Statistics · Eqora guide
Mean vs median: which number survives an outlier?
Calculate both centers, see what a single extreme value changes, handle an even-sized set, and choose a statistic that answers the actual question.
Mean vs median becomes a real decision when one observation is far from the rest. Imagine five small studio commissions worth 4, 5, 5, 6, and 40 units. Their mean is 12, while their median is 5. Both calculations are correct, but a claim that a 'typical' commission is worth 12 would conceal the fact that four of the five are worth 6 or less. The issue is not which formula is fashionable. It is what question the number must answer.
This guide works through the arithmetic without a calculator or app: sum and divide for the mean, order and locate the middle for the median, then change the high value to see the effect of an outlier. We will also handle an even number of values, repeated values, missing observations, and a case where the mean is actually the more useful measure. The photographs use pottery as a visual metaphor; their sizes are not the numerical data. Use the printed numbers when solving an exercise.
OpenStax explains the mean and median as two measures of center and notes that an extreme value can pull the mean while the median depends on order. A Polish national statistics teaching page likewise introduces the median as the central value of an ordered set. These sources support the method, but they do not choose for you: always state what the data represent, which observations are included, and why your chosen center fits the purpose.
Mean vs median starts with the same five observations
Write the values in one consistent unit: 4, 5, 5, 6, 40. There are five observations, not four distinct numbers; the value 5 appears twice and must count twice. The arithmetic mean is the total divided by the number of observations. Here 4 + 5 + 5 + 6 + 40 = 60, and 60 / 5 = 12. The mean is the equal-share amount: if the total 60 could be redistributed evenly, each of the five observations would receive 12. It need not be one of the observed values.
For the median, order the data from least to greatest. They are already ordered, and the third value in a list of five is 5. Two observations lie before it and two after it, counting repeated values in their proper positions. The median is therefore 5. Do not divide the sum by two, and do not choose the value halfway between 4 and 40. Our guide to interpreting word problems can help separate the requested 'middle observation' from an 'equal share' before any arithmetic begins.
See exactly what the high value changes
Replace the 40 with 7 and keep the other four values. The new list is 4, 5, 5, 6, 7. Its total is 27, so its mean is 27 / 5 = 5.4. Its median remains the third value, 5. Changing only the final observation from 7 to 40 adds 33 to the total. With five observations, the mean rises by 33 / 5 = 6.6, from 5.4 to 12. The median stays 5 because the middle position still contains the same value.
This does not make the 40 an error. It might be a genuine large commission, a special order, or a mis-entered 4.0. Verify the source before removing it. An outlier is a conspicuous observation, not a license to delete inconvenient data. If the question is 'What did these five commissions earn per order on average?', the mean of 12 records the full total, including the large order. If the question is 'What size order would a typical client see?', the median of 5 often describes the center of the four small orders better. State which question you are answering.
Find the median when the count is even
An even-sized data set has no single middle observation. Add a value of 7 to the original set and order all six: 4, 5, 5, 6, 7, 40. The middle positions are the third and fourth, containing 5 and 6. Their arithmetic mean is (5 + 6) / 2 = 5.5, which is the median of this six-value set. The whole-set mean is (4 + 5 + 5 + 6 + 7 + 40) / 6 = 67 / 6, or about 11.17. The two numbers remain very different because of the 40.
Do not choose the third value simply because six divided by two is three; the fourth sits equally close to the center. Also do not average the smallest and largest values: (4 + 40) / 2 = 22 is the midrange, a different statistic. The German federal statistics office explains that for an even number of observations, the median is the arithmetic mean of the two central sorted values. This is a rule about positions after sorting, not about averaging the entire data set.
Know what an outlier cannot tell you by itself
A mean much higher than a median is a clue to a right-hand tail, not proof of a bad data record. The five-value list is visibly stretched by 40. With many observations, inspect a table, dot plot, histogram, or suitable summary before explaining why. The difference between the two centers does not identify which individual value is wrong, and a small difference does not prove the absence of outliers: a very low and very high observation can pull in opposite directions.
