Fundamentals of Business Mathematics and Statistics · Measures of Central Tendency and Dispersion

Arithmetic Mean: Formula, Methods and Solved Problems

Updated 10 October 2026 · Fact-checked

The arithmetic mean is the sum of all observations divided by their number. For grouped data you use x̄ = Σfx ÷ Σf with class mid-points. To save effort, take an assumed mean A and class width h, then use x̄ = A + (Σfd ÷ Σf) × h. Weighted and combined means follow the same idea.

Understand Arithmetic Mean

The arithmetic mean (AM) is the most common average. It is the value each item would have if the total were shared equally. If five students score 40, 50, 60, 70 and 80, the total is 300. Shared equally, each gets 60. So the mean is 60.

For raw data, add all values and divide by how many there are. For a frequency table, each value counts as many times as its frequency. So you multiply each value by its frequency, add, and divide by the total frequency. For classes, use the mid-point of each class as the value.

A weighted mean is used when items are not equally important. Each value is multiplied by its weight, not its frequency. For example, a final result may count an exam at 70% and a project at 30%. The simple mean would ignore that difference.

The mean has a useful property: the sum of deviations of items from the mean is always zero. This is why a wrong total gives a wrong mean, and why corrections work by fixing the total. The mean also uses every item, so extreme values pull it up or down.

The assumed mean and step deviation methods are shortcuts. You pick a convenient value near the middle, work with small deviations, and then adjust back. The answer is exactly the same as the direct method. Only the arithmetic is lighter.

Key formulas to remember

Simple mean (raw data): x̄ = Σx ÷ n; n is the number of observations.
Mean of a frequency distribution: x̄ = Σfx ÷ Σf; For classes, x is the mid-point of the class. Σf = N.
Weighted mean: x̄w = Σwx ÷ Σw; Use when items have different importance. w is the weight.
Assumed mean method: x̄ = A + Σfd ÷ Σf, where d = x − A; A is the assumed mean. Any A gives the same answer.
Step deviation method: x̄ = A + (Σfd′ ÷ Σf) × h, where d′ = (x − A) ÷ h; h is the common class width. Use only when mid-points are equally spaced.
Combined mean of two groups: x̄12 = (n1x̄1 + n2x̄2) ÷ (n1 + n2); Weight each group mean by its size. Do not average the two means directly unless n1 = n2.
Corrected mean: Correct Σx = Wrong Σx − Wrong values + Correct values; Correct mean = Correct Σx ÷ n; First get the wrong total as n × wrong mean. If an item is omitted or added, n changes too.
Change of origin and scale: If y = a + bx, then ȳ = a + b x̄; Adding a constant to every item adds it to the mean. Multiplying every item multiplies the mean.

How to solve Arithmetic Mean questions

Use this method for any arithmetic mean question, whether it is raw data, a frequency table, a combined group or a correction.

1Identify the type: raw data, frequency table, weighted, combined groups, or a corrected mean.
2Write down what is given: n or Σf, the old mean, group sizes, weights, or the table.
3For classes, find the mid-point of each class: (lower limit + upper limit) ÷ 2. Convert inclusive classes to continuous form only if the table needs it.
4Choose the method. For small, simple numbers use Σfx ÷ Σf. For large values or classes, pick A near the middle and use deviations or step deviations.
5Compute the column totals carefully. Keep the signs of negative deviations.
6Apply the formula. In the step method, remember to multiply by h at the end and add A.
7For correction or combined questions, work with totals (mean × number), change the total, then divide again by the correct count.
8Check if the answer lies between the smallest and largest value. A mean outside that range is wrong.

Quickest way: Total-first shortcut and mid-class assumed mean

When to use it: Use this under time pressure for MCQs on corrected means, combined means and grouped data.

For correction and combined questions, never recompute the whole table. Convert the mean to a total (n × mean) and adjust only the changed items.
For corrections, add the net difference: (correct value − wrong value) for each item. Then divide by the right n.
For grouped data, take A as a mid-point near the middle and h as the class width. Most columns of d′ then become −2, −1, 0, 1, 2.
Write only Σf and Σfd′. The answer is A + (Σfd′ ÷ Σf) × h.
Test the four options. The mean must lie inside the data range, and often near A, so you can reject options that are far away.

