EconomicsISC Class 11 · Measures of Central Value

Arithmetic Mean: All Series, Short-cut Methods and Weighted Mean (ISC Class 11)

Updated 17 October 2026 · Fact-checked

The arithmetic mean is the sum of all observations divided by their number. For a discrete series mean = Sum fx / Sum f; for a continuous series mean = Sum fm / Sum f (m = mid-point); the step deviation method gives mean = A + h x Sum fd' / Sum f, with d' = (m - A) / h. The weighted mean = Sum wx / Sum w, and the combined mean of two groups = (n1 x mean1 + n2 x mean2) / (n1 + n2).

The mean is the balance point: deviations on the left (-9) cancel those on the right (+9), so their sum is zero.

Illustration drawn for QuestionBuddy.

Understand Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series

Averages. An average is a single value representing a whole series. A good average is rigidly defined, easy to understand and compute, based on all observations, not unduly affected by extremes, and suitable for further algebraic treatment.

Individual series. Mean = Sum X / n. *Assumed mean method:* choose A; d = X - A; mean = A + Sum d / n.

Discrete series. Mean = Sum fx / Sum f (N). Short-cut: mean = A + Sum fd / N.

Continuous series. Find mid-points m of each class. Direct: mean = Sum fm / N. *Step deviation:* d' = (m - A) / h (h = class width), mean = A + h x Sum fd' / N.

Weighted mean. When items have different importance, mean = Sum wx / Sum w.

Combined mean. Two groups with sizes n1, n2 and means x1, x2: combined mean = (n1 x1 + n2 x2) / (n1 + n2).

Properties. The sum of deviations from the mean is zero. If every value is increased by k, the mean increases by k; if every value is multiplied by k, the mean is multiplied by k. The mean is affected by extreme values.

Finding a missing frequency or value. Substitute the given mean into the formula and solve the equation.

Merits. Simple, uses all items, algebraically manageable. Demerits. Affected by extremes; cannot be found for open-ended classes without assumptions.

Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series: key facts

Discrete and continuous: Mean = Sum fx / Sum f (discrete) | Mean = Sum fm / Sum f (continuous), m = mid-point; Direct method.
Step deviation: Mean = A + h x Sum fd' / Sum f | d' = (m - A) / h; Short-cut for classes.
Weighted and combined: Weighted mean = Sum wx / Sum w | Combined mean = (n1 x1 + n2 x2) / (n1 + n2); Two special cases.

How to answer Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series questions

Choose the method that suits the data and set out a table.

1Individual series: add and divide, or use an assumed mean.
2Discrete: multiply each x by f, total and divide by N.
3Continuous: find mid-points, then fm or fd'.
4For a missing value, form an equation with the given mean.

Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Weighted mean

Suppose a student scores 70, 80, 60 and 90 in tests worth 4, 3, 2 and 1 credits respectively.

What is her weighted average and how does it differ from the simple mean?

Try it first, then show the solutionSolution

Sum wx = 280 + 240 + 120 + 90 = 730; Sum w = 10; weighted mean = 73.
Simple mean = 300 / 4 = 75.

Answer: Weighted mean 73 against simple mean 75.

Memory hook: Heavier tests count more.

Tricks to remember Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series

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

Sum over number
Mean = total / count.
Use it when: Basic.
A + h times average d prime
Short-cut: add h x (Sum fd' / N) to A.
Use it when: Step deviation.
Combine by totals
Add the group totals, divide by total size.
Use it when: Combined mean.

Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series: solved examples

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

Example 1: Property · 1 mark · MCQBoard level

The algebraic sum of deviations of items from their arithmetic mean is:
(a) maximum (b) zero (c) one (d) negative

Try it first, then show the solutionSolution

Property of the mean.

Answer: (b) zero

Example 2: Individual and discrete series · 4 marks · Short answerBoard level

(a) Find the mean of 12, 18, 25, 30, 35 by the assumed mean method with A = 25. (b) Find the mean for x = 2, 4, 6, 8, 10 with f = 3, 5, 8, 6, 3. (4 marks)

Try it first, then show the solutionSolution

(a) d = -13, -7, 0, 5, 10; Sum d = -5; mean = 25 - 5/5 = 24.
(b) Sum fx = 6 + 20 + 48 + 48 + 30 = 152; N = 25; mean = 152 / 25 = 6.08.

Answer: (a) 24; (b) 6.08.

Example 3: Continuous series · 6 marks · Long answerHarder board level

Find the mean of the series by the direct and step deviation methods: classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 8, 15, 9, 4. (6 marks)

Try it first, then show the solutionSolution

Mid-points: 5, 15, 25, 35, 45. Sum f = 40.
Direct: Sum fm = 20 + 120 + 375 + 315 + 180 = 1,010; mean = 1,010 / 40 = 25.25.
Step deviation with A = 25, h = 10: d' = -2, -1, 0, 1, 2; fd' = -8, -8, 0, 9, 8; Sum fd' = 1.
Mean = 25 + 10 x 1 / 40 = 25.25.

Answer: Mean = 25.25 by both methods.

