Mean by Step-Deviation Method
The step-deviation method is the most efficient technique for calculating the mean of grouped data, especially when class intervals have a uniform width and the numbers involved are large.
It is a refinement of the assumed mean method. By dividing deviations by the class width (h), the working numbers become small integers, making calculations faster and less error-prone.
This method is preferred in CBSE Class 10 when the frequency distribution has equal class widths and large mid-values.
What is Mean by Step-Deviation Method?
Definition: The step-deviation method calculates the mean of grouped data by using a change of variable that converts mid-values into small integers called step-deviations (uᵢ).
Key terms:
- xᵢ = class mark (mid-value) of the i-th class = (upper limit + lower limit) / 2
- a = assumed mean (typically the class mark of the middle or highest-frequency class)
- h = class width (difference between upper and lower limits of a class)
- uᵢ = step-deviation = (xᵢ − a) / h
- fᵢ = frequency of the i-th class
Why use step-deviation?
- Direct method requires computing Σfᵢxᵢ with large numbers.
- Assumed mean method requires computing Σfᵢdᵢ where dᵢ can still be large.
- Step-deviation method makes uᵢ small integers (…, −2, −1, 0, 1, 2, …), so Σfᵢuᵢ is easy to compute.
Mean by Step-Deviation Method Formula
Step-Deviation Formula for Mean:
Mean (̄x) = a + h × (Σfᵢuᵢ / Σfᵢ)
Where:
- a = assumed mean
- h = class width
- uᵢ = (xᵢ − a) / h
- fᵢ = frequency of i-th class
- Σfᵢuᵢ = sum of products of frequency and step-deviation
- Σfᵢ = total frequency (N)
Comparison of three methods:
| Method | Formula | Best When |
|---|---|---|
| Direct | ̄x = Σfᵢxᵢ / Σfᵢ | xᵢ values are small |
| Assumed Mean | ̄x = a + Σfᵢdᵢ / Σfᵢ | xᵢ values are large |
| Step-Deviation | ̄x = a + h(Σfᵢuᵢ / Σfᵢ) | xᵢ large, equal class widths |
Derivation and Proof
Derivation from the direct method:
- Direct method: ̄x = Σfᵢxᵢ / Σfᵢ
- Let a be the assumed mean. Define dᵢ = xᵢ − a, so xᵢ = a + dᵢ.
- Then: Σfᵢxᵢ = Σfᵢ(a + dᵢ) = aΣfᵢ + Σfᵢdᵢ
- So: ̄x = a + Σfᵢdᵢ / Σfᵢ (this is the assumed mean method).
- Now define uᵢ = dᵢ / h = (xᵢ − a) / h, so dᵢ = h × uᵢ.
- Then: Σfᵢdᵢ = h × Σfᵢuᵢ
- Substituting: ̄x = a + h(Σfᵢuᵢ / Σfᵢ)
The step-deviation method is algebraically equivalent to both the direct method and the assumed mean method. All three give the same answer. The step-deviation method simply makes the arithmetic easier.
Types and Properties
When to use each method:
- Direct method: Use when class marks are small (single or double digits) and frequencies are small. No assumed mean needed.
- Assumed mean method: Use when class marks are large but class widths are unequal.
- Step-deviation method: Use when class marks are large AND class widths are equal. This is the most efficient choice for CBSE problems.
Choosing the assumed mean (a):
- Pick the class mark of the class with the highest frequency (modal class).
- Or pick the class mark nearest to the middle of the distribution.
- The answer is the same regardless of which a you choose — but a good choice makes uᵢ values symmetric around 0, simplifying computation.
Methods
Steps to calculate mean by step-deviation method:
- Find class marks (xᵢ): xᵢ = (lower limit + upper limit) / 2 for each class.
- Choose assumed mean (a): Pick the xᵢ of the class with highest frequency or the middle class.
- Find class width (h): h = upper limit − lower limit of any class (must be uniform).
- Calculate uᵢ: uᵢ = (xᵢ − a) / h for each class. These should be integers.
- Calculate fᵢuᵢ: Multiply frequency by step-deviation for each class.
- Find Σfᵢuᵢ: Sum all fᵢuᵢ values (include sign).
- Find Σfᵢ: Sum all frequencies = N.
- Apply formula: ̄x = a + h × (Σfᵢuᵢ / Σfᵢ).
