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Chapter 13: STATISTICS

Grade 10 Mathematics · Chapter 13

Statistics

Master the three pillars of data analysis — Mean, Median & Mode — for grouped data, cumulative frequency distributions, and ogives. Everything you need to ace your board exam.

📊 Grouped Data 📈 Ogives 🎯 Exam Ready

📋 Contents

1. Introduction
2. Mean of Grouped Data
3. Three Methods for Mean
4. Mode of Grouped Data

5. Median of Grouped Data
6. Cumulative Frequency & Ogive
7. Worked Examples (10+)
8. Practice Sets A–D

📖	Section 1 — Introduction

In your earlier classes, you learned to organise raw data into frequency distributions — both ungrouped and grouped. You also studied how to display data through bar graphs, histograms, and frequency polygons.

You were introduced to three special numbers that summarise a dataset:

Measure	What it tells you	Symbol
Mean	The arithmetic average of all values	x̄
Median	The middle value when data is arranged in order	M
Mode	The value that appears most frequently	Mo

In this chapter, we extend these concepts to grouped data — where data is organised into class intervals rather than listed individually. We also explore the cumulative frequency distribution and how to draw curves called ogives.

💡 Why grouped data?
In real life, datasets can have hundreds or thousands of values. Grouping them into class intervals makes analysis practical and meaningful.

📐	Section 2 — Mean of Grouped Data

What is the Mean?

The mean (or arithmetic average) is the sum of all observed values divided by the total number of observations.

For ungrouped data with values x₁, x₂, …, xₙ having frequencies f₁, f₂, …, fₙ:

x̄ = (Σ fᵢxᵢ) ÷ (Σ fᵢ)

Class Mark (Mid-point)

When data is grouped into class intervals, we cannot use exact values. Instead, we use the class mark — the midpoint of each interval — to represent all values in that class.

Class Mark = (Upper Limit + Lower Limit) ÷ 2

Example: For the class interval 20–30, the class mark = (20 + 30) ÷ 2 = 25

🔢	Section 3 — Three Methods to Calculate Mean

Method 1: Direct Method

Simply multiply each class mark xᵢ by its frequency fᵢ, sum all products, then divide by total frequency.

x̄ = Σ(fᵢ · xᵢ) ÷ Σfᵢ

When to use: When the values of xᵢ and fᵢ are small enough to compute easily.

Method 2: Assumed Mean Method

Choose any class mark as an assumed mean ‘a’ (usually the central one). Calculate deviations dᵢ = xᵢ − a for each class.

x̄ = a + [Σ(fᵢ · dᵢ) ÷ Σfᵢ]
where dᵢ = xᵢ − a

When to use: When class marks are large numbers — deviations are smaller and easier to work with.

Method 3: Step-Deviation Method

A further simplification: divide the deviations by the class size h to get smaller numbers uᵢ.

x̄ = a + h × [Σ(fᵢ · uᵢ) ÷ Σfᵢ]
where uᵢ = (xᵢ − a) ÷ h, and h = class size

When to use: Best when all deviations dᵢ share a common factor (which equals h, the class size).

🔑 Key Fact: All three methods give the same answer. The step-deviation method is usually fastest for board exam calculations.

🏆	Section 4 — Mode of Grouped Data

The mode is the value that occurs most frequently. For grouped data, we first identify the modal class — the class interval with the highest frequency.

Modal Class → Mode Formula

Mode = l + [ (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) ] × h

Symbol	Meaning
l	Lower boundary of the modal class
f₁	Frequency of the modal class (highest frequency)
f₀	Frequency of the class before the modal class
f₂	Frequency of the class after the modal class
h	Width (size) of the class interval

⚠️ Note: The formula works only when all class intervals have equal width. Make sure class intervals are continuous before applying it.

⚖️	Section 5 — Median of Grouped Data

The median divides an ordered dataset into two equal halves. For grouped data, we use cumulative frequencies to locate the median class — the class containing the n/2-th observation.

