Measures of Central Tendency

Written and reviewed by Devendra Dubey, Mathematics & Commerce faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching). Based on NCERT Economics Class XI (Statistics for Economics), Chapter 5.

What is the "average" income of this village?

A village has ten families. Nine earn ₹10,000 a month and one earns ₹1,00,000. A newspaper writes: "average monthly income in the village is ₹19,000." A resident replies: "nobody I know earns that." Who is right?

Both are. Add all the incomes and divide by ten and you get ₹19,000. But the middle family earns ₹10,000, and the most common income is ₹10,000. "Average" is not one idea. It is three, and each answers a different question. This chapter teaches you to compute all three and, more important, to pick the one that tells the truth for the data in front of you.

Three averages, three questions

A measure of central tendency is one number that stands for a whole data set, so that you can compare data sets (is a 7 acre farmer small or large for this village?). The three in this chapter are the arithmetic mean, the median and the mode.

For the village: mean ₹19,000, median ₹10,000, mode ₹10,000. The one rich family drags the mean upward but leaves the other two untouched. That is the single most useful fact in the chapter.

Arithmetic mean: the balance point

The arithmetic mean is the sum of all observations divided by their number. Think of it as the balance point of a see-saw: the deviations (distances) above it exactly cancel the deviations below it.

$$\bar{X} = \dfrac{\sum X}{N}, \qquad \sum (X - \bar{X}) = 0$$

The second equation is the first property of the mean, and the proof is one line: $\sum (X - \bar{X}) = \sum X - N\bar{X} = \sum X - \sum X = 0$. The second property is that the mean is affected by extreme values. The first property holds only for deviations from the mean, not for deviations from the median or the mode. Since deviations from the mean always cancel out, you cannot use their sum to measure spread, which is why later chapters square them.

When the values are large, subtract a convenient assumed mean $A$ from every item first. Each item is $X = A + d$, so its average is $A$ plus the average of the deviations $d$.

$$\bar{X} = A + \dfrac{\sum d}{N}, \qquad d = X - A$$

If the deviations also share a common factor $c$ (equal class widths are the usual case), divide by it to shrink the numbers further. This is the step deviation method, and the one step students forget is multiplying the result back by $c$.

$$\bar{X} = A + \dfrac{\sum f d'}{\sum f} \times c, \qquad d' = \dfrac{m - A}{c}$$

Worked example: mean and median of individual data. Eight families have weekly incomes (₹): 700, 760, 820, 850, 900, 980, 1050 and 1340. Find the mean using an assumed mean of 900, and the median. Then see what happens if the last income is 13,400 instead.

The mean jumped from 925 to 2432.5 because of one number, but the median did not move. That is why the median is preferred when a few extreme values are present.

Common wrong step. Writing $\bar X = \sum d / N = 25$ and stopping there. The deviations only give the shift from $A$; you must add $A$ back.

For discrete data the mean uses $\bar X = \dfrac{\sum f X}{\sum f}$. For continuous data, replace each class by its mid-value $m$ and use $\bar X = \dfrac{\sum f m}{\sum f}$. The procedure is the same whether classes are exclusive, inclusive or unequal.

Worked example: a discrete series. In a quiz out of 5, 40 students scored as below. Find the mean, the median and the mode.

Common wrong step. Taking the middle of the six possible scores 0 to 5, which is 2.5. The median is the score of the middle student (the 20.5th), not the middle of the list of possible scores.

Worked example: a continuous series by step deviation. The daily wages (₹) of 100 workers are given below. Find the mean using $A = 650$ and $c = 100$.

Here $\sum f d' = -54$ and $\sum f = 100$. So $\bar X = 650 + \dfrac{-54}{100} \times 100 = 650 - 54 = 596$. As a check, the direct method gives $\sum f m = 59{,}600$, and $59{,}600 / 100 = 596$.

Common wrong step. Forgetting to multiply by $c$ gives $650 - 0.54 = 649.46$, which is outside the believable range for this data. Another slip is using the lower limits instead of the mid-values, which gives 546, far too low because every worker is treated as earning the bottom of their class.

Weighted mean. When items are not equally important, give each a weight $W$:

$$\bar X_w = \dfrac{\sum W X}{\sum W}$$

Worked example: a weighted mean. A student scores 60, 70 and 80 in three subjects with credits 1, 2 and 3. Find the average that reflects the credits.

Common wrong step. Using the simple mean of 70. It treats a 1 credit subject and a 3 credit subject as equally important, which they are not. The weighted mean is higher because the best score has the largest weight.

Pause and predict: in the village, the richest family's income rises from ₹1,00,000 to ₹10,00,000. What happens to the mean, the median and the mode?

Check your answer: the mean rises to ₹1,09,000, because the total becomes ₹10,90,000 for ten families. The median and the mode stay at ₹10,000, because the middle and the most common family are unchanged.

Median: the middle one

The median divides the ordered data into two equal halves. To use it, arrange the data in order first. Its position depends on the kind of series.

