MathsCBSE Class 10 · Statistics

Relation Between Mean, Median and Mode: Empirical Formula

Updated 14 October 2026 · Fact-checked

The empirical relation says that for a moderately skewed distribution, three times the median equals the mode plus twice the mean. It lets you find one average when the other two are given, in short and long answers and in MCQ-type questions, and it helps you check whether averages computed from a table are consistent.

Understand Relation Between Mean, Median and Mode

You already know three averages for grouped data: the mean (balance point), the median (middle value) and the mode (most common value). For a perfectly symmetric distribution, all three are equal. Most real data is a little lopsided, or skewed, so the three values drift apart. The surprising fact is that they drift apart in a fairly regular way.

Statisticians noticed that for distributions that are only moderately skewed, the averages are linked approximately by $3 Median = Mode + 2 Mean$ . This is called the empirical relation because it comes from observation, not from a proof. In board questions you treat it as exact whenever two averages are given and the third is asked.

Why does it make sense? Subtract $3 Mean$ from both sides and you get $3 (Median - Mean) = Mode - Mean$ . So the gap between median and mean is one-third of the gap between mode and mean. The mean gets pulled furthest by a long tail of extreme values, the mode stays at the peak, and the median sits in between, about one-third of the way from the mean towards the mode. This also tells you the median always lies between the mean and the mode.

The relation can be rearranged three ways: $Mode = 3 Median - 2 Mean$ , $Mean = \frac{3 Median - Mode}{2}$ and $Median = \frac{Mode + 2 Mean}{3}$ . Pick the form that has your unknown alone on one side.

It connects the whole chapter. A long question may ask you to compute two averages from a table using the median and mode formulas, then find the third using the relation. You can also use it as a check: if your computed mean, median and mode are far from satisfying it, recheck your arithmetic. Remember its limits: it is approximate, it works only for moderately skewed, single-peaked data (the syllabus avoids bimodal data), and if a question asks for a value by a named method, you must use that method, not the relation.

Relation Between Mean, Median and Mode: formulas to remember

Empirical relation: $3 Median = Mode + 2 Mean$; Approximate for moderately skewed data; treated as exact in board questions.
Mode from mean and median: $Mode = 3 Median - 2 Mean$; Coefficients: 3 goes with the median, 2 with the mean.
Mean from median and mode: $Mean = \frac{3 Median - Mode}{2}$; Divide the whole of 3 Median minus Mode by 2.
Median from mean and mode: $Median = \frac{Mode + 2 Mean}{3}$; The median is a weighted average: one part mode, two parts mean.
Gap form: $Mode - Mean = 3 (Median - Mean)$; Use when a difference between two averages is given rather than the values.
Mode of grouped data: $Mode = l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h$; The modal class is the class with the highest frequency; f₀ and f₂ are the frequencies of the classes before and after it.
Median of grouped data: $Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$; The median class is the first class whose cumulative frequency reaches n/2; cf is the cumulative frequency of the class before it. It need not be the modal class.
Mean (direct method): $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$; x_i is the class mark of each class.
Mean (step deviation): $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}, u_{i} = \frac{x _{i} - a}{h}$; a is an assumed mean (a central class mark) and h the class size.

How to solve Relation Between Mean, Median and Mode questions

Every question on this topic either gives two averages directly or makes you compute them first. This method covers both.

1Read what is given and what is asked. Note whether you are told to use a particular formula or method.
2If the averages are not given, compute the two you need from the table (median and mode by their formulas, or the mean by a named method), showing the full table.
3Write the empirical relation $3 Median = Mode + 2 Mean$ in full before substituting; this line earns a mark.
4If differences or sums of averages are given, write each as an equation and solve the pair together.
5Rearrange so the unknown stands alone, keeping brackets: for example $Mean = \frac{3 Median - Mode}{2}$ .
6Substitute and simplify carefully with decimals.
7Check: the median must lie between the mean and the mode; substitute back into the relation. Add units.

