Mean of Grouped Data (Class 10 Statistics) — Teaching Slide Deck

Chapter 13 · Statistics

Mean of Grouped Data

Welcome to this comprehensive masterclass. Before jumping into board formulas, let us review the core foundation so that you understand why each method exists.

Today's Roadmap:
1. Prerequisites & Ungrouped Mean Recap
2. Class Marks & Interval Continuity
3. The 3 CBSE Methods (Direct, Assumed Mean, Step-Deviation)
4. Board Exam Missing Frequency Problems (1 and 2 unknowns)

Guaranteed 4 to 6 marks in CBSE Class 10 Board Examinations.
Click Next Step or press Space to reveal each step.

Prerequisites & Methods Blueprint CBSE Class 10

P

Prerequisites RecapRaw Mean, Frequency Tables, Continuous Intervals

1

Direct MethodStandard approach for small numbers: x̄ = Σfᵢxᵢ ÷ N

2

Assumed Mean MethodShift of origin for large values: x̄ = a + (Σfᵢdᵢ ÷ N)

3

Step-Deviation MethodScaling shortcut with step uᵢ: x̄ = a + h·(Σfᵢuᵢ ÷ N)

Prerequisite 1 · Foundation Recap

What is the Mean?

In Class 9, you learned that the Mean (or Arithmetic Average) is the single numerical balance point that represents an entire set of data.

Mean (x̄) = Sum of all observationsTotal number of observations = Σ xn

If you have individual scores: 8, 12, 15, 25
Sum = 8 + 12 + 15 + 25 = 60 · Total count n = 4
Mean x̄ = 60 ÷ 4 = 15.
Important property: Mean is sensitive to extreme outliers!

Visual Balance Concept Center of Mass

8Observation 1

12Observation 2

15Observation 3

25Observation 4

Sum = 8 + 12 + 15 + 25 = 60
Divide by n = 4 ➔ Balance point x̄ = 15

When data grows into hundreds of observations, adding them one by one becomes impossible. That is why we organize data into frequency tables.

Prerequisite 2 · Foundation Recap

Ungrouped Frequency Table

When distinct observations repeat multiple times, we list each distinct value xᵢ along with its frequency fᵢ (how many times it occurs).

x̄ = Σ fᵢ xᵢΣ fᵢ = Σ fᵢ xᵢN

fᵢ (Frequency): Acts as the statistical weight of that observation.
N = Σ fᵢ: The total number of all students/observations.
Instead of adding 5 three times (5+5+5), we multiply: 3 × 5 = 15.

Discrete Frequency Example Score × Count

Score (xᵢ)	Students (fᵢ)	Product (fᵢ xᵢ)
10	4	40
20	6	120
30	10	300
Total	N = 20	Σ fᵢ xᵢ = 460

x̄ = 46020 = 23 marks

Core Transition · Class 9 to Class 10

Why Grouped Data?

If 100 students score marks between 0 and 100, a discrete table would need 100 separate rows! To make large data manageable, we group it into class intervals (e.g. 0–10, 10–20).

The Big Problem: Inside the interval [10 – 20], we do not know the exact score of any individual student!

The Solution: We assume all values cluster around the midpoint!
This midpoint is called the Class Mark (xᵢ).

Class Mark (Midpoint) Formula Central Midpoint

Class Mark (xᵢ) = Lower Limit + Upper Limit2

Ex

For Class [10 – 20]:Lower = 10, Upper = 20

=

xᵢ = (10 + 20) ÷ 2 = 1515 is used as the representative value for this group

h

Class Width (h) = 20 − 10 = 10Difference between Upper and Lower limits

Prerequisite 4 · Interval Preparation

Make Classes Continuous

Continuous classes (0–10, 10–20) share boundaries. If inclusive classes leave a gap (10–19, 20–29), convert them before proceeding.

Gap = 20 − 19 = 1 ➔ Half-Gap = 0.5
Rule: Subtract 0.5 from Lower Limit, Add 0.5 to Upper Limit!

10–19 becomes 9.5 – 19.5 (Midpoint = 14.5).
20–29 becomes 19.5 – 29.5 (Midpoint = 24.5).
Crucial: Class mark is identical whether computed from limits or continuous boundaries!

