MathsCBSE Class 10 · Statistics

Step Deviation Method for the Mean of Grouped Data

Updated 14 October 2026 · Fact-checked

The step deviation method finds the mean of grouped data by choosing an assumed mean a, then dividing each deviation by the common class width h to get small numbers u. The mean is a plus h times the sum of f times u divided by the total frequency. Boards ask it directly, with missing frequencies, and in case studies.

Each class mark is a whole number of steps h from a, giving small u values, so multiply the average of u by h and add a at the end.

Illustration drawn for QuestionBuddy.

Understand Step Deviation Method

The mean of grouped data is $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ , where $x_{i}$ is the class mark. When class marks are large, like 550, 650 and 750, the products $f_{i} x_{i}$ become heavy. The step deviation method is a way to get the same mean with tiny numbers.

First you pick an assumed mean $a$ , usually the class mark of a middle class. Each deviation $d_{i} = x_{i} - a$ is then a multiple of the class width $h$ when all classes have equal width. So you divide once more: $u_{i} = \frac{x _{i} - a}{h}$ . Now the $u_{i}$ are whole numbers like $- 2, - 1, 0, 1, 2$ , and the mean is $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ .

Why it works. Since $x_{i} = a + h u_{i}$ , you get $\sum f_{i} x_{i} = a \sum f_{i} + h \sum f_{i} u_{i}$ . Divide both sides by $\sum f_{i}$ and the formula appears. You shrank every deviation by $h$ , so you must multiply back by $h$ at the end. That is the single most important line in the method.

How it connects. It is the assumed mean method with one extra division. All three methods (direct, assumed mean, step deviation) give exactly the same mean, so one can check another. The choice of $a$ never changes the answer; a middle class mark just keeps the numbers smallest.

When it holds. The method is cleanest when all classes have the same width $h$ . If widths differ, the $u_{i}$ are no longer whole numbers; you can still take $h$ as a common factor of all the deviations, but usually the assumed mean method is simpler then. Also, use class marks, not class limits. With inclusive classes like 25–34, the class marks (29.5 and so on) still differ by 10, so $h = 10$ and you do not need to make the classes continuous for the mean.

Step Deviation Method: formulas to remember

Step deviation: $u_{i} = \frac{x _{i} - a}{h}$; a is the assumed mean (a middle class mark), h the common class width. Middle class gets u = 0.
Mean by step deviation: $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$; Never drop the factor h. Divide by the total frequency, not the number of classes.
Link with assumed mean method: $d_{i} = x_{i} - a = h u_{i}, \sum f_{i} d_{i} = h \sum f_{i} u_{i}$; Shows why both methods give the same mean.
Changing the assumed mean by one class: $a^{'} = a - h \Rightarrow \sum f_{i} u_{i}^{'} = \sum f_{i} u_{i} + \sum f_{i}$; Every u increases by 1, so the sum rises by the total frequency. Useful for checking.

How to solve Step Deviation Method questions

Use this routine for any step deviation question, including ones with a missing frequency.

1Find the class mark $x_{i}$ of every class and check that consecutive class marks differ by the same amount $h$ .
2Choose $a$ as the class mark of a middle class (ideally the one with the largest frequency) and write $a$ and $h$ clearly.
3Make a table with columns: class, $f_{i}$ , $x_{i}$ , $u_{i} = \frac{x _{i} - a}{h}$ , $f_{i} u_{i}$ . Classes below $a$ get negative $u_{i}$ .
4Add the $f_{i}$ column and the $f_{i} u_{i}$ column carefully, keeping signs.
5Write the formula $\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ , then substitute.
6If a frequency is unknown, express both sums in terms of it, equate to the given mean and solve the linear equation.
7Check that the mean lies within the data range and near the busiest classes, and write the unit.

Step Deviation Method 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.

Step deviation method

Suppose a bus depot records the distance (in km) run by each of its 60 buses on one day. The record reads: 200–250 km: 6 buses, 250–300 km: 12, 300–350 km: $k$ , 350–400 km: 14, 400–450 km: 8. One count, $k$ , was not written down, but all 60 buses ran that day.

(a) Find $k$ . (b) Find the mean distance run by a bus using the step deviation method. (c) Find the total distance run by all the buses. (d) The next day every bus runs exactly 10 km more. What is the new mean, and why?