Relation Between Mean, Median and Mode in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Empirical relation between mean, median and mode

Imagine your school runs two bus routes, A and B, and the transport teacher surveys how long students spend on the bus each morning. For Route A she has worked out the mean time as 32 minutes and the median as 34 minutes. For Route B she has only the median, 30 minutes, and a note that the mean is 4.5 minutes more than the mode.

Using the empirical relation, find the mean, median and mode for both routes. Check that each median lies between its mean and mode. Which route has the greater mean time, and by how much? On which route do a few unusually long rides pull the mean above the most common time?

Try it first, then show the solutionSolution

Relation: $3 Median = Mode + 2 Mean$ .
Route A: $Mode = 3 (34) - 2 (32) = 102 - 64 = 38$ minutes.
Route B, gap form: $Mean - Mode = 3 (Mean - Median)$ .
$4.5 = 3 (Mean - 30)$ , so $Mean - 30 = 1.5$ and Mean $= 31.5$ minutes.
Mode for B $= 31.5 - 4.5 = 27$ minutes.
Check A: $3 (34) = 102 = 38 + 64$ , and 34 lies between 32 and 38.
Check B: $3 (30) = 90 = 27 + 63$ , and 30 lies between 27 and 31.5.
Compare means: $32 - 31.5 = 0.5$ minute, so Route A is greater.
On Route B the mean (31.5) is above the mode (27), so a few long rides pull the mean up; on Route A the mean is below the mode, so a few short rides pull it down.

Answer: Route A: mean 32, median 34, mode 38 minutes; Route B: mean 31.5, median 30, mode 27 minutes. Route A has the greater mean, by 0.5 minute. On Route B, a few long rides pull the mean above the most common time.

Memory hook: Bus timetable trick: the median rides between the mean and the mode, one-third of the way from the mean.

Tricks to remember Relation Between Mean, Median and Mode

Short memory aids. Say each one aloud once, then use it on a question the same day.

3-1-2 order
Say 3 Median = Mode + 2 Mean. The median (the middle value) carries the 3. The mode has 1 and the mean has 2, and 1 + 2 = 3.
Use it when: Writing the relation correctly in the first line of any answer.
Middle child
The median is the middle child: it always sits between the mean and the mode, nearer the mean (one-third of the way).
Use it when: Quick check of any answer; also eliminates wrong MCQ options.
Weights add up
In Median = (Mode + 2 Mean)/3, the weights 1 and 2 add to the divisor 3, just like any weighted average.
Use it when: Recalling the median form without rearranging.

Relation Between Mean, Median and Mode: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mode from mean and median (MCQ) · 1 mark · MCQBoard level

For a moderately skewed distribution, the mean is 24 and the median is 26. The mode is
(a) 28
(b) 30
(c) 32
(d) 26

Try it first, then show the solutionSolution

Use $Mode = 3 Median - 2 Mean$ .
$Mode = 3 (26) - 2 (24) = 78 - 48 = 30$ .
Check: median 26 lies between mean 24 and mode 30.

Answer: (b) 30

Example 2: A 2-mark gap question · 2 marks · Very short answerBoard level

In a moderately skewed distribution, the mode exceeds the mean by 12 and the mean is 40. Find the median.

Try it first, then show the solutionSolution

Subtract $3 Mean$ from both sides of $3 Median = Mode + 2 Mean$ : $3 (Median - Mean) = Mode - Mean = 12$ . (1 mark)
So $Median - Mean = 4$ and Median $= 44$ (mode 52). Check: $3 (44) = 132 = 52 + 80$ . (1 mark)

Answer: Median = 44 (and the mode is 52).

Example 3: A ratio with the relation, and a conclusion · 3 marks · Short answerHarder board level

In a moderately skewed distribution, the mean and the median are in the ratio $9 : 10$ and the mode is 36. Find the mean and the median. Towards which side, lower or higher values, does the longer tail of the data lie?

