Mean by Assumed Mean Method
The assumed mean method (also called the short-cut method) is a technique for calculating the mean of grouped data when the class marks and frequencies are large numbers.
Instead of multiplying large values of xᵢ and fᵢ, we choose a convenient value a (the assumed mean) and work with smaller deviations dᵢ = xᵢ − a. This reduces calculation effort significantly.
The assumed mean method gives the exact same answer as the direct method — it is not an approximation. It is simply a computational shortcut.
What is Assumed Mean Method?
Definition: The assumed mean method calculates the mean of grouped data by choosing a convenient value a (assumed mean), computing deviations of each class mark from this value, and adjusting the result.
Key terms:
- xᵢ = class mark (midpoint) of the i-th class = (upper limit + lower limit) / 2
- a = assumed mean — any convenient value, usually the class mark of the middle class or the class with highest frequency
- dᵢ = deviation = xᵢ − a
- fᵢ = frequency of the i-th class
- Σfᵢdᵢ = sum of the products of frequency and deviation
- Σfᵢ = N = total frequency
When to use:
- When class marks are large (e.g., 155, 165, 175...)
- When the direct method involves tedious multiplication
- The assumed mean method works for all grouped data — equal or unequal class widths
Mean by Assumed Mean Method Formula
Formula:
Mean (x̄) = a + Σfᵢdᵢ / Σfᵢ
Where:
- x̄ = mean of the data
- a = assumed mean
- dᵢ = xᵢ − a (deviation of each class mark from assumed mean)
- fᵢ = frequency of the i-th class
- Σfᵢdᵢ = sum of all fᵢ × dᵢ values
- Σfᵢ = sum of all frequencies = N
Derivation and Proof
Why the assumed mean method works:
- The direct method formula is: x̄ = Σfᵢxᵢ / Σfᵢ
- Write xᵢ = a + dᵢ, where dᵢ = xᵢ − a
- Substitute: x̄ = Σfᵢ(a + dᵢ) / Σfᵢ
- Expand: x̄ = (Σfᵢ·a + Σfᵢdᵢ) / Σfᵢ
- Split: x̄ = a·Σfᵢ/Σfᵢ + Σfᵢdᵢ/Σfᵢ
- Simplify: x̄ = a + Σfᵢdᵢ/Σfᵢ
Key insight: The choice of a does NOT affect the final answer. Any value of a gives the same mean. However, choosing a value near the centre of the data makes the deviations small and calculations easier.
Types and Properties
Choosing the Assumed Mean:
- Method 1: Use the class mark of the class with the highest frequency (modal class).
- Method 2: Use the class mark of the middle class.
- Method 3: Use any round number near the centre of the data.
Comparison with other methods:
| Method | Formula | When to Use |
|---|---|---|
| Direct Method | x̄ = Σfᵢxᵢ / Σfᵢ | Small values of xᵢ and fᵢ |
| Assumed Mean Method | x̄ = a + Σfᵢdᵢ / Σfᵢ | Large values of xᵢ |
| Step-Deviation Method | x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h | Large xᵢ with equal class widths |
Methods
Step-by-step procedure:
- Find class marks: xᵢ = (upper limit + lower limit) / 2 for each class.
- Choose assumed mean: Pick a = class mark of the middle or most frequent class.
- Calculate deviations: dᵢ = xᵢ − a for each class.
- Multiply: Find fᵢdᵢ for each class.
- Sum up: Calculate Σfᵢdᵢ and Σfᵢ.
- Apply formula: x̄ = a + Σfᵢdᵢ / Σfᵢ.
Important checks:
- Some dᵢ values will be negative (when xᵢ < a) — handle signs carefully.
- Σfᵢdᵢ can be positive, negative, or zero.
- The answer should be the same regardless of which value is chosen as a.
Solved Examples
Example 1: Basic Assumed Mean Calculation
Problem: Find the mean of the following data using the assumed mean method.
| Class | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 10 | 5 |
Solution:
Step 1: Class marks: 15, 25, 35, 45, 55
Step 2: Assumed mean a = 35 (middle class mark)
Step 3: Deviations: dᵢ = xᵢ − 35 → −20, −10, 0, 10, 20
Step 4: Products fᵢdᵢ: 5(−20), 8(−10), 12(0), 10(10), 5(20) = −100, −80, 0, 100, 100
Step 5: Σfᵢdᵢ = −100 − 80 + 0 + 100 + 100 = 20. Σfᵢ = 40.
Step 6: x̄ = 35 + 20/40 = 35 + 0.5 = 35.5
Answer: Mean = 35.5.
