Exercise 7.2 — Grouped Data

Organisation of grouped data and cumulative frequency.

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Exercise 7.2 – Grouped Frequency Distribution Tables

Exercise 7.2 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) focuses on grouped frequency distributions — a powerful way to summarise large datasets into class intervals so that patterns become visible at a glance. This exercise builds directly on the ungrouped frequency tables studied in Exercise 7.1 and is essential for understanding histograms, frequency polygons, and cumulative frequency curves introduced later in the chapter.

The eight problems in this exercise cover: constructing grouped frequency tables from raw data, working with inclusive and exclusive (continuous) class intervals, converting class limits to class boundaries, finding class marks, and building both less-than and greater-than cumulative frequency tables.

Key Concepts You Must Know Before Solving

Inclusive vs. Exclusive Class Intervals

Data can be grouped in two ways. Inclusive (discontinuous) classes have a gap between the upper limit of one class and the lower limit of the next — for example 1–10, 11–20, 21–30. Both end values belong to the class. Exclusive (continuous) classes have no gap — for example 0–10, 10–20, 20–30 — where the upper limit of each class is the same as the lower limit of the next, and by convention the upper limit is not included in that class.

Inclusive Classes (discontinuous)

Class Interval (Marks)	No. of Students
1 – 10	2
11 – 20	5
21 – 30	3

Both 1 and 10 belong to the first class.

Exclusive Classes (continuous)

Class Interval (Marks)	No. of Students
0 – 10	2
10 – 20	5
20 – 30	3

10 belongs to 10–20, not 0–10.

Class Limits, Boundaries, and Class Marks

Every class interval has a lower limit and an upper limit. To convert inclusive class intervals into continuous boundaries — needed for drawing histograms — we subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Inclusive → Boundaries

Lower	Upper	Lower Bdy.	Upper Bdy.
1	10	0.5	10.5
11	20	10.5	20.5
21	30	20.5	30.5

Exclusive → Boundaries (same)

Lower	Upper	Lower Bdy.	Upper Bdy.
0	10	0	10
10	20	10	20
20	30	20	30

Length of Class Interval

= Upper boundary − Lower boundary

Range of Data

= Maximum value − Minimum value

Length of Class Interval (formula)

= Range ÷ Number of classes

Class Mark (Mid-point)

= (Lower limit + Upper limit) ÷ 2

💡 Important: When Range ÷ Number of classes is not a whole number, always round up to the next integer. For example, 9.66 rounds up to 10. This ensures no data value falls outside the last class.

Worked Solutions — Exercise 7.2

Problem 1

Ages of 45 people in a colony — Construct a grouped frequency distribution with 6 class intervals

The raw data of 45 ages is:

33, 8, 7, 25, 31, 26, 5, 50, 25, 48, 56, 33, 28, 22, 15, 62, 59, 16, 14, 19, 24, 35, 26, 9, 12, 46, 15, 42, 63, 32, 5, 22, 11, 42, 23, 52, 48, 62, 10, 24, 43, 51, 37, 48, 36

Step-by-Step Solution

Step 1 — Range = Maximum value − Minimum value = 63 − 5 = 58

Step 2 — Number of classes = 6 (given)

Step 3 — Length of class interval = Range ÷ Classes = 58 ÷ 6 = 9.66 ≈ 10 (rounded up)

Step 4 — First class starts at minimum value = 5, so classes are 5–14, 15–24, 25–34, …

Step 5 — Tally each data value into the correct class interval.

After tallying all 45 values, the completed grouped frequency distribution table is:

Class Interval (Age)	Tally Marks	Frequency
5 – 14	𝍸𝍸 \|\|\|\|	9
15 – 24	𝍸𝍸 \|\|\|\|	9
25 – 34	𝍸𝍸 \|\|\|\|	9
35 – 44	𝍸 \|	6
45 – 54	𝍸 \|\|	7
55 – 64	𝍸	5
Total		45

✅ Verify: Always check that all frequencies add up to the total number of data values. Here 9+9+9+6+7+5 = 45. ✓

The first three classes (5–14, 15–24, 25–34) each have 9 people, indicating an even spread of younger to middle-aged residents. The count drops for older age groups, which is a typical pattern in many colonies.