Try it first, then show the solutionSolution

Let Mean $= 9 k$ and Median $= 10 k$ . (½ mark)
Relation: $3 (10 k) = 36 + 2 (9 k)$ , so $30 k = 36 + 18 k$ and $k = 3$ . (1 mark)
Mean $= 27$ , Median $= 30$ . Check: $3 (30) = 90 = 36 + 54$ . (½ mark)
Mean (27) < Median (30) < Mode (36). The mean is pulled below the most common value by a few low values, so the longer tail lies towards lower values. (1 mark)

Answer: Mean = 27, Median = 30; the longer tail is towards lower values (mean < median < mode).

Example 4: Gap and sum given, then a transformation · 3 marks · Short answerHarder board level

In a moderately skewed distribution, the median is 4 more than the mean, and the sum of the mean and the mode is 100.
(i) Find the mean, median and mode.
(ii) Every observation is now doubled and then increased by 3. Find the new mean, median and mode, and verify the empirical relation for them.

Try it first, then show the solutionSolution

Gap form: $Mode - Mean = 3 (Median - Mean) = 3 (4) = 12$ . (½ mark)
With $Mode + Mean = 100$ : adding, $2 Mode = 112$ , so Mode $= 56$ and Mean $= 44$ . (½ mark)
Median $= 44 + 4 = 48$ . Check: $3 (48) = 144$ and $56 + 2 (44) = 144$ . (½ mark)
Doubling and adding 3 changes every observation $x$ to $2 x + 3$ , so every class mark and every class boundary changes the same way. Each average therefore becomes $2 \times old + 3$ . (½ mark)
New mean $= 2 (44) + 3 = 91$ , new median $= 2 (48) + 3 = 99$ , new mode $= 2 (56) + 3 = 115$ . (½ mark)
Verify: $3 (99) = 297$ and $115 + 2 (91) = 115 + 182 = 297$ . Relation holds. (½ mark)

Answer: (i) Mean = 44, Median = 48, Mode = 56 (ii) Mean = 91, Median = 99, Mode = 115; 297 = 297, relation holds

Example 5: Median and mode from a table, mean by the relation · 5 marks · Long answerHarder board level

The daily distance (in km) cycled by 60 members of a cycling club is given below.
Distance (km): 0–10, 10–20, 20–30, 30–40, 40–50
Members: 6, 14, 20, 16, 4
Find the mode and the median. Hence, using the empirical relation, estimate the mean. Compare it with the mean found by the direct method.

Try it first, then show the solutionSolution

Modal class (highest frequency 20) is 20–30: $l = 20$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = 14$ , $f_{2} = 16$ . (½ mark)
$Mode = 20 + \frac{20 - 14}{40 - 14 - 16} \times 10 = 20 + \frac{6}{10} \times 10 = 26$ km. (1 mark)
Cumulative frequencies: 6, 20, 40, 56, 60; $n = 60$ , $\frac{n}{2} = 30$ . (½ mark)
Median class is 20–30 (first cf ≥ 30): $l = 20$ , $c f = 20$ , $f = 20$ , $h = 10$ . (½ mark)
$Median = 20 + \frac{30 - 20}{20} \times 10 = 25$ km. (1 mark)
$Mean = \frac{3 ( 25 ) - 26}{2} = \frac{49}{2} = 24.5$ km. (1 mark)
Direct: $\sum f_{i} x_{i} = 30 + 210 + 500 + 560 + 180 = 1480$ , mean $= \frac{1480}{60} \approx 24.67$ km; close to 24.5, as expected for an approximate relation. (½ mark)

Answer: Mode = 26 km, Median = 25 km, estimated Mean = 24.5 km (direct method gives about 24.67 km)

Example 6: Missing frequency from the median, then mode and mean · 5 marks · Long answerHarder board level

The marks (out of 50) scored by a group of students in a class test are given below. The median is 23.5.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Students: 6, 12, $p$ , 8, 4
(i) Find $p$ . (ii) Find the mode using the mode formula. (iii) Using the empirical relation, find the mean, and compare it with the mean by the direct method.