Example 4: Missing frequency · 6 marks · Long answerHarder board level

The mean of the distribution is 25.25 and the frequencies for classes 0-10, 10-20, 20-30, 30-40, 40-50 are 4, 8, x, 9, 4. Find x. Then find the weighted mean of marks 70, 80, 60, 90 with weights 4, 3, 2, 1 and the combined mean of 30 students with mean 40 and 20 students with mean 55. (6 marks)

Try it first, then show the solutionSolution

Sum fm = 20 + 120 + 25x + 315 + 180 = 635 + 25x; N = 25 + x.
(635 + 25x) / (25 + x) = 25.25 gives 635 + 25x = 631.25 + 25.25x, so 3.75 = 0.25x and x = 15.
Weighted mean = (280 + 240 + 120 + 90) / 10 = 73.
Combined mean = (30 x 40 + 20 x 55) / 50 = (1,200 + 1,100) / 50 = 46.

Answer: x = 15; weighted mean 73; combined mean 46.

Example 5: Effect of an error · 4 marks · Short answerHOTS / competency

The mean of 50 observations was found to be 36. Later it was found that an observation 48 was wrongly read as 84. Find the correct mean. If every observation is then increased by 10% and Rs 5 is subtracted, find the new mean. (4 marks)

Try it first, then show the solutionSolution

Wrong total = 50 x 36 = 1,800.
Correct total = 1,800 - 84 + 48 = 1,764; correct mean = 1,764 / 50 = 35.28.
New mean = 35.28 x 1.10 - 5 = 38.808 - 5 = 33.808.

Answer: Correct mean 35.28; new mean 33.808.

Where students lose marks in Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series

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

Mistake: Dividing Sum fx by the number of classes
Wrong divisor.
Wrong working
152 / 5
Right working
152 / 25
Do this: Divide by Sum f.
Usually costs: 2 marks.
Mistake: Using class limits instead of mid-points
Careless.
Wrong working
Using 10, 20, ...
Right working
Using 5, 15, ...
Do this: Find mid-points.
Usually costs: 2 marks.
Mistake: Forgetting to multiply by h in the step deviation
Formula slip.
Wrong working
25 + 1/40
Right working
25 + 10 x 1/40
Do this: Multiply by h.
Usually costs: 2 marks.

Exam tips

Show the table with columns f, m, fm or d' and fd'.
Write N = Sum f and check the total.
Use short-cut methods to reduce arithmetic.

Practice questions on Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series

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 · MCQBoard level
The mean of 6, 8, 10, 12, 14 is:
(a) 9 (b) 10 (c) 11 (d) 12
Hint
Sum = 50.
Answer
(b) 10
Q2 · 3 marks · Short answerBoard level
State any two merits and two demerits of the arithmetic mean. (3 marks)
Hint
Uses all items; affected by extremes.
Answer
Merits: based on all observations and easy to calculate. Demerits: affected by extreme values and cannot be located graphically.
Q3 · 5 marks · Long answerHarder board level
Find the mean of classes 10-20, 20-30, 30-40, 40-50, 50-60 with frequencies 5, 10, 20, 10, 5. Then find the mean if the class 50-60 had frequency 15 instead. (5 marks)
Hint
Mid-points 15, 25, 35, 45, 55.
Answer
First: Sum fm = 75 + 250 + 700 + 450 + 275 = 1,750; N = 50; mean = 35. Second: Sum fm = 1,750 - 275 + 825 = 2,300; N = 60; mean = 38.33.
Q4 · 5 marks · Long answerHarder board level
Class 1 has 40 students with an average mark of 55 and class 2 has 60 students with an average of 70. Find the combined mean. If 10 students with an average of 40 join the class 1, find the new mean of class 1 and the new combined mean. (5 marks)
Hint
2,200 and 4,200.
Answer
Combined = (2,200 + 4,200) / 100 = 64. Class 1 new = (2,200 + 400) / 50 = 52. New combined = (2,600 + 4,200) / 110 = 61.82.
Q5 · 4 marks · Short answerHOTS / competency
A firm pays workers in three grades: 20 workers at Rs 400, 30 at Rs 500 and 10 at Rs 800 a day. The owner says "the average wage is Rs 566.67 (the simple mean of the three rates)." Evaluate, find the correct average and the total daily wage bill. (4 marks)
Hint
Weights are numbers of workers.
Answer
The simple mean ignores the number of workers. Weighted mean = (8,000 + 15,000 + 8,000) / 60 = 31,000 / 60 = Rs 516.67. Total wage bill = Rs 31,000.

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.Discrete mean formula?Show answer
Sum fx / Sum f.
2.Step deviation formula?Show answer
A + h Sum fd' / N.
3.Sum of deviations from mean?Show answer
Zero.
4.Weighted mean?Show answer
Sum wx / Sum w.
5.Mid-point of 20-30?Show answer
25.

Arithmetic Mean: Individual, Discrete, Continuous and Weighted Series: frequently asked questions

Why use an assumed mean?

To reduce the size of the numbers.

Can the answer depend on A?

No.

Is the mean always a data value?

No.

What if a class is open-ended?

Assume a width.