Step-by-Step Process

Step	What to do
1	Calculate the cumulative frequency (cf) for each class
2	Find n/2 (where n = total number of observations)
3	Identify the class whose cumulative frequency first exceeds n/2 → this is the median class
4	Apply the median formula

Median = l + [ (n/2 − cf) ÷ f ] × h

Symbol	Meaning
l	Lower boundary of the median class
n	Total number of observations
cf	Cumulative frequency of the class before the median class
f	Frequency of the median class
h	Class width (size of the interval)

📉	Section 6 — Cumulative Frequency & Ogive

A cumulative frequency is a running total — it tells you how many observations fall below a certain value.

Less-Than Ogive
Plot points: (upper class limit, cumulative frequency)
→ Rising curve from bottom-left to top-right

More-Than Ogive
Plot points: (lower class limit, cumulative frequency)
→ Falling curve from top-left to bottom-right

Ogive Diagram (Schematic)

cf
↑

Values (Class Boundaries) →

Darker shading = more cumulative frequency (Less-Than Ogive)

📌 Finding Median from Ogive:
Draw the two ogives on the same graph. The x-coordinate of their intersection point gives the median.

✏️	Section 7 — Worked Examples (10 Solved Problems)

EXAMPLE 1 Direct Method — Mean from Grouped Data

Question: Find the mean score from this frequency distribution of 40 students:

Score (Class)	Frequency (fᵢ)	Class Mark (xᵢ)	fᵢ × xᵢ
10 – 20	4	15	60
20 – 30	8	25	200
30 – 40	14	35	490
40 – 50	10	45	450
50 – 60	4	55	220
Total	40	—	1420

👁 Show Solution

x̄ = Σ(fᵢxᵢ) ÷ Σfᵢ = 1420 ÷ 40 = 35.5

The mean score is 35.5 marks.

EXAMPLE 2 Assumed Mean Method

Question: Using the assumed mean method, find the mean height (cm) of 50 plants:

Height (cm)	fᵢ	xᵢ	dᵢ = xᵢ − 75	fᵢdᵢ
50–60	5	55	−20	−100
60–70	8	65	−10	−80
70–80	15	75	0	0
80–90	14	85	+10	140
90–100	8	95	+20	160
Total	50	—		120

👁 Show Solution

Assumed mean a = 75, Σfᵢdᵢ = 120, Σfᵢ = 50

x̄ = a + (Σfᵢdᵢ ÷ Σfᵢ) = 75 + (120 ÷ 50) = 75 + 2.4 = 77.4 cm

EXAMPLE 3 Step-Deviation Method

Question: Find the mean age (years) of 60 participants using the step-deviation method (h = 5):

Age	fᵢ	xᵢ	uᵢ=(xᵢ−32.5)/5	fᵢuᵢ
15–20	6	17.5	−3	−18
20–25	10	22.5	−2	−20
25–30	14	27.5	−1	−14
30–35	16	32.5	0	0
35–40	10	37.5	+1	10
40–45	4	42.5	+2	8
Total	60	—		−34

👁 Show Solution

a = 32.5, h = 5, Σfᵢuᵢ = −34, Σfᵢ = 60

x̄ = 32.5 + 5 × (−34 ÷ 60) = 32.5 + (−2.83) = 29.67 years

EXAMPLE 4 Mode of Grouped Data

Question: Find the mode for this distribution of daily wages of 80 workers:

Wages (₹)	Number of Workers
300–350	8
350–400	16
400–450 ← Modal Class	28
450–500	18
500–550	10

👁 Show Solution

Modal class = 400–450 (highest frequency = 28)

l = 400, f₁ = 28, f₀ = 16, f₂ = 18, h = 50

Mode = 400 + [(28 − 16) ÷ (2×28 − 16 − 18)] × 50

= 400 + [12 ÷ 22] × 50 = 400 + 27.27 = ₹427.27

EXAMPLE 5 Median of Grouped Data

Question: Find the median from the following distribution (n = 80):

Class	f	Cumulative f (cf)
0–10	5	5
10–20	8	13
20–30	20	33
30–40 ← Median Class	24	57
40–50	15	72
50–60	8	80

👁 Show Solution

n/2 = 40. The class whose cf first exceeds 40 is 30–40 (cf = 57).

l = 30, cf (before) = 33, f = 24, h = 10

Median = 30 + [(40 − 33) ÷ 24] × 10 = 30 + [7/24] × 10 = 30 + 2.92 = 32.92

EXAMPLE 6 Find Missing Frequency Using Mean

Question: The mean of the following distribution is 50. Find the missing frequency p.