Try it first, then show the solutionSolution

(a) $6 + 12 + k + 14 + 8 = 60$ , so $k = 20$ .
(b) Class marks: 225, 275, 325, 375, 425. Take $a = 325$ (the class with the largest frequency) and $h = 50$ .
$u_{i} = \frac{x _{i} - 325}{50}$ : $- 2, - 1, 0, 1, 2$ . $f_{i} u_{i}$ : $- 12, - 12, 0, 14, 16$ , so $\sum f_{i} u_{i} = 6$ .
Mean $= 325 + 50 \times \frac{6}{60} = 325 + 5 = 330$ km.
(c) Total distance $= 60 \times 330 = 19800$ km.
(d) Adding 10 km to every value adds 10 km to the mean: $330 + 10 = 340$ km. Every class mark shifts by 10, so the new mean is the old mean plus 10.

Memory hook: Big numbers shrink by h, then stretch back: shift every value and the mean shifts with it.

Tricks to remember Step Deviation Method

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

Shrink, then stretch
You shrank deviations by h to make u; stretch the answer back by multiplying by h.
Use it when: Stops you writing a + Σfu/Σf, the commonest lost mark.
Zero in the middle
The class you choose as a gets u = 0; above it count 1, 2, 3; below it count −1, −2, −3.
Use it when: Write u values in seconds without dividing each one.
A-H-U order
Assume a, find h, then fill u. Write a and h above the table.
Use it when: Examiners see your choices clearly and you avoid mixing values.

Step Deviation Method: solved examples

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

Example 1: Basic step deviation · 2 marks · Very short answerBoard level

Find the mean of the distribution using the step deviation method.
Classes: 10–20, 20–30, 30–40, 40–50, 50–60
Frequencies: 5, 8, 12, 10, 5

Try it first, then show the solutionSolution

Class marks 15, 25, 35, 45, 55; take $a = 35$ , $h = 10$ ; $u_{i} = - 2, - 1, 0, 1, 2$ . (½ mark)
$f_{i} u_{i} = - 10, - 8, 0, 10, 10$ ; $\sum f_{i} u_{i} = 2$ , $\sum f_{i} = 40$ . (½ mark)
$\overset{x}{ˉ} = 35 + 10 \times \frac{2}{40} = 35 + 0.5$ . (½ mark)
$\overset{x}{ˉ} = 35.5$ . (½ mark)

Answer: Mean = 35.5

Example 2: Large class marks · 3 marks · Short answerBoard level

The daily earnings (in Rs) of 50 shopkeepers in a market are given. Find the mean daily earning by the step deviation method.
Earnings (Rs): 500–600, 600–700, 700–800, 800–900, 900–1000
Shopkeepers: 6, 10, 15, 12, 7

Try it first, then show the solutionSolution

Class marks 550, 650, 750, 850, 950; $a = 750$ , $h = 100$ . (½ mark)
$u_{i} = - 2, - 1, 0, 1, 2$ . (½ mark)
$f_{i} u_{i} = - 12, - 10, 0, 12, 14$ ; $\sum f_{i} u_{i} = 4$ , $\sum f_{i} = 50$ . (1 mark)
$\overset{x}{ˉ} = 750 + 100 \times \frac{4}{50} = 750 + 8$ . (½ mark)
$\overset{x}{ˉ} = 758$ . (½ mark)

Answer: Mean daily earning = Rs 758

Example 3: Missing frequency from the mean · 3 marks · Short answerHarder board level

The mean of the distribution below is 52. Using the step deviation method, find $f$ .
Classes: 0–20, 20–40, 40–60, 60–80, 80–100
Frequencies: 7, $f$ , 12, 13, 8

Try it first, then show the solutionSolution

Class marks 10, 30, 50, 70, 90; $a = 50$ , $h = 20$ ; $u_{i} = - 2, - 1, 0, 1, 2$ . (½ mark)
$f_{i} u_{i} = - 14, - f, 0, 13, 16$ ; $\sum f_{i} u_{i} = 15 - f$ ; $\sum f_{i} = 40 + f$ . (1 mark)
$52 = 50 + 20 \times \frac{15 - f}{40 + f}$ , so $2 (40 + f) = 20 (15 - f)$ . (½ mark)
$80 + 2 f = 300 - 20 f \Rightarrow 22 f = 220 \Rightarrow f = 10$ . (½ mark)
Check: $\sum f_{i} u_{i} = 5$ , $\sum f_{i} = 50$ , $50 + 20 \times \frac{5}{50} = 52$ . (½ mark)