For a continuous series, find the median class from $N/2$, then interpolate inside it.

$$\text{Median} = L + \dfrac{\tfrac{N}{2} - c.f.}{f} \times h$$

Here $L$ is the lower limit of the median class, $c.f.$ is the cumulative frequency of the class before it, $f$ is its own frequency and $h$ is its width. The logic is simple: you need $N/2$ observations, the classes before it supply $c.f.$ of them, so you need $(N/2 - c.f.)$ more from the median class. Those are taken as a fraction $(N/2 - c.f.)/f$ of the way across it.

Worked example: median of the wage data. Use the 100 worker wage distribution above.

Common wrong step. Using the cumulative frequency of the median class itself (56) instead of the class before it (26). That gives $500 + \frac{50-56}{30}\times 100 = 480$, which falls below the lower limit of the class it was meant to lie in. Whenever a median lands outside its own class, you have used the wrong $c.f.$

Quartiles and percentiles

If the median cuts the data into two halves, quartiles cut it into four and percentiles into 100. In an ordered series, $Q_1$ has 25 per cent of items below it, $Q_2$ is the median, and $Q_3$ has 75 per cent below it. The 50th percentile is also the median.

$$Q_1 = \left(\dfrac{N+1}{4}\right)\text{th item}, \qquad Q_3 = \left(\dfrac{3(N+1)}{4}\right)\text{th item}$$

For the ordered series 12, 18, 22, 25, 28, 32, 35, 38, 41, 44, 46 we have $N = 11$. So $Q_1$ is the 3rd item (22), the median is the 6th (32) and $Q_3$ is the 9th (41). Scoring in the 82nd percentile means 82 per cent of candidates scored below you, so 18 per cent scored above you. It does not mean you got 82 marks.

Mode: the most common

The mode is the value that occurs most often. A series with one peak is unimodal, with two peaks bimodal, and if every value occurs equally often there is no mode. For a continuous series, find the modal class (the class with the highest frequency) and place the mode inside it by looking at the neighbouring frequencies:

$$\text{Mode} = L + \dfrac{D_1}{D_1 + D_2} \times h, \qquad D_1 = f_1 - f_0, \quad D_2 = f_1 - f_2$$

Here $f_1$ is the modal frequency, $f_0$ the frequency before and $f_2$ the one after. If the class before is smaller than the class after, $D_1$ is the larger difference and the mode sits in the upper part of the modal class, pulled toward the bigger neighbour. The classes must be of equal width and exclusive. A "less than" cumulative table cannot give a mode directly, so convert it to an ordinary frequency table first.

Worked example: mode of the wage data. Find the mode of the 100 worker wage distribution.

Common wrong step. Swapping $D_1$ and $D_2$, which gives $500 + \frac{8}{20} \times 100 = 540$. $D_1$, the difference with the class before, goes on the top because it pulls the mode toward the upper side. A quick sanity check: the neighbour after (22) is bigger than the neighbour before (18), so the mode should sit above the middle of the class, i.e. above 550. The answer 540 fails this check, 560 passes.

The three together: skewness

For the wage data the three averages are: mean 596, median 580 and mode 560. They are in order, and the median is in the middle. This is no accident.

The memory aid: write them alphabetically, Mean, Median, Mode. The values run steadily up or steadily down in that order, so the median is always the one between the other two. The mean is pulled hardest toward the tail.

Pause and predict: in a certain distribution the mean is 50 and the mode is 40. Which of 38, 44 and 55 can be the median?

Check your answer: only 44. The median must lie between the mean and the mode, so between 40 and 50.

Which average should you use?

No single measure is best in every situation. That is exactly why there are three.

Beyond the textbook. For the exam: use the median for incomes and other data with extreme values. In reality: official figures such as per capita income are means. Because a small number of very high incomes pull the mean upward, the income of the typical household, the median, is lower than the mean. Why it matters: when you read that "average income rose", ask whether the mean or the median rose, and whether the gains reached the middle household or only the top.

Beyond the textbook. For the exam: the chapter develops three averages, the mean, median and mode. In reality: the geometric mean, which multiplies instead of adds, is the correct average for growth rates. If a quantity grows by 10 per cent in one year and 20 per cent in the next, it has multiplied by $1.10 \times 1.20 = 1.32$ in total, so the steady yearly rate is $\sqrt{1.32} - 1 \approx 14.9$ per cent, not the arithmetic 15. Why it matters: the mean is not always the right average, even when it is easy to compute.

Beyond the textbook. For the exam: use the mode formula to find the mode of a continuous series. In reality: for moderately skewed data there is a rough shortcut, $\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean}$. For the wage data it gives $3(580) - 2(596) = 548$, close to the 560 from the formula. Why it matters: it is a quick way to catch a slip, but the formula answer is what the exam expects.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Presentation of Data shows the mode and median read from a histogram and ogives, which this chapter calculates exactly.
• Earlier foundation: Organisation of Data defines the class marks and cumulative frequencies used in every grouped formula here.
• Next: Correlation uses means of two variables as the starting point for deviations, and Index Numbers uses weighted averages of price changes.