Problem 2

Number of students in 30 classrooms — Exclusive class interval of 4

The data (number of students in each of 30 classrooms) is:

25, 30, 24, 18, 21, 24, 32, 34, 22, 20, 22, 32, 40, 28, 30, 22, 26, 31, 34, 15, 38, 28, 20, 16, 15, 20, 24, 30, 25, 18

Step-by-Step Solution

Step 1 — Range = 40 − 15 = 25

Step 2 — Length of class interval = 4 (given as exclusive)

Step 3 — Number of classes = Range ÷ Class length = 25 ÷ 4 = 6.25 ≈ 7 (rounded up)

Step 4 — Classes start from minimum value 15: 15–19, 19–23, 23–27, … (exclusive type)

🔵 Exclusive classes: In the class 15–19, the value 19 does NOT belong here — it goes into 19–23. This is the key difference from inclusive classes. The upper limit is always excluded.

No. of Students (Class)	Tally Marks	Frequency
15 – 19	\|\|\|\|	5
19 – 23	𝍸 \|\|	7
23 – 27	𝍸 \|	6
27 – 31	𝍸	5
31 – 35	𝍸	5
35 – 39	\|	1
39 – 43	\|	1
Total		30

Most classrooms have between 19 and 27 students, which forms the peak of this distribution. Very few classrooms (just 2) have more than 35 students.

Problem 3

Given class intervals 4–11, 12–19, 20–27, 28–35, 36–43 — find length, boundaries, class marks, and next two classes

Part (i) — Length of Each Class Interval

The given classes are inclusive (discontinuous). The length is calculated from the boundaries. Lower boundary of 4–11 = 3.5, upper boundary = 11.5.

Length = Upper boundary − Lower boundary = 11.5 − 3.5 = 8

Alternatively, since these are inclusive classes: Length = (Upper limit − Lower limit) + 1 = (11 − 4) + 1 = 8.

Part (i continued) — Next Two Class Intervals

Each class spans 8 values (inclusive). Continuing the pattern after 36–43:

Next two classes: 44 – 51 and 52 – 59

Part (ii) — Class Boundaries of All Classes

To convert inclusive limits to continuous boundaries, subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class:

Class Interval	Lower Limit	Upper Limit	Lower Boundary	Upper Boundary
4 – 11	4	11	3.5	11.5
12 – 19	12	19	11.5	19.5
20 – 27	20	27	19.5	27.5
28 – 35	28	35	27.5	35.5
36 – 43	36	43	35.5	43.5
44 – 51	44	51	43.5	51.5
52 – 59	52	59	51.5	59.5

Notice that consecutive boundaries share the same value (11.5 appears as both the upper boundary of the first class and the lower boundary of the second). This makes the classes continuous — a requirement for drawing histograms.

Part (iii) — Class Marks of Each Class

The class mark (or mid-value) is the average of the two limits of a class:

Class mark = (Lower limit + Upper limit) ÷ 2

Class Interval	Calculation	Class Mark
4 – 11	(4 + 11) ÷ 2	7.5
12 – 19	(12 + 19) ÷ 2	15.5
20 – 27	(20 + 27) ÷ 2	23.5
28 – 35	(28 + 35) ÷ 2	31.5
36 – 43	(36 + 43) ÷ 2	39.5
44 – 51	(44 + 51) ÷ 2	47.5
52 – 59	(52 + 59) ÷ 2	55.5

💡 Class marks are used when calculating the mean of grouped data and when plotting frequency polygons. They represent all values in that class with a single point.

Understanding Cumulative Frequency

A cumulative frequency is a running total of frequencies up to (or from) a certain class. There are two types, and both are important for board exams in Telangana, Andhra Pradesh, and CBSE Class 8.

Less Than Cumulative Frequency

Add frequencies from the top downward.
Tells how many values are LESS than the upper limit of a class.

Greater Than (More Than) Cumulative Frequency

Add frequencies from the bottom upward.
Tells how many values are MORE than the lower limit of a class.

Using the simple dataset (10 students, marks in ranges 0–10, 10–20, 20–30) from the lesson:

Less Than Cumulative Frequency

Class (Marks)	Frequency	Less Than CF
0 – 10	2	2
10 – 20	5	2 + 5 = 7
20 – 30	3	7 + 3 = 10

7 students scored less than 20 marks.