Answer: f = 10

Example 4: Two missing frequencies · 5 marks · Long answerHarder board level

The distribution below has total frequency 80 and mean 31. Find $p$ and $q$ using the step deviation method.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Frequencies: 8, $p$ , 20, $q$ , 14, 10

Try it first, then show the solutionSolution

Total: $52 + p + q = 80 \Rightarrow p + q = 28$ . (1 mark)
Class marks 5, 15, 25, 35, 45, 55; $a = 25$ , $h = 10$ ; $u_{i} = - 2, - 1, 0, 1, 2, 3$ . (1 mark)
$f_{i} u_{i} = - 16, - p, 0, q, 28, 30$ ; $\sum f_{i} u_{i} = 42 - p + q$ . (1 mark)
$31 = 25 + 10 \times \frac{42 - p + q}{80} \Rightarrow 42 - p + q = 48 \Rightarrow q - p = 6$ . (1 mark)
Adding with $p + q = 28$ : $2 q = 34$ , $q = 17$ , $p = 11$ . (½ mark)
Check: $\sum f_{i} u_{i} = - 16 - 11 + 17 + 28 + 30 = 48$ ; $25 + 10 \times \frac{48}{80} = 31$ . (½ mark)

Answer: p = 11, q = 17

Example 5: Case study: sprint timings · 4 marks · Case-basedHOTS / competency

Suppose a PE teacher times 40 students in a 100 m race.
Time (s): 12–14, 14–16, 16–18, 18–20, 20–22
Students: 4, 10, 14, 8, 4
She uses the step deviation method with $a = 17$ .
(i) Write $h$ and the value of $u$ for the class 14–16. (1 mark)
(ii) Find $\sum f_{i} u_{i}$ . (1 mark)
(iii) A student says the mean is 16.95 s. Find the correct mean, state the student's error, and say whether the class mean is under 17 s. (2 marks)

Try it first, then show the solutionSolution

(i) Class marks 13, 15, 17, 19, 21, so $h = 2$ ; $u = \frac{15 - 17}{2} = - 1$ . (1 mark)
(ii) $u_{i} = - 2, - 1, 0, 1, 2$ ; $f_{i} u_{i} = - 8, - 10, 0, 8, 8$ ; $\sum f_{i} u_{i} = - 2$ . (1 mark)
(iii) $\overset{x}{ˉ} = 17 + 2 \times \frac{- 2}{40} = 17 - 0.1 = 16.9$ s. (1 mark)
The student computed $17 - \frac{2}{40} = 16.95$ , forgetting to multiply by $h = 2$ . The correct mean 16.9 s is under 17 s. (1 mark)

Answer: (i) h = 2, u = −1 (ii) −2 (iii) Mean = 16.9 s; the student forgot to multiply by h; yes, the mean is under 17 s.

Where students lose marks in Step Deviation Method

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

Mistake: Forgetting to multiply by h
The u values were divided by h, so the average of u is h times too small.
Wrong working
$\overset{x}{ˉ} = 750 + \frac{4}{50} = 750.08$
Right working
$\overset{x}{ˉ} = 750 + 100 \times \frac{4}{50} = 758$
Do this: Write the full formula with h before substituting.
Usually costs: Final 1 mark, often more as the answer is far off.
Mistake: Wrong signs of u below a
Classes below the assumed mean have negative deviations.
Wrong working
$u_{i} = 2, 1, 0, 1, 2$ , giving $\sum f_{i} u_{i} = 12 + 10 + 0 + 12 + 14 = 48$
Right working
$u_{i} = - 2, - 1, 0, 1, 2$ , giving $\sum f_{i} u_{i} = 4$
Do this: Classes with smaller class marks than a get negative u: −1, −2, and so on.
Usually costs: 1 to 2 marks.
Mistake: Using class limits instead of class marks
The method needs the midpoint of each class.
Wrong working
Lower limits 500, 600, ..., 900 taken as $x_{i}$ with $a = 700$ : mean $= 700 + 100 \times \frac{4}{50} = 708$ , which is 50 too low
Right working
Class marks 550, 650, ..., 950 with $a = 750$ : mean $= 750 + 100 \times \frac{4}{50} = 758$
Do this: Always add an $x_{i}$ column first.
Usually costs: Table mark and final answer.
Mistake: Dividing by the number of classes
The mean divides by the total frequency.
Wrong working
$35 + 10 \times \frac{2}{5} = 39$
Right working
$35 + 10 \times \frac{2}{40} = 35.5$
Do this: Total the f column and use that sum.
Usually costs: 1 mark.

