MathsICSE Class 9 · Statistics

Grouped Data: Class Intervals, Frequency Tables and the Frequency Polygon (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Raw data are arranged in an array and then in a frequency table using tally marks. A discrete variable takes isolated values; a continuous variable takes any value in an interval. For a class interval note the class limits, class boundaries (halfway between adjacent limits), class size and class mark (the mid-value). A frequency polygon joins points (class mark, frequency).

Plot each class mark against its frequency, join the points with straight lines, and add zero-frequency class marks at both ends so the polygon closes.

Illustration drawn for QuestionBuddy.

Understand Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon

Types of data. *Raw data* are the observations as collected; an *array* lists them in order; a *frequency distribution* shows each value (or class) with how often it occurs, counted with tally marks in bunches of five.

Variables. A discrete variable takes separate values (number of siblings, marks out of 10). A continuous variable can take any value in a range (height, time, mass).

Class intervals. For continuous data (or large discrete data) group values into classes.

*Inclusive classes* like $10 - 19$ , $20 - 29$ have class limits (lower 10, upper 19). To make them continuous, the class boundaries are halfway between the upper limit of one class and the lower limit of the next: $9.5 - 19.5$ , $19.5 - 29.5$ .
*Exclusive classes* like $10 - 20$ , $20 - 30$ already touch; a value equal to the upper limit goes into the next class.
Class size $=$ upper boundary $-$ lower boundary (for $10 - 19$ : $19.5 - 9.5 = 10$ ).
Class mark $= \frac{lower limit + upper limit}{2}$ (for $5 - 9$ : $7$ ).

Cumulative frequency. The running total of frequencies up to a class. The last value equals the total, and the differences recover the frequencies.

Frequency polygon. Mark class marks on the x-axis and frequencies on the y-axis, plot (class mark, frequency), and join by straight lines. Add class marks with zero frequency before the first and after the last class so that the polygon closes on the x-axis. For classes $10 - 20, \dots, 40 - 50$ with frequencies $4, 9, 12, 5$ the points are $(5, 0), (15, 4), (25, 9), (35, 12), (45, 5), (55, 0)$ .

Check. The sum of frequencies must equal the number of observations; cumulative frequencies must never decrease.

Level. Class IX is a school-exam class. Most items here are routine; missing-frequency and polygon-point questions are at the hardest school-exam level.

Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon: formulas to remember

Class mark: $\frac{lower limit + upper limit}{2}$; Mid-value of the class.
Class size: $upper boundary - lower boundary$; Equal for all classes.
Boundaries of inclusive classes: $limit \mp 0.5$; For gap of 1 between classes.
Cumulative frequency: $c f_{k} = f_{1} + f_{2} + \dots + f_{k}$; Last cf = total.
Frequency polygon points: $(class mark, f)$; Add zero ends.

How to solve Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon questions

Tally, tabulate, check the total, then work out marks, boundaries and cumulative frequencies.

1Decide the classes (equal size, covering all data).
2Tally each observation exactly once and count.
3Check that the frequencies add to the number of observations.
4Compute class marks, boundaries, class size, and cumulative frequencies as asked.
5For a polygon, plot (class mark, f) and add the two zero-frequency ends.
6Join the points with straight lines using a ruler.

Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Survey of travel time

Suppose 30 students report their travel time to school (minutes). Grouped into classes 0–10, 10–20, 20–30, 30–40 the frequencies are 6, 11, 9, 4.

How many travel under 20 minutes? Which class is most common? What are the class marks?

Try it first, then show the solutionSolution

Under 20 minutes: $6 + 11 = 17$ students.
Most common (highest frequency 11): the class $10 - 20$ .
Class marks: $5, 15, 25, 35$ .

Answer: 17 students; modal class 10–20; marks 5, 15, 25, 35.

Memory hook: Cumulative frequency answers "fewer than".

Counting shoe sizes

Imagine 20 customers whose shoe sizes are recorded as 4, 7, 5, 4, 6, 7, 5, 4, 5, 6, 4, 5, 7, 6, 5, 4, 5, 6, 7, 5.

Make a frequency table and find the most common size.

Try it first, then show the solutionSolution

Size 4: 5; size 5: 7; size 6: 4; size 7: 4 (total 20).
Most common: size 5.

Answer: Frequencies 5, 7, 4, 4; the mode is size 5.

Memory hook: Tally in fives and check the total.

Tricks to remember Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon

Short memory aids. Say each one aloud once, then use it on a question the same day.

Five-bar gate
Tally four strokes and cross the fifth.
Use it when: Fast, accurate counting.
Plus or minus a half
For inclusive classes with gap 1, boundaries are limits ∓ 0.5.
Use it when: Class boundaries.
Close the polygon
Add a zero-frequency class at each end.
Use it when: Frequency polygon.

Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Class boundaries · 1 mark · MCQBoard level

For the class interval $10 - 19$ (inclusive classes), the class boundaries are:
(a) $9$ and $19$
(b) $9.5$ and $19.5$
(c) $10.5$ and $18.5$
(d) $10$ and $20$

Try it first, then show the solutionSolution

The gap between $19$ and $20$ is $1$ , so subtract and add $0.5$ : $9.5$ and $19.5$ .

Answer: (b)

Example 2: Frequency table · 3 marks · Short answerBoard level

The shoe sizes of 20 customers are $4, 7, 5, 4, 6, 7, 5, 4, 5, 6, 4, 5, 7, 6, 5, 4, 5, 6, 7, 5$ . Prepare a frequency table and state the size most customers wear.