Common mistakes:
- Forgetting the sign of uᵢ (negative for classes below a, positive for classes above a).
- Using unequal class widths — step-deviation requires equal class widths.
- Choosing a value that is NOT a class mark — this makes uᵢ non-integer and defeats the purpose.
- Forgetting to multiply Σfᵢuᵢ / Σfᵢ by h in the final step.
Solved Examples
Example 1: Basic Step-Deviation Calculation
Problem: Find the mean of the following distribution:
| Class | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 |
|---|---|---|---|---|---|
| Frequency | 7 | 8 | 12 | 10 | 3 |
Solution:
Given: h = 20. Choose a = 50 (class mark of 40–60, highest frequency).
| Class | fᵢ | xᵢ | uᵢ = (xᵢ−50)/20 | fᵢuᵢ |
|---|---|---|---|---|
| 0–20 | 7 | 10 | −2 | −14 |
| 20–40 | 8 | 30 | −1 | −8 |
| 40–60 | 12 | 50 | 0 | 0 |
| 60–80 | 10 | 70 | 1 | 10 |
| 80–100 | 3 | 90 | 2 | 6 |
| Total | 40 | −6 |
Applying the formula:
- ̄x = 50 + 20 × (−6/40)
- ̄x = 50 + 20 × (−0.15)
- ̄x = 50 − 3 = 47
Answer: Mean = 47
Example 2: Marks Distribution
Problem: Find the mean marks:
| Marks | 10–25 | 25–40 | 40–55 | 55–70 | 70–85 |
|---|---|---|---|---|---|
| Students | 6 | 10 | 14 | 8 | 2 |
Solution:
Given: h = 15. Choose a = 47.5 (class mark of 40–55, highest frequency).
| Marks | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 10–25 | 6 | 17.5 | −2 | −12 |
| 25–40 | 10 | 32.5 | −1 | −10 |
| 40–55 | 14 | 47.5 | 0 | 0 |
| 55–70 | 8 | 62.5 | 1 | 8 |
| 70–85 | 2 | 77.5 | 2 | 4 |
| Total | 40 | −10 |
Applying the formula:
- ̄x = 47.5 + 15 × (−10/40)
- ̄x = 47.5 + 15 × (−0.25)
- ̄x = 47.5 − 3.75 = 43.75
Answer: Mean marks = 43.75
Example 3: Daily Wages of Workers
Problem: Find the mean daily wage:
| Wage (Rs) | 200–250 | 250–300 | 300–350 | 350–400 | 400–450 |
|---|---|---|---|---|---|
| Workers | 5 | 12 | 18 | 10 | 5 |
Solution:
Given: h = 50. Choose a = 325 (class mark of 300–350).
| Wage | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 200–250 | 5 | 225 | −2 | −10 |
| 250–300 | 12 | 275 | −1 | −12 |
| 300–350 | 18 | 325 | 0 | 0 |
| 350–400 | 10 | 375 | 1 | 10 |
| 400–450 | 5 | 425 | 2 | 10 |
| Total | 50 | −2 |
Applying the formula:
- ̄x = 325 + 50 × (−2/50)
- ̄x = 325 + 50 × (−0.04)
- ̄x = 325 − 2 = 323
Answer: Mean daily wage = Rs 323
Example 4: Age Distribution
Problem: Find the mean age of the following group:
| Age (years) | 15–25 | 25–35 | 35–45 | 45–55 | 55–65 |
|---|---|---|---|---|---|
| Persons | 4 | 14 | 22 | 16 | 4 |
Solution:
Given: h = 10. Choose a = 40 (class mark of 35–45, highest f).
| Age | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 15–25 | 4 | 20 | −2 | −8 |
| 25–35 | 14 | 30 | −1 | −14 |
| 35–45 | 22 | 40 | 0 | 0 |
| 45–55 | 16 | 50 | 1 | 16 |
| 55–65 | 4 | 60 | 2 | 8 |
| Total | 60 | 2 |
Applying the formula:
- ̄x = 40 + 10 × (2/60)
- ̄x = 40 + 10 × (1/30)
- ̄x = 40 + 0.333 = 40.33
Answer: Mean age = 40.33 years
Example 5: Electricity Consumption
Problem: Find the average monthly electricity consumption:
| Units | 50–100 | 100–150 | 150–200 | 200–250 | 250–300 | 300–350 |
|---|---|---|---|---|---|---|
| Households | 8 | 15 | 20 | 12 | 10 | 5 |
Solution:
Given: h = 50. Choose a = 175 (class mark of 150–200, highest f).
