must sum to exactly zero. Consequently, only n−1n - 1 deviations are free to vary.
  • Dividing by nn would systematically underestimate the true population variance (a phenomenon known as downward bias). Dividing by n−1n - 1 (Bessel's correction) provides an unbiased estimator of population variance.

Sample Size Considerations in Valuation

  • Small Samples (n<30n < 30): Residential appraisers frequently work with small samples (e.g., 5 to 15 closed sales in a specific neighborhood over 6 months). In small samples, individual outliers exert dramatic distortion, standard errors are larger, and parametric statistical assumptions must be applied cautiously.
  • Paired Sales Verification: Paired sales analysis relies on isolating a single variance between two properties. However, if market noise (random price variation) exceeds the contributory value of the feature, the extracted adjustment is invalid. Use repeated evidence and an appropriate statistical model when making an inferential claim. A difference between two sales is not automatically significant merely because it exceeds a broadly calculated standard error.

Step-by-Step Numerical Walkthrough: Central Tendency & Dispersion

An appraiser collects eight recent closed sales of detached single-family homes in the subject's competitive subdivision:

Closed Sales: $310,000,  $315,000,  $320,000,  $320,000,  $330,000,  $335,000,  $340,000,  $450,000\text{Closed Sales: } \$310,000, \; \$315,000, \; \$320,000, \; \$320,000, \; \$330,000, \; \$335,000, \; \$340,000, \; \$450,000

Step 1: Arrange the Ranked Array

Order the sales from lowest to highest:

  1. $310,000
  2. $315,000
  3. $320,000
  4. $320,000
  5. $330,000
  6. $335,000
  7. $340,000
  8. $450,000 (Sale 8 represents a fully renovated home with an in-ground swimming pool)

Step 2: Calculate the Measures of Central Tendency

  • Sum: ∑x=310,000+315,000+320,000+320,000+330,000+335,000+340,000+450,000=\sum x = 310,000 + 315,000 + 320,000 + 320,000 + 330,000 + 335,000 + 340,000 + 450,000 = $2,720,000
  • Arithmetic Mean: xˉ=$2,720,0008=$340,000\bar{x} = \frac{\$2,720,000}{8} = \mathbf{\$340,000}
  • Median: Because n=8n = 8 is even, average the 4th and 5th values: Median=$320,000+$330,0002=$325,000\text{Median} = \frac{\$320,000 + \$330,000}{2} = \mathbf{\$325,000}
  • Mode: The value $320,000 appears twice; all other values appear once. Mode=$320,000\text{Mode} = \mathbf{\$320,000}

Note

Observe the skewness: Mean ($340,000) > Median ($325,000) > Mode ($320,000). The single $450,000 outlier inflated the mean so severely that the calculated average ($340,000) exceeds six of the eight transactions and equals one! Relying on the mean would substantially overvalue a typical home in this neighborhood. The median of $325,000 describes the center of these ordered sample prices; it is not automatically the subject property’s value.

Step 3: Calculate the Range

R=$450,000−$310,000=$140,000R = \$450,000 - \$310,000 = \mathbf{\$140,000}

Step 4: Calculate Deviations, Variance, and Standard Deviation

Sale (ii)Sale Price (xix_i)Mean (xˉ\bar{x})Deviation (xi−xˉx_i - \bar{x})Squared Deviation (xi−xˉ)2(x_i - \bar{x})^2
Sale 1$310,000$340,000-$30,000900,000,000
Sale 2$315,000$340,000-$25,000625,000,000
Sale 3$320,000$340,000-$20,000400,000,000
Sale 4$320,000$340,000-$20,000400,000,000
Sale 5$330,000$340,000-$10,000100,000,000
Sale 6$335,000$340,000-$5,00025,000,000
Sale 7$340,000$340,000$00
Sale 8$450,000$340,000+$110,00012,100,000,000
Total$2,720,000$014,550,000,000
  1. Sum of Squared Deviations: 14,550,000,00014,550,000,000
  2. Sample Variance (s2s^2): s2=14,550,000,0008−1=14,550,000,0007≈2,078,571,429s^2 = \frac{14,550,000,000}{8 - 1} = \frac{14,550,000,000}{7} \approx \mathbf{2,078,571,429}
  3. Sample Standard Deviation (ss): s=2,078,571,429≈$45,591s = \sqrt{2,078,571,429} \approx \mathbf{\$45,591}
  4. Coefficient of Variation (CV): CV=$45,591$340,000×100%≈13.41%\text{CV} = \frac{\$45,591}{\$340,000} \times 100\% \approx \mathbf{13.41\%}

Step 5: Impact of Removing the Outlier

If the appraiser investigates Sale 8 and determines that its premium pool and custom renovation make it non-comparable to typical neighborhood sales, removing it leaves n=7n = 7 sales:

  • Sum = $2,270,000
  • New Mean xˉ\bar{x} = $2,270,000 ÷ 7 = $324,286
  • New Median = $320,000
  • New Standard Deviation ss ≈ $10,965
  • New CV=10,965324,286×100%≈3.38%\text{CV} = \frac{10,965}{324,286} \times 100\% \approx \mathbf{3.38\%}

Removing the non-comparable outlier reduces relative market dispersion from 13.41% to a narrow 3.38%, describing a narrower price distribution. Whether the retained sales are competitive and representative still requires independent analysis; do not remove a valid sale simply to reduce dispersion.

Test Your Knowledge

An appraiser is evaluating two distinct residential market areas to determine relative pricing stability and comparability. Submarket A has an arithmetic mean sale price of $250,000 and a sample standard deviation of $30,000. Submarket B has an arithmetic mean sale price of $750,000 and a sample standard deviation of $45,000. Using the Coefficient of Variation (CV), which statement correctly evaluates the relative dispersion of these two markets?

A

Submarket B exhibits greater relative price volatility because its absolute standard deviation ($45,000) is $15,000 higher than Submarket A.

B

Both submarkets display equal relative dispersion because both standard deviations fall within standard secondary mortgage market tolerance thresholds.

C

Submarket A displays superior pricing consistency because starter-home neighborhoods naturally produce lower absolute dollar risk.

D

Submarket A exhibits twice the relative price dispersion of Submarket B because its Coefficient of Variation is 12.0% compared to 6.0% for Submarket B.

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