Try it first, then show the solutionSolution

Median 23.5 lies in 20–30: $l = 20$ , $h = 10$ , $f = p$ , $c f = 6 + 12 = 18$ , $n = 30 + p$ . (½ mark)
$23.5 = 20 + \frac{\frac{30 + p}{2} - 18}{p} \times 10$ , so $3.5 = \frac{10 ( p - 6 )}{2 p} = \frac{5 ( p - 6 )}{p}$ . (1 mark)
$3.5 p = 5 p - 30 \Rightarrow 1.5 p = 30 \Rightarrow p = 20$ , so $n = 50$ . (½ mark)
Modal class 20–30 (highest frequency 20): $l = 20$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = 12$ , $f_{2} = 8$ . (½ mark)
$Mode = 20 + \frac{20 - 12}{40 - 12 - 8} \times 10 = 20 + \frac{8}{20} \times 10 = 24$ marks. (1 mark)
$Mean = \frac{3 ( 23.5 ) - 24}{2} = \frac{70.5 - 24}{2} = \frac{46.5}{2} = 23.25$ marks. (1 mark)
Direct: class marks 5, 15, 25, 35, 45; $\sum f_{i} x_{i} = 30 + 180 + 500 + 280 + 180 = 1170$ , mean $= \frac{1170}{50} = 23.4$ marks, within 0.15 of 23.25. (½ mark)

Answer: (i) p = 20 (ii) Mode = 24 marks (iii) Mean by the relation = 23.25 marks; the direct method gives 23.4 marks, very close

Example 7: Case study: is the claimed mean possible? · 4 marks · Case-basedHOTS / competency

Suppose a pizza shop studies the delivery times of its orders over a month. Its records give a median of 30 minutes and a modal time of 26 minutes. Treat the data as moderately skewed.
(i) Estimate the mean delivery time using the empirical relation. (1 mark)
(ii) Is the longer tail of the data towards shorter or longer times? Give a reason. (1 mark)
(iii) A manager claims the mean delivery time is 28 minutes. Without finding any other value, explain why this cannot be right. If the mean really were 28 minutes with the same median, what would the relation give for the mode, and why does that clash with the records? (2 marks)

Try it first, then show the solutionSolution

(i) $2 Mean = 3 (30) - 26 = 64$ , so Mean $= 32$ minutes. (1 mark)
(ii) Mode (26) < Median (30) < Mean (32). A few very late deliveries pull the mean up, so the longer tail is towards longer times. (1 mark)
(iii) In a moderately skewed distribution the median lies between the mean and the mode. With mode 26 below the median 30, the mean must be above 30, so 28 is impossible. (1 mark)
With mean 28: Mode $= 3 (30) - 2 (28) = 34$ , which is above the median, while the records show a mode of 26. So the claim contradicts the data. (1 mark)

Answer: (i) 32 minutes (ii) Towards longer times, since mode < median < mean (iii) The mean must exceed the median 30; a mean of 28 would force a mode of 34, not the recorded 26.

Where students lose marks in Relation Between Mean, Median and Mode

The slips examiners see most often, and what to write instead.

Mistake: Swapping the coefficients
Students remember '3 and 2' but not which average each goes with.
Wrong working
Mean 40, median 44: Mode $= 3 (40) - 2 (44) = 32$
Right working
Mode $= 3 (44) - 2 (40) = 132 - 80 = 52$
Do this: Write $3 Median = Mode + 2 Mean$ in full first; check the median lies between the other two.
Usually costs: The whole question, usually 1 to 3 marks.
Mistake: Dropping the bracket when finding the mean
Only the mode gets divided by 2 in a careless line.
Wrong working
Mean $= 3 (25) - \frac{26}{2} = 75 - 13 = 62$
Right working
Mean $= \frac{3 ( 25 ) - 26}{2} = \frac{49}{2} = 24.5$
Do this: Write the fraction bar under the whole numerator before substituting.
Usually costs: 1 to 2 marks, and an absurd answer outside the data range.
Mistake: Using the relation when a method is named
The relation looks quicker than a full table.
Wrong working
'Find the mean by the step deviation method': student writes Mean $= \frac{3 Median - Mode}{2}$
Right working
Build the $u_{i}$ table and use $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$
Do this: Use the empirical relation only when asked, or when two averages are given and nothing else.
Usually costs: Most of a 3 to 5-mark answer.
Mistake: Dividing by the wrong number for the median
Students divide by 2 out of habit.
Wrong working
Median $= \frac{38 + 2 ( 32 )}{2} = 51$
Right working
Median $= \frac{38 + 64}{3} = 34$
Do this: Weights 1 and 2 add to 3, so divide by 3.
Usually costs: 1 mark in an MCQ, 2 in a short answer.