Example 2: Large Class Marks
Problem: Find the mean using assumed mean method.
| Class | 100–120 | 120–140 | 140–160 | 160–180 | 180–200 |
|---|---|---|---|---|---|
| f | 6 | 10 | 14 | 8 | 2 |
Solution:
Given: Class marks: 110, 130, 150, 170, 190. Let a = 150.
dᵢ = xᵢ − 150: −40, −20, 0, 20, 40
fᵢdᵢ: 6(−40), 10(−20), 14(0), 8(20), 2(40) = −240, −200, 0, 160, 80
Σfᵢdᵢ = −240 − 200 + 0 + 160 + 80 = −200. Σfᵢ = 40.
x̄ = 150 + (−200)/40 = 150 − 5 = 145
Answer: Mean = 145.
Example 3: Mean of Marks Distribution
Problem: The marks distribution of 50 students is given below. Find the mean marks using the assumed mean method.
| Marks | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 |
|---|---|---|---|---|---|
| Students | 4 | 10 | 16 | 14 | 6 |
Solution:
Given: Class marks: 10, 30, 50, 70, 90. Let a = 50.
dᵢ: −40, −20, 0, 20, 40
fᵢdᵢ: 4(−40), 10(−20), 16(0), 14(20), 6(40) = −160, −200, 0, 280, 240
Σfᵢdᵢ = −160 − 200 + 0 + 280 + 240 = 160. Σfᵢ = 50.
x̄ = 50 + 160/50 = 50 + 3.2 = 53.2
Answer: Mean marks = 53.2.
Example 4: Choosing a Different Assumed Mean
Problem: Using the data from Example 1 (classes 10–20 to 50–60), verify that choosing a = 25 gives the same mean.
Solution:
Given: Class marks: 15, 25, 35, 45, 55. Let a = 25.
dᵢ = xᵢ − 25: −10, 0, 10, 20, 30
fᵢdᵢ: 5(−10), 8(0), 12(10), 10(20), 5(30) = −50, 0, 120, 200, 150
Σfᵢdᵢ = −50 + 0 + 120 + 200 + 150 = 420. Σfᵢ = 40.
x̄ = 25 + 420/40 = 25 + 10.5 = 35.5
Answer: Mean = 35.5 — same as before, confirming the choice of a does not affect the result.
Example 5: Daily Wages Data
Problem: Find the mean daily wages (in ₹) of 60 workers.
| Wages (₹) | 200–250 | 250–300 | 300–350 | 350–400 | 400–450 |
|---|---|---|---|---|---|
| Workers | 8 | 12 | 18 | 15 | 7 |
Solution:
Given: Class marks: 225, 275, 325, 375, 425. Let a = 325.
dᵢ: −100, −50, 0, 50, 100
fᵢdᵢ: 8(−100), 12(−50), 18(0), 15(50), 7(100) = −800, −600, 0, 750, 700
Σfᵢdᵢ = −800 − 600 + 0 + 750 + 700 = 50. Σfᵢ = 60.
x̄ = 325 + 50/60 = 325 + 0.833 = 325.83
Answer: Mean daily wage = ₹325.83.
Example 6: Negative Result for Σfᵢdᵢ
Problem: Find the mean height of 40 plants.
| Height (cm) | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|
| Plants | 8 | 14 | 12 | 6 |
Solution:
Given: Class marks: 25, 35, 45, 55. Let a = 45.
dᵢ: −20, −10, 0, 10
fᵢdᵢ: 8(−20), 14(−10), 12(0), 6(10) = −160, −140, 0, 60
Σfᵢdᵢ = −160 − 140 + 0 + 60 = −240. Σfᵢ = 40.
x̄ = 45 + (−240)/40 = 45 − 6 = 39
Answer: Mean height = 39 cm.
Example 7: Finding Missing Frequency
Problem: The mean of the following data is 28. Find the missing frequency p.
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| f | 5 | p | 10 | 8 | 7 |
Solution:
Given: Mean = 28. Class marks: 5, 15, 25, 35, 45. Let a = 25.
dᵢ: −20, −10, 0, 10, 20
fᵢdᵢ: 5(−20), p(−10), 10(0), 8(10), 7(20) = −100, −10p, 0, 80, 140
Σfᵢdᵢ = −100 − 10p + 0 + 80 + 140 = 120 − 10p
Σfᵢ = 5 + p + 10 + 8 + 7 = 30 + p
x̄ = a + Σfᵢdᵢ/Σfᵢ → 28 = 25 + (120 − 10p)/(30 + p)
3 = (120 − 10p)/(30 + p)
3(30 + p) = 120 − 10p → 90 + 3p = 120 − 10p → 13p = 30 → p = 30/13 ≈ 2.3
Answer: p ≈ 2.3. (If p must be a whole number, check the data — the problem may use p = 2 or adjust the mean.)