More Than Cumulative Frequency

Class (Marks)	Frequency	More Than CF
0 – 10	2	8 + 2 = 10
10 – 20	5	3 + 5 = 8
20 – 30	3	3

8 students scored more than 10 marks.

Problem 4

Class marks are given — Construct class intervals, less than CF, and greater than CF

Given class marks and frequencies:

Class Marks	10	22	34	46	58	70
Frequency	6	14	20	21	9	5

Step 1 — Find the Class Interval Length (h)

Difference between consecutive class marks = 22 − 10 = 12

h = 12, so h/2 = 6

Lower boundary of each class = class mark − 6

Upper boundary of each class = class mark + 6

Step 2 — Build the Class Intervals


      Class mark 10 → (10−6) to (10+6) → 4 – 16

      Class mark 22 → (22−6) to (22+6) → 16 – 28

      Class mark 34 → (34−6) to (34+6) → 28 – 40

      Class mark 46 → (46−6) to (46+6) → 40 – 52

      Class mark 58 → (58−6) to (58+6) → 52 – 64

      Class mark 70 → (70−6) to (70+6) → 64 – 76

Step 3 — Complete Table with Both Cumulative Frequencies

Class Mark	Class Interval	Frequency	Less Than CF	More Than CF
10	4 – 16	6	6	69 + 6 = 75
22	16 – 28	14	6 + 14 = 20	55 + 14 = 69
34	28 – 40	20	20 + 20 = 40	35 + 20 = 55
46	40 – 52	21	40 + 21 = 61	14 + 21 = 35
58	52 – 64	9	61 + 9 = 70	5 + 9 = 14
70	64 – 76	5	70 + 5 = 75	5

The last value of less than CF (75) equals the first value of more than CF (75) — and both equal the total number of observations. This is a quick check to confirm your cumulative frequency table is correct.

Problem 5

Marks of 35 students in a Statistics test (out of 50) — Construct exclusive frequency table with class interval 10–20

The raw marks data is:

35, 1, 15, 35, 45, 23, 31, 40, 21, 13, 15, 20, 47, 48, 42, 34, 43, 45, 33, 37, 11, 13, 27, 18, 12, 37, 39, 38, 16, 13, 18, 5, 41, 47, 43

Step-by-Step Solution

Range = 48 − 1 = 47

Given class: 10–20 (exclusive, i.e., 20 is NOT included in this class)

Length of class interval = 20 − 10 = 10

Number of classes = 47 ÷ 10 = 4.7 ≈ 5 (rounded up)

Classes: 0–10, 10–20, 20–30, 30–40, 40–50 (exclusive throughout)

Class (Marks)	Frequency
0 – 10	2
10 – 20	10
20 – 30	4
30 – 40	9
40 – 50	10
Total	35

The marks are bimodal — one peak in the 10–20 range and another in 40–50. Very few students scored in the middle range (20–30), suggesting the class had a mixed level of preparation.

Problem 6

Ages of children (inclusive classes) — Find boundaries and construct both cumulative frequency tables

Given distribution:

Ages	1–3	4–6	7–9	10–12	13–15
No. of Children	10	12	15	13	9

These are inclusive class intervals (the gap between 3 and 4 is bridged by converting to boundaries). Subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Class Interval (Age)	Frequency	Class Boundaries	Less Than CF	More Than CF
1 – 3	10	0.5 – 3.5	10	49 + 10 = 59
4 – 6	12	3.5 – 6.5	10 + 12 = 22	37 + 12 = 49
7 – 9	15	6.5 – 9.5	22 + 15 = 37	22 + 15 = 37
10 – 12	13	9.5 – 12.5	37 + 13 = 50	9 + 13 = 22
13 – 15	9	12.5 – 15.5	50 + 9 = 59	9

✅ Check: Last value of less than CF = 59 = First value of more than CF = 59 = Total children. ✓

Problem 7

Given cumulative frequency table of cricketers' runs — Identify type and find individual class frequencies

Given data:

Runs (Class)	0–10	10–20	20–30	30–40	40–50
No. of Cricketers	3	8	19	25	30

Step 1 — Identify the Type

The values in the "number of cricketers" row are increasing from top to bottom (3 → 8 → 19 → 25 → 30). This is the pattern of a Less Than Cumulative Frequency distribution, where each entry tells you how many cricketers scored less than the upper limit of that class.