Exam tips

Write the relation $3 Median = Mode + 2 Mean$ as the first line; it carries method marks even if arithmetic slips.
In MCQs, eliminate options at once using the rule that the median lies between the mean and the mode.
In long answers that ask for all three averages, you may compute two from the table and the third by the relation unless the question names a method for it.
Keep decimals exact rather than rounding early.
If you compute all three averages, test them against the relation; a large mismatch means an arithmetic error.

Practice questions on Relation Between Mean, Median and Mode

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the mean of a moderately skewed distribution is 30 and its mode is 36, then its median is 34, and the longer tail of the data is towards lower values.
Reason (R): For a moderately skewed distribution, $3 Median = Mode + 2 Mean$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find the median first, then check both parts of A.
Answer
(d). By R, Median $= \frac{36 + 60}{3} = 32$ , not 34, so A is false (its tail statement is right, since mean < median < mode, but the median is wrong). R is true.
Q2 · 3 marks · Short answerHarder board level
The mean of the distribution below is 26.6.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 4, 8, $f$ , 12, 6
Find $f$ and the mode. Then use the empirical relation to estimate the median.
Hint
Class marks are 5, 15, 25, 35, 45.
Answer
$\sum f_{i} = 30 + f$ and $\sum f_{i} x_{i} = 20 + 120 + 25 f + 420 + 270 = 830 + 25 f$ . $830 + 25 f = 26.6 (30 + f) = 798 + 26.6 f$ , so $1.6 f = 32$ and $f = 20$ . Modal class 20–30: $Mode = 20 + \frac{20 - 8}{40 - 8 - 12} \times 10 = 26$ . Median $\approx \frac{26 + 2 ( 26.6 )}{3} = \frac{79.2}{3} = 26.4$ . (The formula gives $20 + \frac{25 - 12}{20} \times 10 = 26.5$ , very close.)
Q3 · 5 marks · Long answerHarder board level
The daily travel time (in minutes) of 50 office workers is given below.
Time (min): 0–10, 10–20, 20–30, 30–40, 40–50
Workers: 4, 20, 16, 6, 4
(i) Find the mean by the step-deviation method. (ii) Find the median and the mode, noting that they lie in different classes. (iii) Check the empirical relation, and say which way the data is skewed.
Hint
Take $a = 25$ and $h = 10$ . Find the median class from $\frac{n}{2} = 25$ and the modal class from the highest frequency separately.
Answer
(i) $u_{i} = - 2, - 1, 0, 1, 2$ ; $f_{i} u_{i} = - 8, - 20, 0, 6, 8$ , so $\sum f_{i} u_{i} = - 14$ . Mean $= 25 + 10 \times \frac{- 14}{50} = 22.2$ min. (1½ marks)
(ii) cf: 4, 24, 40, 46, 50. The first cf reaching 25 is 40, so the median class is 20–30: Median $= 20 + \frac{25 - 24}{16} \times 10 = 20.625$ min. The modal class is 10–20 (frequency 20): Mode $= 10 + \frac{20 - 4}{40 - 4 - 16} \times 10 = 18$ min. (2 marks)
(iii) $3 Median = 61.875$ and $Mode + 2 Mean = 18 + 44.4 = 62.4$ : close, as expected. Mode (18) < Median (20.6) < Mean (22.2), so a few long journeys pull the mean up; the longer tail is towards longer times. (1½ marks)
Q4 · 5 marks · Long answerHarder board level
The mode of the distribution below is 25, and 20–30 is the modal class.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 4, 10, 16, $f$ , 6
(i) Find $f$ . (ii) Find the median. (iii) Use the empirical relation to find the mean, and compare it with the mean by the direct method.
Hint
In the mode formula, $f_{0} = 10$ and $f_{2} = f$ .