Try it first, then show the solutionSolution

Size $4$ : 5; size $5$ : 7; size $6$ : 4; size $7$ : 4.
Total $= 5 + 7 + 4 + 4 = 20$ . ✓
The highest frequency is $7$ , so size $5$ is the most common.

Answer: Frequencies 5, 7, 4, 4; size 5.

Example 3: Class mark, boundaries and cumulative frequency · 4 marks · Short answerHarder board level

Classes (inclusive) $5 - 9$ , $10 - 14$ , $15 - 19$ have frequencies $3, 8, 4$ . Find the class marks, class boundaries, class size and cumulative frequencies.

Try it first, then show the solutionSolution

Class marks: $7, 12, 17$ .
Boundaries: $4.5 - 9.5$ , $9.5 - 14.5$ , $14.5 - 19.5$ .
Class size $= 9.5 - 4.5 = 5$ .
Cumulative frequencies: $3, 11, 15$ .

Answer: Marks 7, 12, 17; boundaries as above; size 5; cf 3, 11, 15

Example 4: Plotting a frequency polygon · 4 marks · Short answerHarder board level

The classes $10 - 20$ , $20 - 30$ , $30 - 40$ , $40 - 50$ have frequencies $4, 9, 12, 5$ . Write the coordinates needed to draw the frequency polygon and state the number of observations.

Try it first, then show the solutionSolution

Class marks: $15, 25, 35, 45$ .
Points: $(15, 4), (25, 9), (35, 12), (45, 5)$ .
Add the zero ends: $(5, 0)$ and $(55, 0)$ .
Total $= 4 + 9 + 12 + 5 = 30$ observations.

Answer: (5,0), (15,4), (25,9), (35,12), (45,5), (55,0); 30 observations

Example 5: Missing frequency · 5 marks · Long answerHOTS / competency

A distribution with classes $0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50$ has frequencies $5, 8, x, 12, 6$ and $40$ observations in total. Find $x$ , the cumulative frequencies, the class in which the 20th observation lies, and the percentage of observations below 30.

Try it first, then show the solutionSolution

$5 + 8 + x + 12 + 6 = 40$ , so $x = 9$ .
Cumulative frequencies: $5, 13, 22, 34, 40$ .
The 20th observation: the first cumulative frequency $\geq 20$ is $22$ , so it lies in $20 - 30$ .
Below 30: $22$ observations, $\frac{22}{40} \times 100 = 55%$ .

Answer: x = 9; cf 5, 13, 22, 34, 40; class 20–30; 55%

Where students lose marks in Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon

The slips examiners see most often, and what to write instead.

Mistake: Using limits as boundaries
Inclusive classes need the half-unit adjustment for continuity.
Wrong working
Boundary $= 19$ .
Right working
Boundary $= 19.5$ .
Do this: Check the gap between classes.
Usually costs: Whole answer.
Mistake: Not checking the total
Tally errors show up in the total.
Wrong working
Frequencies add to 19.
Right working
They should add to 20: recount.
Do this: Add and compare.
Usually costs: Whole answer.
Mistake: Leaving the polygon open
The polygon should touch the x-axis.
Wrong working
Join only the four plotted points.
Right working
Add (5, 0) and (55, 0).
Do this: Add zero-frequency ends.
Usually costs: Graph mark.

Exam tips

Class IX is examined in school papers; frequency tables are always checked against the total.
State the scale on both axes of a graph.
Write class marks in a separate column.
Use a ruler for the polygon.

Practice questions on Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For the classes $20 - 29$ and $30 - 39$ , the common class boundary is:
(a) $29$
(b) $29.5$
(c) $30$
(d) $30.5$
Hint
Halfway between 29 and 30.
Answer
(b) 29.5
Q2 · 3 marks · Short answerHarder board level
Find the class marks of the classes $0 - 10$ , $10 - 20$ , $20 - 30$ and the class size.
Hint
Mean of the limits.
Answer
Marks 5, 15, 25; class size 10.
Q3 · 3 marks · Short answerHarder board level
The cumulative frequencies of five classes are $4, 11, 20, 28, 30$ . Find the frequency of each class.
Hint
Successive differences.
Answer
4, 7, 9, 8, 2 (total 30).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The height of students is a continuous variable.
Reason (R): Height can take any value in an interval.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Definition of continuous.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: 30 students report their travel time (minutes). The classes $0 - 10, 10 - 20, 20 - 30, 30 - 40$ have frequencies $6, 11, 9, 4$ .
(a) Find the class size.
(b) Find the class marks.
(c) Name the class with the highest frequency.
(d) How many travel under 20 minutes?
Hint
Add the first two frequencies for part (d).
Answer
(a) 10 (b) 5, 15, 25, 35 (c) 10–20 (d) 17

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the class mark of 20–30?Show answer
25.
2.What is a continuous variable?Show answer
One that can take any value in a range.
3.Class size of boundaries 9.5–19.5?Show answer
10.
4.What does the last cumulative frequency equal?Show answer
The total number of observations.

Grouped Data, Class Intervals, Frequency Tables and the Frequency Polygon: frequently asked questions

Do I need a histogram for a frequency polygon?

No; plot class marks against frequencies directly.

What is a discrete variable?

One taking separate values, such as the number of children.

Why are class boundaries needed?

They make classes continuous.