Class	20–30	30–40	40–50	50–60	60–70	70–80
f	5	8	20	p	6	2

👁 Show Solution

Total frequency = 41 + p. Σfᵢxᵢ = 5(25) + 8(35) + 20(45) + p(55) + 6(65) + 2(75)

= 125 + 280 + 900 + 55p + 390 + 150 = 1845 + 55p

x̄ = 50 → (1845 + 55p) ÷ (41 + p) = 50

1845 + 55p = 2050 + 50p → 5p = 205 → p = 41

EXAMPLE 7 Cumulative Frequency Table

Question: From the data below, prepare a less-than type cumulative frequency table:

Marks	0–10	10–20	20–30	30–40	40–50
f	3	9	15	12	6

👁 Show Answer

Marks Less Than	Cumulative Frequency
10	3
20	12
30	27
40	39
50	45

EXAMPLE 8 Empirical Relationship: Mean, Median, Mode

Question: The mean of a dataset is 26 and the mode is 23. Using the empirical relationship, find the median.

👁 Show Solution

Empirical formula: 3 Median = Mode + 2 Mean

3 Median = 23 + 2(26) = 23 + 52 = 75

Median = 75 ÷ 3 = 25

EXAMPLE 9 Find x and y Given Median and Total Frequency

Question: The following distribution has median = 46 and total frequency = 100. Find x and y.

Class	20–30	30–40	40–50	50–60	60–70	70–80
f	12	30	x	y	6	10

👁 Show Solution

Σf = 100 → 12 + 30 + x + y + 6 + 10 = 100 → x + y = 42 … (i)

Median = 46 lies in class 40–50. cf before = 42, f = x, l = 40, h = 10

46 = 40 + [(50 − 42) ÷ x] × 10 → 6 = (80 ÷ x) → x = 80/6 ≈ however using exactly:

6x = 80 → This gives x = 80/6… re-checking with cf: n/2 = 50

cf before 40–50 = 42. 46 = 40 + [(50 − 42)/x] × 10 → 6 = 80/x → x = 80/6. For integer: x ≈ 13, then from (i): y = 29

EXAMPLE 10 Compare Mean, Median & Mode

Question: For a dataset with grouped distribution, we found Mean = 54.8, Median = 52.5, Mode = 47.1. Verify the empirical relationship.

👁 Show Verification

Empirical formula: 3 Median = Mode + 2 Mean

LHS: 3 × 52.5 = 157.5

RHS: 47.1 + 2(54.8) = 47.1 + 109.6 = 156.7 ≈ 157.5 ✓

The small difference is due to rounding — the relationship holds approximately for real data.

📝	Section 8 — Practice Sets A, B, C & D

Practice Set A — Mean (Direct Method)

1. Find the mean of the following distribution:

Class	0–8	8–16	16–24	24–32	32–40
f	6	10	14	8	2

2. The mean score is 35. The frequencies for class 20–30 and 30–40 are 10 and f respectively, with total = 50. Find f.

3. Calculate the mean electricity consumption (units) from the table: 0–50: 10 families, 50–100: 20, 100–150: 15, 150–200: 5.

👁 Answers

1. x̄ = (6×4 + 10×12 + 14×20 + 8×28 + 2×36) ÷ 40 = (24+120+280+224+72)÷40 = 720÷40 = 18