| Units | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 50–100 | 8 | 75 | −2 | −16 |
| 100–150 | 15 | 125 | −1 | −15 |
| 150–200 | 20 | 175 | 0 | 0 |
| 200–250 | 12 | 225 | 1 | 12 |
| 250–300 | 10 | 275 | 2 | 20 |
| 300–350 | 5 | 325 | 3 | 15 |
| Total | 70 | 16 |
Applying the formula:
- ̄x = 175 + 50 × (16/70)
- ̄x = 175 + 50 × 0.2286
- ̄x = 175 + 11.43 = 186.43
Answer: Mean consumption = 186.43 units
Example 6: Height of Students
Problem: Find the mean height:
| Height (cm) | 140–145 | 145–150 | 150–155 | 155–160 | 160–165 |
|---|---|---|---|---|---|
| Students | 5 | 15 | 25 | 10 | 5 |
Solution:
Given: h = 5. Choose a = 152.5 (class mark of 150–155, highest f).
| Height | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 140–145 | 5 | 142.5 | −2 | −10 |
| 145–150 | 15 | 147.5 | −1 | −15 |
| 150–155 | 25 | 152.5 | 0 | 0 |
| 155–160 | 10 | 157.5 | 1 | 10 |
| 160–165 | 5 | 162.5 | 2 | 10 |
| Total | 60 | −5 |
Applying the formula:
- ̄x = 152.5 + 5 × (−5/60)
- ̄x = 152.5 + 5 × (−0.0833)
- ̄x = 152.5 − 0.42 = 152.08
Answer: Mean height = 152.08 cm
Example 7: Comparison with Direct Method
Problem: Verify that the step-deviation method gives the same answer as the direct method for:
| Class | 0–10 | 10–20 | 20–30 |
|---|---|---|---|
| f | 3 | 5 | 2 |
Solution:
Direct method:
- xᵢ: 5, 15, 25
- Σfᵢxᵢ = 3(5) + 5(15) + 2(25) = 15 + 75 + 50 = 140
- ̄x = 140/10 = 14
Step-deviation method: a = 15, h = 10
- uᵢ: −1, 0, 1
- Σfᵢuᵢ = 3(−1) + 5(0) + 2(1) = −3 + 0 + 2 = −1
- ̄x = 15 + 10(−1/10) = 15 − 1 = 14
Answer: Both methods give ̄x = 14. Verified.
Example 8: Finding Missing Frequency
Problem: The mean of the following data is 50. Find the missing frequency p:
| Class | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 |
|---|---|---|---|---|---|
| f | 7 | p | 10 | 9 | 4 |
Solution:
Given: Mean = 50, h = 20. Choose a = 50.
| Class | fᵢ | xᵢ | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 0–20 | 7 | 10 | −2 | −14 |
| 20–40 | p | 30 | −1 | −p |
| 40–60 | 10 | 50 | 0 | 0 |
| 60–80 | 9 | 70 | 1 | 9 |
| 80–100 | 4 | 90 | 2 | 8 |
- Σfᵢ = 30 + p
- Σfᵢuᵢ = −14 − p + 0 + 9 + 8 = 3 − p
Using the formula:
- 50 = 50 + 20 × [(3 − p)/(30 + p)]
- 0 = 20(3 − p)/(30 + p)
- 3 − p = 0
- p = 3
Answer: Missing frequency p = 3
Real-World Applications
Government Census Data:
- Population age groups, income brackets, and household sizes are summarised as grouped data. Step-deviation method efficiently computes mean values for policy planning.
Education:
- Computing average marks of students across classes, schools, or districts from grouped frequency distributions.
Industry:
- Quality control departments calculate average weight, length, or lifespan of products from grouped inspection data.
Economics:
- Calculating average income, expenditure, or savings from income-group data published by statistical agencies.
Key Points to Remember
- Step-deviation formula: ̄x = a + h(Σfᵢuᵢ / Σfᵢ).
- uᵢ = (xᵢ − a) / h converts class marks to small integers.