Compare 1, 5, 5, 5, 9. Its mean is 25 / 5 = 5 and its median is also 5, despite the spread on both sides. A single center hides that variation. Add the range, minimum and maximum, or another suitable spread measure if the question concerns consistency. In the studio example, saying only 'average 12' invites a different mental picture from showing 4, 5, 5, 6, 40. A responsible summary gives the sample size, unit, relevant spread, and the reason the center was selected.
Choose the mean when totals and balance matter
The median resists an extreme value, but the mean is not inferior by default. If a workshop earned 60 units across five actual orders, the mean revenue per order is exactly 12. Multiply mean by count to recover the total: 12 × 5 = 60. The median of 5 cannot reconstruct that total. Budget planning, total consumption, and a fair equal-share model often require the mean because every quantity contributes to the sum. The right choice follows the task, not a universal ranking of statistics.
Be careful when combining groups. Suppose a first group has five orders averaging 12, and a second group has ten averaging 6. The combined mean is (5 × 12 + 10 × 6) / 15 = 120 / 15 = 8, not (12 + 6) / 2 = 9. The groups have different sizes, so their means need different weights. Without the group counts, two reported means are insufficient for an exact combined mean. This is another reason to keep the number of observations beside any reported average.
Choose the median for a typical position in a skewed set
If your question is where the middle observation lies, the median is the direct answer. The German statistical office notes that it is less sensitive to outliers and therefore useful for uneven distributions such as incomes. That is not a claim that exactly half of people earn precisely the median or that everyone below it has the same experience. It means the ordered data are split around a central threshold, subject to ties and the precise definition used by the publisher.
The median also has limits. In 4, 5, 5, 6, 40, it ignores how far 40 is from 6. If 40 becomes 400, the median remains 5 even though the total changes dramatically. A median alone can therefore hide economically important totals. If you need both a typical order and overall revenue, report both median and mean, plus count and possibly range. Explain the discrepancy instead of calling one number false. Our answer-checking guide offers a general framework for checking whether a numerical summary answers the stated question.
Protect the calculation from missing and mixed data
Before calculating, check that every entry has the same meaning and unit. Five prices in euros and one in cents cannot be summed directly; convert first. A blank response is not zero. If one of five observations is missing, you have four observed values and must not quietly divide by five. Decide whether the task asks for a summary of observed entries or requires a separate method for missing data. Do not invent the absent value just to preserve the original count.
For repeated values, every occurrence counts. A frequency table can compress the display but not the number of observations: if 5 appears twice, its contribution to the total is 2 × 5 and it occupies two positions in the ordered list. If the data measure categories such as colours rather than numerical quantities, a numerical mean is meaningless and a median may be undefined without a meaningful order. Always inspect the variable before applying familiar formulas. Clear definitions matter more than fast arithmetic.
Practise with a fresh set and verify both centers
Try 3, 4, 4, 5, 9, 11 without referring back to the worked example. There are six values. Their sum is 36, so the mean is 36 / 6 = 6. The two central positions are third and fourth, holding 4 and 5; the median is (4 + 5) / 2 = 4.5. Both summaries are valid. The mean of 6 reflects the two larger values, while the median identifies the midpoint of the ordered observations. Do not infer that a typical observation must equal either result exactly.
Check the mean by multiplying 6 × 6 to recover the total 36. Check the median by counting three positions on either side of the gap between the two center values. If the largest value 11 became 21, the sum would rise by 10 and the mean by 10 / 6, but the two central values would remain 4 and 5. Finally write a sentence stating whether your intended question concerns total-per-observation or middle position. This last sentence is part of the mathematical answer, not decoration.
Put it into practice now
Choose one problem from your current homework or review set. Attempt it before opening Eqora, then use the app only at the point where your own reasoning stops.
- State what the problem is asking before you solve it
- Identify the first step you cannot justify
- Ask Eqora one focused follow-up about that step
- Finish with a similar problem and no solution in view
The session is complete when the method is clearer, not simply when the worksheet has one more answer.