Common mistakes in Arithmetic Mean

Using class limits instead of mid-points in a grouped table.
Students multiply frequency by the lower or upper limit because it is printed in the table.
Fix: Always compute the mid-point first and write it as a separate column.
Forgetting to multiply by h and add A in the step deviation method.
Students stop after finding Σfd′ ÷ Σf and treat it as the mean.
Fix: Write the full formula A + (Σfd′ ÷ Σf) × h before substituting. Check that the answer is near A.
Averaging two group means directly for the combined mean.
It looks like a simple average of two numbers.
Fix: Use (n1x̄1 + n2x̄2) ÷ (n1 + n2). The larger group pulls the combined mean toward its own mean.
Dividing the corrected total by the old n when an item was added or dropped.
Students focus on the values and forget the count changes.
Fix: Decide if n changes. Removing a wrongly included item reduces n by one. Adding a missed item increases it.
Subtracting the correct value and adding the wrong one in a corrected mean.
The direction of the correction is reversed under pressure.
Fix: Remember: remove what was wrongly used, add what is correct. Net change = correct − wrong.
Using the simple mean when weights are given.
Students overlook the weights or think they are frequencies of a different kind.
Fix: If importance differs, use Σwx ÷ Σw. Write weights in a column next to the values.

Worked examples

Example 1

Find the arithmetic mean of the following distribution using the step deviation method. Classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.

Show the solution

Total frequency Σf = 5 + 8 + 15 + 16 + 6 = 50.
Mid-points are 5, 15, 25, 35, 45. Take A = 25 and h = 10.
d′ = (x − 25) ÷ 10 gives −2, −1, 0, 1, 2.
fd′ = 5×(−2) = −10; 8×(−1) = −8; 15×0 = 0; 16×1 = 16; 6×2 = 12.
Σfd′ = −10 − 8 + 0 + 16 + 12 = 10.
Mean = 25 + (10 ÷ 50) × 10 = 25 + 2 = 27.
Check by the direct method: Σfx = 25 + 120 + 375 + 560 + 270 = 1350, and 1350 ÷ 50 = 27.

Answer: Arithmetic mean = 27

Example 2

The mean of 40 observations was found to be 50. Later it was found that 84 was misread as 48, and 29 was misread as 25. Find the correct mean.

Show the solution

Wrong total = 40 × 50 = 2,000.
Correction for the first item = 84 − 48 = +36.
Correction for the second item = 29 − 25 = +4.
Correct total = 2,000 + 36 + 4 = 2,040.
The number of observations is unchanged at 40.
Correct mean = 2,040 ÷ 40 = 51.

Answer: Correct mean = 51

Exam tips

Most questions are one of four types: grouped mean, combined mean, weighted mean or corrected mean. Spot the type first and pick the matching formula.
In correction questions, work only with totals. It is faster and avoids slips.
With no negative marking, always mark an option. If stuck on a grouped table, estimate the mean near the middle class and eliminate options outside the data range.
For combined mean, the answer must lie between the two group means and closer to the bigger group's mean. Use this to eliminate options quickly.
Check whether the classes are inclusive (like 10–19) or continuous. Mid-points are the same either way, but a mismatch in width h will spoil the step deviation.

Practice questions from Measures of Central Tendency and Dispersion

Arithmetic Mean in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Arithmetic Mean: frequently asked questions

What is the arithmetic mean formula for CMA Foundation?

For raw data it is x̄ = Σx ÷ n. For a frequency distribution it is x̄ = Σfx ÷ Σf, using class mid-points for grouped data. The shortcut forms are the assumed mean and step deviation methods.

How do I calculate the arithmetic mean by the step deviation method?

Find mid-points, choose A near the middle and h as the class width. Compute d′ = (x − A) ÷ h and then Σfd′. The mean is A + (Σfd′ ÷ Σf) × h.

What is the difference between simple and weighted arithmetic mean?

The simple mean treats every item as equally important. The weighted mean multiplies each item by a weight that shows its importance, then divides by the total weight. Use it for things like marks with different weightage.

How do I solve corrected mean problems?

Find the wrong total as n × wrong mean. Subtract the wrong values, add the correct values, and divide by the correct number of items. Check whether any item was added or dropped, since that changes n.

Does the choice of assumed mean change the answer?

No. Any assumed mean gives the same final mean, because the adjustment term corrects for the difference. A value near the middle just makes the deviations smaller and easier to handle.