In-depth learning guide

Five Number Summary, IQR and Outliers Guide

The five-number summary describes position and spread; the common 1.5×IQR rule then provides explicit fences for a first-pass box-plot outlier check.

3 min readStep-by-step methodReviewed 1 Oct 2026Published by LynaraOpen Five Number Summary & Outlier Calculator →

Method

  1. Sort observations and find the median.
  2. Use the median of the lower and upper halves for Q1 and Q3.
  3. Calculate IQR = Q3 − Q1.
  4. Calculate Q1 − 1.5×IQR and Q3 + 1.5×IQR, then flag observations beyond those fences and identify the furthest in-range whiskers.

Worked example

For 1, 2, 3, 4, 5, 100, the median-of-halves method gives Q1 = 2 and Q3 = 5; the upper fence is 9.5, so 100 is flagged and 5 is the upper whisker.

Notes

Open Five Number Summary & Outlier Calculator

Quartile conventions must be stated because software can disagree

The five-number summary contains minimum, first quartile, median, third quartile and maximum. Different statistical packages use different interpolation rules for quartiles, especially in small datasets. Lynara uses a stated median-of-halves convention so the result can be reproduced.

When comparing results with another tool, check the quartile method before assuming one is wrong. Small differences can come entirely from the convention.

Use IQR fences as an outlier flag, not an automatic deletion rule

The interquartile range is Q3 minus Q1. Tukey-style fences at Q1 - 1.5×IQR and Q3 + 1.5×IQR flag unusually distant observations for review. They do not prove that the values are data errors.

Investigate flagged values in context. An extreme measurement can be a valid and important observation, and removing it without a defensible reason can distort the analysis.

Worked support-ticket example: one long case changes the maximum

Take eight fictional support-ticket handling times, already sorted in minutes: 10, 12, 13, 15, 18, 20, 21 and 35. With the median-of-halves convention, the overall median is (15 + 18) ÷ 2 = 16.5. The lower-half median is Q1 = (12 + 13) ÷ 2 = 12.5, and the upper-half median is Q3 = (20 + 21) ÷ 2 = 20.5. Therefore IQR = 20.5 − 12.5 = 8 minutes. The 1.5×IQR fences are 12.5 − 12 = 0.5 and 20.5 + 12 = 32.5 minutes.

The 35-minute case is beyond the upper fence, so a Tukey-style plot flags it for review. The largest observation inside the fences is 21 minutes, which becomes the upper whisker under the common convention; the data maximum remains 35. The five-number summary itself is minimum 10, Q1 12.5, median 16.5, Q3 20.5 and maximum 35. Those are different ideas from the whisker endpoints. A long, complex customer issue may be valid and should not be deleted just because the plot flags it.

Odd-length samples reveal differences between quartile methods

Consider seven ordered values: 2, 4, 5, 7, 9, 10 and 20. Their median is 7. Under the exclude-the-median median-of-halves rule, the lower half is 2, 4, 5 and the upper half is 9, 10, 20. Thus Q1 = 4, Q3 = 10 and IQR = 6. The upper inner fence is 10 + 1.5 × 6 = 19, so 20 is flagged by this particular method. Another software package using interpolation or including the median in both halves may produce different quartiles and a different fence.

When sharing a report, name the quartile rule, include the ordered data for small examples and confirm that the linked calculator and chart use the same convention. A disagreement of a fraction of a unit is not necessarily a programming bug. For skewed business processes, pair the outlier flags with the record's actual cause rather than automatically removing slow cases and reporting an artificially improved average.

Editorial review and limitations

Review status: reviewed under Lynara's editorial policy on 1 Oct 2026. Institution-specific grading rules, assignment instructions and publisher requirements can override a general example.
Verification: see the methodology and sources & verification pages. To report a reproducible error, use the contact page with the page URL, inputs and expected result.

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