2. Using x̄ = 35: Σfᵢxᵢ = 1750. Work out to find f = 15

3. x̄ = (10×25 + 20×75 + 15×125 + 5×175) ÷ 50 = 4500÷50 = 90 units

Practice Set B — Mode

1. Find the mode: Class intervals 10–20, 20–30, 30–40, 40–50, 50–60 with frequencies 4, 10, 22, 12, 6.

2. For a data set, the modal class is 25–35 with f₁ = 18, f₀ = 12, f₂ = 10, h = 10. Find the mode.

3. In a shoe size survey, modal class is size 7–8, f₁ = 40, f₀ = 20, f₂ = 15, h = 1. Find the modal size.

👁 Answers

1. Modal class = 30–40 (f₁=22, f₀=10, f₂=12, h=10, l=30). Mode = 30 + [(22−10)÷(44−10−12)]×10 = 30 + (12÷22)×10 = 35.45

2. Mode = 25 + [(18−12)÷(36−12−10)]×10 = 25 + (6÷14)×10 = 25 + 4.29 = 29.29

3. Mode = 7 + [(40−20)÷(80−20−15)]×1 = 7 + 20/45 = 7.44

Practice Set C — Median

1. Find the median for: Classes 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 6, 14, 16, 10, 4 (n = 50).

2. The median of a distribution is 35.4. The cf before the median class is 20, f = 25, h = 10, l = 30. Verify.

3. From the ogive, median is found where both curves intersect at x = 42. State the median.

👁 Answers

1. n/2 = 25. cf: 6, 20, 36→ Median class = 40–60. Median = 40 + [(25−20)÷16]×20 = 40 + 6.25 = 46.25

2. Median = 30 + [(25−20)÷25]×10 = 30 + 2 = 32 ≠ 35.4 (check data values)

3. Median = 42 (read directly from ogive intersection)

Practice Set D — Mixed Problems

1. If Mean = 48 and Mode = 45, use the empirical formula to find Median.

2. A data set has Median = 31.5 and Mean = 33. Find the Mode.

3. In a class of 30 students, the frequency distribution of heights (cm) is: 140–145: 5, 145–150: 8, 150–155: 10, 155–160: 5, 160–165: 2. Find (a) mean height (b) modal height.

👁 Answers

1. 3M = 45 + 2(48) = 141 → Median = 47

2. Mode = 3(31.5) − 2(33) = 94.5 − 66 = 28.5

3a. x̄ = [5(142.5)+8(147.5)+10(152.5)+5(157.5)+2(162.5)]÷30 = 4530÷30 = 151 cm
3b. Modal class = 150–155 (f=10, f₀=8, f₂=5, h=5). Mode = 150 + [(10−8)÷(20−8−5)]×5 = 150 + (2/7)×5 = 151.43 cm

📚	Chapter Summary — Statistics at a Glance

📊 Mean Formulas

Direct: x̄ = Σfᵢxᵢ ÷ Σfᵢ
Assumed Mean: x̄ = a + Σfᵢdᵢ ÷ Σfᵢ
Step-Deviation: x̄ = a + h(Σfᵢuᵢ ÷ Σfᵢ)

📈 Ogive

Less-than: plot (upper limit, cf)
More-than: plot (lower limit, cf)
Intersection → Median

🏆 Mode Formula

Mode = l + [(f₁−f₀) ÷ (2f₁−f₀−f₂)] × h

Modal class = class with highest frequency

⚖️ Median Formula

Median = l + [(n/2 − cf) ÷ f] × h

Empirical: 3 Median = Mode + 2 Mean

🎯	Exam Quick-Check — 8 Must-Know Points

1	The class mark is always the midpoint of a class interval = (upper + lower) ÷ 2
2	All three methods (direct, assumed mean, step-deviation) give the same mean
3	For mode: f₁ is the modal class frequency, f₀ is the one before it, f₂ is after it
4	For median: find n/2, locate the class whose cumulative frequency first exceeds n/2
5	Empirical relationship: 3 Median = Mode + 2 Mean (holds approximately)
6	For ogive: less-than plots upper limits; more-than plots lower limits — both on y-axis cumulative frequency
7	Mode formula requires equal class widths — always check before applying
8	Mean is best for comparing distributions; Median is best for skewed data; Mode shows the most common value

⚠️ Common Mistakes to Avoid

✗	Using actual data values instead of class marks in the mean formula for grouped data
✗	Forgetting to check if class intervals are continuous before finding mode/median
✗	Using n instead of n/2 to locate the median class
✗	Confusing “cf” in median formula — it is cumulative frequency of the class BEFORE the median class, not the median class itself

About This Chapter: This page covers Grade 10 Statistics — a key chapter in secondary mathematics. Students learn to calculate the mean of grouped data using three methods: direct, assumed mean, and step-deviation. The chapter explains how to find the mode using the modal class formula and the median using cumulative frequency. Students also learn to construct and interpret less-than and more-than ogives. The empirical relationship between mean, median and mode (3 Median = Mode + 2 Mean) is a frequently tested concept. This topic is essential for board exams and builds the foundation for higher-level statistics. Regular practice with worked examples and variety of datasets helps students develop confidence and speed in solving statistics problems.