Continuity Adjustment Step The ± 0.5 Rule

A

Given Discontinuous Classes (Gaps)10–19 | 20–29 | 30–39

±

Apply Half-Gap Correction (0.5)Lower − 0.5 and Upper + 0.5

B

Continuous True Class Boundaries9.5–19.5 | 19.5–29.5 | 29.5–39.5

CBSE Method 1 of 3

Direct Method: Concept

The most straightforward method. Find midpoint xᵢ for every class, multiply directly by frequency fᵢ, sum all products, and divide by total N.

x̄ = Σ fᵢ xᵢΣ fᵢ = Σ fᵢ xᵢN

When to use: When class marks xᵢ and frequencies fᵢ are small integers.
Requires 4 standard columns in board exams: Class Interval, fᵢ, xᵢ, fᵢ xᵢ.

Direct Method Algorithm Execution Steps

1

Step 1: Compute Class Marks (xᵢ)xᵢ = (Lower limit + Upper limit) ÷ 2

2

Step 2: Multiply Row-wise (fᵢ · xᵢ)Weight each class mark by its frequency

3

Step 3: Sum and Divide by NFind Σfᵢ and Σfᵢxᵢ, then apply x̄ = Σfᵢxᵢ ÷ N

Method 1 · Worked Example

Direct Method: Solution

Problem: A survey records the daily reading time (in minutes) of 20 students. Calculate the mean daily reading time.

x̄ = 44020 = 22 minutes

Board Conclusion: The mean reading time of the students is 22 minutes.

Board Exam Table Work Total Students N = 20

Class Interval	fᵢ	xᵢ	fᵢ xᵢ
0 – 10	3	5	15
10 – 20	5	15	75
20 – 30	7	25	175
30 – 40	5	35	175
Total	N = 20	—	Σ fᵢ xᵢ = 440

Motivation for Method 2

Why Not Direct Always?

Imagine if class intervals were daily wages: ₹200–₹250, ₹250–₹300. Then class marks are 225, 275, 325 and frequencies are 18, 24, 32.

Multiplying 275 × 24 by hand takes 5 minutes and invites multiplication slips during board exams!

The Genius Idea: Shift the Origin!
Pick a convenient middle number 'a' (Assumed Mean).
Subtract 'a' from all class marks: dᵢ = xᵢ − a. Large numbers shrink down to small manageable deviations!

Origin Shift Principle Shrinking Multi-Digit Numbers

xᵢ = 275Raw midpoint

a = 250Assumed origin

dᵢ = +25Tiny deviation!

fᵢ dᵢSuper fast to multiply

Instead of multiplying 275 × 24 = 6600,
we only multiply 25 × 24 = 600!

CBSE Method 2 of 3

Assumed Mean Method

Choose an assumed mean 'a' near the center of the xᵢ values. Compute signed deviations dᵢ = xᵢ − a, weight them by frequency, and add back 'a'.

dᵢ = xᵢ − a · x̄ = a + Σ fᵢ dᵢΣ fᵢ

dᵢ = xᵢ − a: Signed deviation from assumed mean a.
Values below a give negative deviations. Keep the signs!
Average deviation (Σ fᵢ dᵢ ÷ N) shifts 'a' to the exact true mean.

Balancing Around Reference 'a' Signed Deviations

a

Center Reference (e.g. a = 25)Middle class mark chosen as origin

−

Negative Deviations (Below a)5 − 25 = −20 | 15 − 25 = −10

+

Positive Deviations (Above a)35 − 25 = +10

Method 2 · Worked Example

Assumed Mean: Solution

Solve the reading time table using Assumed Mean a = 25 (the central class mark). Notice how small the numbers become!

x̄ = 25 + −6020 = 25 − 3 = 22 minutes

Negative correction (−3) pulls the assumed mean 25 down to the true mean 22. Exact match with Direct Method!

Assumed Mean Table (a = 25) dᵢ = xᵢ − 25

xᵢ	fᵢ	dᵢ = xᵢ − 25	fᵢ dᵢ
5	3	−20	−60
15	5	−10	−50
25 (a)	7	0	0
35	5	+10	+50
—	N = 20	—	Σ fᵢ dᵢ = −60

CBSE Method 3 of 3

Step-Deviation Method

When all class intervals share a common width h, the deviations are all multiples of h. Divide by h to turn deviations into tiny single integers!

uᵢ = xᵢ − ah · x̄ = a + h · Σ fᵢ uᵢΣ fᵢ

Step variable: uᵢ = dᵢ ÷ h. Turns −20, −10, 0, 10 into −2, −1, 0, 1.
Golden Rule: You divided by h, so you MUST multiply the correction bracket by h!
Do NOT use step-deviation if class widths are unequal.