- This method works only when class widths are equal.
- Choose a = class mark of the class with highest frequency for best results.
- uᵢ values are …, −2, −1, 0, 1, 2, … when a is a class mark.
- All three methods (direct, assumed mean, step-deviation) give the same answer.
- Step-deviation is the fastest method when class marks are large and class widths are uniform.
- Watch the signs of uᵢ carefully — classes below a give negative uᵢ.
- Σfᵢuᵢ can be positive, negative, or zero depending on the data distribution.
- This is the preferred method in CBSE exams for grouped data mean calculations.
Practice Problems
- Find the mean using step-deviation: Classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 6, 10, 7, 3.
- The following table shows the weights (kg) of 50 students. Find the mean weight: 30-35 (4), 35-40 (8), 40-45 (16), 45-50 (13), 50-55 (6), 55-60 (3).
- Find the mean of: 100-120 (10), 120-140 (15), 140-160 (20), 160-180 (12), 180-200 (3) using the step-deviation method.
- The mean of a distribution is 35. Classes are 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 with frequencies 2, 3, p, 6, 4, 1. Find p.
- Calculate the mean using both the assumed mean method and step-deviation method for: 25-30 (3), 30-35 (5), 35-40 (12), 40-45 (6), 45-50 (4). Verify both give the same answer.
- Find the mean daily income from: Rs 100-200 (12), Rs 200-300 (18), Rs 300-400 (25), Rs 400-500 (15), Rs 500-600 (10).
Frequently Asked Questions
Q1. What is the step-deviation method?
The step-deviation method is a technique for finding the mean of grouped data. It simplifies calculations by converting class marks into small integers using uᵢ = (xᵢ − a)/h, where a is the assumed mean and h is the class width. Formula: ̄x = a + h(Σfᵢuᵢ/Σfᵢ).
Q2. When should I use the step-deviation method instead of the direct method?
Use step-deviation when the class marks (mid-values) are large numbers and the class widths are equal. It converts large numbers into small integers, making arithmetic faster and less error-prone.
Q3. Does the choice of assumed mean affect the answer?
No. The mean is the same regardless of which assumed mean you choose. However, choosing the class mark of the highest-frequency class or the middle class makes uᵢ values smaller and computation easier.
Q4. Can I use step-deviation for unequal class widths?
No. Step-deviation requires equal class widths because h must be a constant. For unequal class widths, use the assumed mean method or the direct method.
Q5. What does uᵢ represent?
uᵢ = (xᵢ − a)/h is the step-deviation. It represents how many class widths the class mark xᵢ is away from the assumed mean a. If uᵢ = 2, the class mark is 2 class widths above a.
Q6. Why is uᵢ always an integer?
Because the difference xᵢ − a between any two class marks is always a multiple of h (the class width). Since both xᵢ and a are class marks in a uniform distribution, their difference divided by h is always a whole number.
Q7. Do all three methods give the same answer?
Yes. The direct method, assumed mean method, and step-deviation method are algebraically equivalent. They always give exactly the same mean. The step-deviation method is just computationally simpler.
Q8. How is this method tested in CBSE board exams?
Questions typically give a grouped frequency distribution (5-6 classes) and ask to find the mean using the step-deviation method. Sometimes they ask to find a missing frequency given the mean. This carries 3-5 marks.
Q9. What if Σfᵢuᵢ equals zero?
If Σfᵢuᵢ = 0, then ̄x = a + h(0/Σfᵢ) = a. The mean equals the assumed mean. This happens when the data is perfectly symmetric around a.
Q10. Can the step-deviation method be used for ungrouped data?
No. It is designed specifically for grouped (continuous) data with equal class intervals. For ungrouped data, use the direct formula: ̄x = Σxᵢ/n.
Related Topics
- Mean of Grouped Data
- Mean by Assumed Mean Method
- Median of Grouped Data
- Mode of Grouped Data
- Statistics - Collection and Presentation
- Frequency Distribution Table
- Histogram of Grouped Data
- Frequency Polygon
- Mean of Ungrouped Data
- Median of Ungrouped Data
- Cumulative Frequency Distribution
- Ogive (Cumulative Frequency Curve)
- Empirical Relationship Between Mean, Median, Mode
- Statistics in Real Life