Example 8: Seven Classes
Problem: Find the mean of the following distribution using the assumed mean method.
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 |
|---|---|---|---|---|---|---|---|
| f | 4 | 6 | 10 | 12 | 8 | 7 | 3 |
Solution:
Given: Class marks: 5, 15, 25, 35, 45, 55, 65. Let a = 35 (highest frequency class).
dᵢ: −30, −20, −10, 0, 10, 20, 30
fᵢdᵢ: −120, −120, −100, 0, 80, 140, 90
Σfᵢdᵢ = −120 − 120 − 100 + 0 + 80 + 140 + 90 = −30. Σfᵢ = 50.
x̄ = 35 + (−30)/50 = 35 − 0.6 = 34.4
Answer: Mean = 34.4.
Real-World Applications
Where the assumed mean method is useful:
- CBSE board exams: A common 3-mark or 5-mark question. Students must show the table with xᵢ, dᵢ, fᵢdᵢ columns.
- Census and surveys: When summarising grouped age, income, or distance data with large class marks.
- Quality control: Computing the mean of measurements where values cluster around a known standard.
- Any large-number dataset: The method reduces arithmetic errors by working with smaller numbers.
Key Points to Remember
- The assumed mean method formula is: x̄ = a + Σfᵢdᵢ / Σfᵢ.
- a is the assumed mean — choose the class mark near the centre of the distribution.
- dᵢ = xᵢ − a is the deviation of each class mark from the assumed mean.
- The choice of a does NOT affect the final answer — any value gives the same mean.
- Some deviations dᵢ will be negative. Handle signs carefully in fᵢdᵢ products.
- Σfᵢdᵢ can be positive, negative, or zero.
- This method gives the exact mean, not an approximation.
- It reduces calculation effort compared to the direct method when class marks are large.
- In exams, always present the work in a table with columns: Class, fᵢ, xᵢ, dᵢ, fᵢdᵢ.
- The assumed mean method works for both equal and unequal class intervals.
Practice Problems
- Find the mean using the assumed mean method: 0–10 (f=7), 10–20 (f=10), 20–30 (f=15), 30–40 (f=8), 40–50 (f=10).
- The following data gives the marks of 100 students. Find the mean by assumed mean method: 0–20 (f=12), 20–40 (f=18), 40–60 (f=27), 60–80 (f=25), 80–100 (f=18).
- Find the mean of the following distribution using two different values of assumed mean and verify the answers are equal: 5–15 (f=6), 15–25 (f=12), 25–35 (f=14), 35–45 (f=8).
- The mean of the following data is 42. Find the missing frequency: 10–20 (f=5), 20–30 (f=x), 30–40 (f=12), 40–50 (f=15), 50–60 (f=8).
- Find the mean monthly expenditure (in ₹) of 35 families: 1000–1500 (f=3), 1500–2000 (f=8), 2000–2500 (f=12), 2500–3000 (f=7), 3000–3500 (f=5).
- Using the assumed mean method, find the mean of: 150–155 (f=8), 155–160 (f=14), 160–165 (f=20), 165–170 (f=10), 170–175 (f=8).
Frequently Asked Questions
Q1. What is the assumed mean method?
It is a shortcut method for finding the mean of grouped data. Instead of using class marks directly, we use deviations from a chosen value (assumed mean) to simplify calculations. Formula: x̄ = a + Σfᵢdᵢ/Σfᵢ.
Q2. Does the choice of assumed mean affect the final answer?
No. The final mean is the same regardless of which value is chosen as the assumed mean. The formula adjusts for the difference.
Q3. How do you choose the best assumed mean?
Choose the class mark of the class with the highest frequency, or the middle class, or any convenient round number. A good choice makes the deviations small and balanced.
Q4. What is the difference between assumed mean and step-deviation method?
The assumed mean method uses dᵢ = xᵢ − a. The step-deviation method further divides by the class width: uᵢ = (xᵢ − a)/h. Step-deviation is faster when all class widths are equal.
Q5. Can the assumed mean method be used for unequal class intervals?
Yes. The assumed mean method works for any grouped data — equal or unequal class widths. Only the step-deviation method requires equal class widths.
Q6. What does a negative Σfᵢdᵢ mean?
A negative Σfᵢdᵢ means the actual mean is less than the assumed mean. The formula x̄ = a + (negative value) gives a mean below a.
Q7. Is the assumed mean method an approximation?
No. It gives the exact same result as the direct method. It is mathematically equivalent — just computationally simpler.
Q8. Why is this method called the short-cut method?
Because it reduces the size of numbers involved in multiplication. Working with deviations (often single or double-digit numbers) is faster than multiplying large class marks by frequencies.
Related Topics
- Mean of Grouped Data
- Mean by Step-Deviation 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