Answer
(i) $25 = 20 + \frac{16 - 10}{32 - 10 - f} \times 10$ , so $5 (22 - f) = 60$ and $f = 10$ ; $n = 46$ . (ii) cf: 4, 14, 30, 40, 46; $\frac{n}{2} = 23$ , median class 20–30: Median $= 20 + \frac{23 - 14}{16} \times 10 = 25.625$ . (iii) Mean $\approx \frac{3 ( 25.625 ) - 25}{2} \approx 25.94$ . Direct: $\frac{20 + 150 + 400 + 350 + 270}{46} = \frac{1190}{46} \approx 25.87$ , very close.
Q5 · 4 marks · Case-basedHOTS / competency
Suppose a school music club records the daily practice time (in minutes) of its 50 members.
Time (min): 10–20, 20–30, 30–40, 40–50, 50–60
Members: 2, 8, 20, 12, 8
(i) Find the median practice time. (1 mark)
(ii) Find the modal practice time. (1 mark)
(iii) Use the empirical relation to find the mean. Does the data have a longer tail towards shorter or longer times? A member claims the mean is 34 minutes; explain without further calculation why this cannot be right. (2 marks)
Hint
The median class has the first cf ≥ 25. The median must lie between the mean and the mode.
Answer
(i) cf: 2, 10, 30, 42, 50; median class 30–40: Median $= 30 + \frac{25 - 10}{20} \times 10 = 37.5$ min. (ii) Modal class 30–40: Mode $= 30 + \frac{20 - 8}{40 - 8 - 12} \times 10 = 36$ min. (iii) Mean $= \frac{3 ( 37.5 ) - 36}{2} = \frac{76.5}{2} = 38.25$ min. Since mean > median > mode, a few long practice times pull the mean up, so the longer tail is towards longer times. The median 37.5 must lie between the mean and the mode 36, so the mean must be above 37.5; a mean of 34 minutes is impossible. (The direct method gives $\frac{1910}{50} = 38.2$ min, confirming this.)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the empirical relation.Show answer
$3 Median = Mode + 2 Mean$
2.Mode in terms of mean and median?Show answer
$Mode = 3 Median - 2 Mean$
3.Mean in terms of median and mode?Show answer
$Mean = \frac{3 Median - Mode}{2}$
4.Mean 20, mode 26. Median?Show answer
$\frac{26 + 40}{3} = 22$
5.For which distributions does the relation hold?Show answer
Moderately skewed, single-peaked distributions (approximately).
6.In a symmetric distribution, how are the three averages related?Show answer
They are all equal.

Relation Between Mean, Median and Mode: frequently asked questions

Is the empirical relation exact?

No, it is an approximation observed for moderately skewed data. In board questions, when two averages are given, you treat it as exact and use it to find the third.

How do I remember which average gets 3 and which gets 2?

Remember that the median is the middle value, so it is the one alone on the left with the 3. The mode and twice the mean are on the right, and their weights 1 and 2 add up to 3.

Can I use the relation instead of the mean formula in a long answer?

Only if the question allows it or asks for it. If the question says find the mean by a particular method, you must use that method with a full table.

Why must the median lie between the mean and the mode?

The relation can be rewritten as Median = (Mode + 2 Mean)/3, which is a weighted average of the mode and mean. A weighted average always lies between the two values, closer to the one with the bigger weight, which is the mean.

What if the data has two classes with the same highest frequency?

Such bimodal data is avoided in the syllabus, because the mode formula needs a single modal class. In this topic you will always have one clear modal class.