Step Compression (h = 10) Single-Digit Arithmetic

d₁ = −20u₁ = −2

d₂ = −10u₂ = −1

d₃ = 0u₃ = 0

d₄ = +10u₄ = +1

Divide by h ➔ Multiply small single digits ➔ Multiply back by h at the end!

Method 3 · Worked Example

Step-Deviation: Solution

Take a = 25 and common width h = 10. The products fᵢuᵢ are tiny single digits that you can calculate in your head!

x̄ = 25 + 10 · −620 = 25 − 3 = 22 minutes

★ All 3 Methods Converge:
Direct = Assumed Mean = Step-Deviation = 22 minutes.

Step-Deviation Table (a = 25, h = 10) uᵢ = (xᵢ − 25) ÷ 10

xᵢ	fᵢ	uᵢ = (xᵢ−25)/10	fᵢ uᵢ
5	3	−2	−6
15	5	−1	−5
25 (a)	7	0	0
35	5	+1	+5
—	N = 20	—	Σ fᵢ uᵢ = −6

Decision Matrix

Which Method to Choose?

All three methods give the exact same answer. Choose the one that minimizes calculation risk in your exam paper.

Direct Method: When numbers are small (single/double digits). No extra columns needed.
Assumed Mean Method: When class marks are large, but interval widths are unequal.
Step-Deviation Method: Best for large numbers with uniform class width h.

Golden Rule: If the question specifies a method by name, you MUST use that method for full marks!

Method Comparison Matrix Exam Blueprint

1

Direct MethodSmall integers · 4 columns · Fast multiplication

2

Assumed Mean MethodLarge xᵢ · Unequal widths · Deviations dᵢ

3

Step-Deviation MethodEqual widths · Simplest numbers · Step uᵢ

Board Special Question Type 1

One Missing Frequency (m)

Problem: Given xᵢ = [5, 15, 25, 35], frequencies = [4, 6, m, 5], and Mean x̄ = 22. Find unknown frequency m.

1

Total Frequency: N = 15 + mDo NOT forget to include m in N!

2

Σ fᵢ xᵢ = 4(5) + 6(15) + 25m + 5(35) = 285 + 25mSum of products with variable m

Most common error: Leaving m out of the total denominator N!

Algebraic Derivation 1 Unknown ➔ 1 Equation

22 = 285 + 25m15 + m

3

22(15 + m) = 285 + 25mCross-multiply denominator

4

330 + 22m = 285 + 25m ➔ 3m = 45Group like terms

m = 15 (Check: 660 ÷ 30 = 22 ✓)

Board Special Question Type 2

Two Missing Frequencies

Given total N = 30, Mean x̄ = 22, and frequencies 4, p, q, 5. Two unknowns require two simultaneous equations.

Eq 1

Total Frequency: 4 + p + q + 5 = 30➔ p + q = 21

Eq 2

Σ fᵢ xᵢ = 30 × 22 = 66020 + 15p + 25q + 175 = 660

Simultaneous Linear Solution 2 Unknowns ➔ 2 Equations

15p + 25q = 465

Sub

Substitute p = 21 − q:15(21 − q) + 25q = 465

➔

315 + 10q = 465 ➔ 10q = 150Solve for q

q = 15 and p = 6

CBSE Examiner Insights

Full-Mark Presentation

CBSE step marking awards marks for formula, table construction, correct substitutions, and units.

Always state formula at the top (½ mark).
Construct columns with neat labels (1½ marks).
State assumptions clearly (a = 25, h = 10).
Box final answer with units (minutes, kg, ₹).

Top 3 Danger Traps Avoid Common Slips

✗

Using Class Limits instead of MidpointsNever use boundaries as xᵢ! Always use midpoint.

✗

Dividing by Number of ClassesDivide by total frequency N = Σ fᵢ, NOT number of rows.

✗

Forgetting h in Step-DeviationMust multiply the correction bracket by h.

Final Mastery Check

Rapid Formula Recap

All three methods converge to the exact same grouped mean.

1

Direct Method:x̄ = Σ fᵢ xᵢ ÷ Σ fᵢ

2

Assumed Mean Method:x̄ = a + (Σ fᵢ dᵢ ÷ Σ fᵢ)

3

Step-Deviation Method:x̄ = a + h · (Σ fᵢ uᵢ ÷ Σ fᵢ)

Live Classroom Challenge Interactive Test

Given N = 50 and Σ fᵢ xᵢ = 1250. Find x̄.

x̄ = 1250 ÷ 50 = 25 units