Step 2 — Recover Individual Frequencies

To go from cumulative frequency back to individual frequency, subtract consecutive cumulative values:

Class (Runs)	Less Than CF	Frequency (Individual)
0 – 10	3	3
10 – 20	8	8 − 3 = 5
20 – 30	19	19 − 8 = 11
30 – 40	25	25 − 19 = 6
40 – 50	30	30 − 25 = 5

The highest number of cricketers (11) scored in the 20–30 runs range. Frequencies were recovered by simple subtraction — a key technique for board exam questions where only cumulative data is provided.

Problem 8

Library readers — Given greater-than CF, find individual frequencies and construct less-than CF table

Given data (number of books borrowed by library readers):

No. of Books	1–10	11–20	21–30	31–40	41–50
Greater Than CF	42	36	23	14	6

Step 1 — Identify the Type

The cumulative values decrease from top to bottom (42 → 36 → 23 → 14 → 6). This confirms it is a Greater Than Cumulative Frequency table. Each value tells you how many readers borrowed at least as many books as the lower limit of that class.

Step 2 — Recover Individual Frequencies

For greater-than CF, subtract consecutive values going downward. The last class gives the frequency directly.


      Frequency of 1–10 = 42 − 36 = 6

      Frequency of 11–20 = 36 − 23 = 13

      Frequency of 21–30 = 23 − 14 = 9

      Frequency of 31–40 = 14 − 6 = 8

      Frequency of 41–50 = 6 (last class, given directly)

Step 3 — Construct the Less Than CF Table

Class (No. of Books)	Greater Than CF	Frequency	Less Than CF
1 – 10	42	6	6
11 – 20	36	13	6 + 13 = 19
21 – 30	23	9	19 + 9 = 28
31 – 40	14	8	28 + 8 = 36
41 – 50	6	6	36 + 6 = 42

✅ Check: The final less than CF value (42) matches the first greater than CF value (42) = Total readers. ✓

Common Mistakes to Avoid in Exercise 7.2

Forgetting to round up: When class length = Range ÷ No. of classes gives a decimal (like 9.66), always round UP to the next integer (10). Rounding down leaves some data values without a class.
Confusing inclusive and exclusive classes: In inclusive classes (e.g., 1–10, 11–20), the value 10 goes into the first class. In exclusive classes (e.g., 0–10, 10–20), the value 10 goes into the second class.
Wrong boundary conversion: For inclusive classes, boundaries are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. For exclusive classes, the limits themselves are the boundaries.
Wrong cumulative direction: Less than CF is built from the top down (adding). More than CF is built from the bottom up (adding backwards). Mixing these up is a very common board exam mistake.
Not verifying totals: Always confirm that the sum of all frequencies equals the total number of data values and that the final/first cumulative frequency values also match this total.
Skipping the tally step: When constructing tables from raw data, draw tally marks first rather than directly counting. This prevents missing or double-counting values in large datasets.

📋 Board Exam Tip (Telangana & AP): In the Telangana SSC and AP SSC examinations, questions from this exercise typically ask you to construct frequency tables, find class boundaries, class marks, or cumulative frequency tables — often combining two or three tasks in a single problem for 4–5 marks. Practise writing all steps clearly with proper column headings.

At a Glance — Types of Class Intervals Compared

Inclusive vs Exclusive class intervals — key differences

What This Exercise Prepares You For

Mastering grouped frequency tables and cumulative frequencies in Exercise 7.2 is the essential foundation for the rest of Chapter 7. In Exercise 7.3, you will draw histograms and frequency polygons directly from the grouped frequency tables you built here — using class boundaries on the x-axis and frequencies on the y-axis.

Cumulative frequency tables (both less-than and more-than types) come back in Class 9 Statistics, where you will plot ogive curves (cumulative frequency graphs) to find median, quartiles, and percentiles graphically — a major topic in both Telangana and CBSE Class 9 and 10 board exams.

The skill of reading and reconstructing frequencies from cumulative data (as practised in Problems 7 and 8) also appears in Class 10 Statistics, where cumulative frequency is used to calculate the median of grouped data using the formula method.