Statistical
Calculate Midrange and Tukey Trimean: extreme vs outlier-resistant centers
Compute the Midrange (MIN + MAX) / 2 and contrast it with Tukey's outlier-resistant Trimean (Q1 + 2*Median + Q3) / 4.
Midrange vs Mean vs Tukey Trimean on skewed data
The Midrange (MIN(Range) + MAX(Range)) / 2 is the exact midpoint between the lowest and highest observations—useful in temperature logs ((DailyLow + DailyHigh) / 2) and machining tolerance centers, but highly sensitive to a single extreme spike.
Tukey's Trimean (Q1 + 2 * Median + Q3) / 4 blends the median with the 25th and 75th quartiles, capturing distribution asymmetry and usually reducing sensitivity to extreme observations compared with the mean.
| Estimator | Formula on [12, 15, 18, 22, 93] | Result |
|---|---|---|
| Midrange | =(MIN + MAX) / 2 = (12 + 93) / 2 | 52.50 (pulled by 93) |
| Arithmetic Mean | =AVERAGE(A2:A6) = 160 / 5 | 32.00 |
| Tukey Trimean | =(15 + 2*18 + 22) / 4 | 18.25 (robust) |
| Median | =MEDIAN(A2:A6) | 18.00 |
Midrange and Tukey Trimean formulas
Use Midrange when extreme bounds define physical limits; use Trimean when summarizing skewed salaries, cycle times, or housing prices.
=(MIN(A2:A6) + MAX(A2:A6)) / 2Tukey's quartile-weighted Trimean
Weights the median twice alongside Q1 and Q3.
=(QUARTILE.INC(A2:A6, 1) + 2 * MEDIAN(A2:A6) + QUARTILE.INC(A2:A6, 3)) / 4Inputs and interpretation
Use a nonempty numeric range. Quartile interpolation and sample size affect sensitivity; extreme observations can affect small-sample quartiles. The midpoint of a daily low and high is not the time-weighted daily average. Choose the statistic to match the question rather than assuming one center is always best.
Try the example
Use the original values or change them to test your understanding.