Mean, median and mode

Nine households live on one street. Their monthly incomes, in thousands of rupees, are 12, 15, 18, 18, 20, 22, 25, 30 and 65. What is the income of the typical household? Three honest answers:

• The mean is 25. Share the total (225) equally among nine households and each gets 25.
• The median is 20. Line the households up from poorest to richest, and the fifth, the one in the middle, has 20.
• The mode is 18. It is the one income that two households share.

None of them is wrong. They answer three different questions, and with skewed data like incomes they differ a lot. Learning which question each one answers is most of this lesson.

The arithmetic mean adds the values and divides by how many there are:

The best picture of it is a seesaw. Put each household on a number line as a weight. The mean is the one point where the line would balance, the fulcrum. Drag the dots below and the blue marker follows. Then press A rich household moves in.

The mean jumped from 25 to 37.5. The median moved from 20 to 21, and only because there were now ten households and the middle fell between two of them. One very high income pulled the balance point a long way, because the mean counts how large every value is. The median counts only which side each value lies on. That is the mean’s strength, since it uses all the information, and its weakness, since it is hostage to extreme values.

Four properties worth knowing cold

1. The deviations from the mean add up to zero: $\sum (x_i-\bar{x})=0$. That is what “balance point” means. JNU 2021 and Hyderabad 2012 both ask it.
2. Add k to every value and the mean rises by k. Multiply by c and the mean is multiplied by c. So $\overline{cx+k}=c\bar{x}+k$.
3. The mean is the value that makes the sum of squared deviations smallest. For any other number c:
   $$\sum (x_i-c{)}^2=\sum (x_i-\bar{x}{)}^2+n(\bar{x}-c{)}^2$$
   The extra term is never negative, so c = x̄ wins. JAM 2021 asks exactly this: S₁ (about the mean) is less than S₂ (about any other k).
4. The mean can be corrected without starting over. If one value was recorded as 33 but should be 88, the total is 55 higher, so the mean rises by 55 ÷ n.

Sometimes some values should count for more. The weighted mean multiplies each value by a weight w and divides by the total weight:

When the weights are group sizes, this is the combined mean of several groups: n₁x̄₁ + n₂x̄₂ over n₁ + n₂. It is why you can never average two averages without knowing how many were behind each.

Averages that move the wrong way

JNU 2013 poses a puzzle. The average income of province A is higher than that of province B. Some people move from A to B, and the average of both provinces rises. Can that happen?

Yes. Let A have incomes 10, 20, 30, 40 (mean 25) and B have 5 and 15 (mean 10). Move the household earning 20 from A to B. A is left with 10, 30, 40, a mean of 26.67. B now has 5, 15, 20, a mean of 13.33. Both rose. The mover earned less than A’s average, so A improves by losing them, and more than B’s average, so B improves by gaining them. The answer is not “when the richer people move”, which would lower A’s mean. It needs someone in between.

The median is the middle value once the data are sorted. With an odd number of values it is the one in position (n + 1) ÷ 2. With an even number it is halfway between the two in the middle. CUET asks both: seven wages 160, 180, 200, 280, 300, 320, 400 have a median of 280, the fourth, and the ten numbers 2, 4, 3, 6, 1, 8, 9, 2, 5, 7 sort to 1, 2, 2, 3, 4, 5, 6, 7, 8, 9, so the median is halfway between 4 and 5, which is 4.5.

The median pays no attention to how far out the extremes are. JNU 2015: six scores have a median of 21; the highest score rises by three points. The median stays 21, because the highest was and still is above the middle.

Where the mean minimises the sum of squared distances, the median minimises the sum of absolute distances $\sum |x_i-c|$. To see why, slide c to the right by one. Every point to its left gets one unit further away and every point to its right gets one unit nearer. That helps while more points are on the right than on the left, so you stop at the middle. This is why JNU 2021 says the mean deviation is least when taken about the median.

The mode is the value that occurs most often. A data set can have no mode (every value different), one mode, or several: two peaks make it bimodal. It is the only average that works for nominal data: the most common religion in a village, the most common occupation. It is also the right answer when you want the commonest size of something, such as a shoe.

For a continuous variable the raw mode is nearly useless. In Ashanagar, ₹12,000 and ₹22,000 each occur three times in the sixty incomes, and that says nothing. Group the data first and the answer appears: the modal class is the one with the highest frequency, and the next section estimates a single figure from it.

Often you only have the table, not the raw data. Here are the Ashanagar incomes grouped into classes ₹10,000 wide, which is how a survey report would print them.

Every value in a class is treated as if it sat at the class mark. That gives the mean, and two formulas give the median and the mode.

Quartiles from an ogive

The median splits the data in half. Quartiles split it into quarters: Q₁ has a quarter of the data below it, Q₃ three quarters. Percentiles do the same in hundredths, so the 90th percentile has 90% below it. All of them are read the same way. Go up the cumulative axis of the ogive to the right share of N, across to the curve, then down. The picture is the formula: within a class you assume the values are evenly spread, so the curve is a straight line, and the formula just reads along it. Try it.

Two conventions are easy to mix up. For raw data, textbooks put Q₁ at position (n + 1) ÷ 4 in the sorted list. For grouped data you work with N ÷ 4 and N ÷ 2, as above. CUET 2020 uses the second: with 500 observations, Q₃ lies in the 3 × 500 ÷ 4 = 375th item. Statistical software uses other rules again, so a small data set can give a slightly different Q₁ in a spreadsheet.

Six situations. Pick the average and read why.