Exam tips

If the question names the step deviation method, you must show a u column; another method may lose the method marks.
Write a and h above your table so the examiner can follow your choices.
In missing-frequency questions, verify by substituting your value back; it takes under a minute.
Check the mean lies near the class with the highest frequency; a mean outside the data range means an error.
For inclusive classes like 25–34, take h as the difference of class marks; no need to make the classes continuous for the mean.

Practice questions on Step Deviation Method

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
If $u_{i} = \frac{x _{i} - 25}{10}$ , $\sum f_{i} u_{i} = 12$ and $\sum f_{i} = 40$ , the mean is
(a) 28
(b) 25.3
(c) 37
(d) 37.5
Hint
Read a and h from the formula for $u_{i}$ .
Answer
(a) 28. Here $a = 25$ and $h = 10$ , so mean $= 25 + 10 \times \frac{12}{40} = 25 + 3 = 28$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): For a given table, the mean found by the step deviation method is the same whichever class mark is taken as $a$ .
Reason (R): Since $x_{i} = a + h u_{i}$ , the expression $a + h \frac{\sum f _{i} u _{i}}{\sum f _{i}}$ is just $\frac{\sum f _{i} x _{i}}{\sum f _{i}}$ written in another way.
(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
Substitute $x_{i} = a + h u_{i}$ into $\sum f_{i} x_{i}$ .
Answer
(a). $\sum f_{i} x_{i} = a \sum f_{i} + h \sum f_{i} u_{i}$ , so dividing by $\sum f_{i}$ gives exactly the step deviation formula. The right side is the true mean for any choice of a, which is why A holds.
Q3 · 3 marks · Short answerBoard level
The lengths (in mm) of 40 leaves are: 100–120: 6, 120–140: 9, 140–160: 12, 160–180: 8, 180–200: 5. Find the mean length by the step deviation method.
Hint
Take a = 150 and h = 20.
Answer
u = −2, −1, 0, 1, 2; Σfu = −12 − 9 + 0 + 8 + 10 = −3; mean = 150 + 20 × (−3/40) = 150 − 1.5 = 148.5 mm.
Q4 · 3 marks · Short answerHarder board level
The mean of the distribution is 25.2. Find k using the step deviation method.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 4, k, 15, 11, 6
Hint
With a = 25, h = 10, Σfu = 15 − k and Σf = 36 + k.
Answer
25.2 = 25 + 10(15 − k)/(36 + k) gives 0.2(36 + k) = 150 − 10k, so 7.2 + 0.2k = 150 − 10k, 10.2k = 142.8, k = 14. Check: Σfu = 1, Σf = 50, mean = 25 + 0.2 = 25.2.
Q5 · 5 marks · Long answerHarder board level
The total frequency is 50 and the mean is 51.6. Find f₁ and f₂.
Classes: 0–20, 20–40, 40–60, 60–80, 80–100
Frequencies: 5, f₁, 15, f₂, 8
Hint
One equation from the total, one from the mean with a = 50, h = 20.
Answer
f₁ + f₂ = 22. Σfu = −10 − f₁ + f₂ + 16 = 6 − f₁ + f₂. 51.6 = 50 + 20(6 − f₁ + f₂)/50 gives 6 − f₁ + f₂ = 4, so f₁ − f₂ = 2. Hence f₁ = 12, f₂ = 10.
Q6 · 3 marks · Short answerBoard level
The ages (in years) of 50 members of a club are: 25–34: 8, 35–44: 12, 45–54: 15, 55–64: 10, 65–74: 5. Find the mean age by the step deviation method.
Hint
Class marks are 29.5, 39.5, ...; they differ by 10.
Answer
a = 49.5, h = 10, u = −2, −1, 0, 1, 2; Σfu = −16 − 12 + 0 + 10 + 10 = −8; mean = 49.5 + 10 × (−8/50) = 49.5 − 1.6 = 47.9 years.
Q7 · 3 marks · Short answerHOTS / competency
For a distribution with equal class width 10 and Σf = 60, Riya took a = 45 and found Σfu = −12. She wrote the mean as 44.8. (i) Is she right? (ii) Her friend uses a = 35 for the same data. What value of Σfu should the friend get?
Hint
Lowering a by one class adds 1 to every u.
Answer
(i) No. Mean = 45 + 10 × (−12/60) = 45 − 2 = 43; she forgot to multiply by h. (ii) Each u rises by 1, so Σfu = −12 + 60 = 48. Check: 35 + 10 × 48/60 = 43.
Q8 · 4 marks · Long answerHarder board level
The marks of 50 students are given as a 'less than' table, with one entry smudged:
Less than 20: 4; less than 40: $p$ ; less than 60: 30; less than 80: 44; less than 100: 50.
The mean mark is 52.4.
(i) Write the frequencies of the classes 0–20, 20–40, 40–60, 60–80 and 80–100. (ii) Using the step deviation method, find $p$ . (iii) Hence write the complete frequency table and check the mean.
Hint
Each frequency is the difference of two consecutive 'less than' totals, so two of them involve p.
Answer
(i) Frequencies: 4, $p - 4$ , $30 - p$ , 14, 6. (1)
(ii) Class marks 10, 30, 50, 70, 90; take $a = 50$ , $h = 20$ , $u_{i}$ : $- 2, - 1, 0, 1, 2$ . $\sum f_{i} u_{i} = - 8 - (p - 4) + 0 + 14 + 12 = 22 - p$ . Mean $= 50 + 20 \times \frac{22 - p}{50} = 52.4$ , so $22 - p = 6$ and $p = 16$ . (2)
(iii) Frequencies 4, 12, 14, 14, 6 (total 50). Check: $\sum f_{i} u_{i} = - 8 - 12 + 0 + 14 + 12 = 6$ , mean $= 50 + 20 \times \frac{6}{50} = 52.4$ . (1)
Q9 · 4 marks · Case-basedHOTS / competency
Case study: A town's water board records the daily water use (in litres) of 60 households.
Use (L): 100–120, 120–140, 140–160, 160–180, 180–200, 200–220
Households: 5, 9, 14, 16, 10, 6
(i) Choose a suitable $a$ and $h$ and write the $u_{i}$ column. (1 mark)
(ii) Find $\sum f_{i} u_{i}$ and the mean daily use. (2 marks)
(iii) A notice claims that the average household here uses more than 165 litres a day. Is the claim supported? If every household cuts its use by 10%, what will the new mean be? (1 mark)
Hint
Take a as the class mark with the largest frequency; a 10% cut multiplies every value, and so the mean, by 0.9.
Answer
(i) Class marks 110, 130, 150, 170, 190, 210; take $a = 170$ , $h = 20$ ; $u_{i}$ : $- 3, - 2, - 1, 0, 1, 2$ . (1)
(ii) $f_{i} u_{i}$ : $- 15, - 18, - 14, 0, 10, 12$ , so $\sum f_{i} u_{i} = - 25$ . Mean $= 170 + 20 \times \frac{- 25}{60} = 170 - 8.33 \approx 161.67$ L. (2)
(iii) No: the mean is about 161.67 L, below 165 L. After a 10% cut every value is multiplied by 0.9, so the mean becomes $0.9 \times 161.67 \approx 145.5$ L. (1)

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.Formula for u in step deviation?Show answer
$u_{i} = \frac{x _{i} - a}{h}$
2.Mean by step deviation?Show answer
$\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{i}}$
3.Which class usually gives a?Show answer
A middle class mark, often of the class with the highest frequency.
4.Does the choice of a change the mean?Show answer
No, every choice gives the same mean.
5.What is h?Show answer
The common class width, the gap between consecutive class marks.

Step Deviation Method: frequently asked questions

What is the difference between the assumed mean and step deviation methods?

Both use an assumed mean a. The step deviation method also divides each deviation by the class width h, making the numbers smaller, and multiplies by h at the end. Both give the same answer.

Can I use the step deviation method if class widths are unequal?

Only if all deviations share a common factor you can use as h. Usually it is simpler to use the assumed mean method for unequal widths.

Which method should I use if the question does not name one?

Any method earns full marks. Step deviation is fastest when class marks are large and widths are equal.

Which class mark should I take as a?

Take a class mark near the middle of the table, ideally the one with the largest frequency. Then the u values stay small (−2 to 2) and one of them is 0, so there are fewer chances of an arithmetic slip. Any class mark gives the same mean.