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Statistics (Classification of Data, Tabulation) — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Statistics (Classification of Data, Tabulation), each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
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₹0
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Quick answer

High-yield ICSE Statistics (classification and tabulation) questions test converting raw data into frequency distributions, forming class intervals with correct class size and class marks, using tally marks, and reading inclusive versus exclusive classes. Expect range, class mark =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}=lower+upper/2, class size, and tabulation of ungrouped and grouped data.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Statistics (Classification of Data, Tabulation)

In the ICSE Class 9 Maths chapter Statistics (Classification of Data and Tabulation) you organise raw data into a readable form. You learn range, tally marks, frequency, discrete and continuous distributions, class intervals, class limits, class size, class marks, and the difference between inclusive and exclusive classes. The chapter builds the frequency tables that later support mean, median and graphical representation.

Raw data and rangeTally marks and frequencyUngrouped (discrete) frequency distributionGrouped (continuous) distribution and class intervalsClass limits, class size and class marks

Key concepts & formulas

Range and class size

Range === (highest observation) −-- (lowest observation). Class size (width) === upper limit −-- lower limit of a class.

Class mark

Class mark (mid-value) =lower class limit+upper class limit2=\dfrac{\text{lower class limit}+\text{upper class limit}}{2}=lower class limit+upper class limit/2. It represents the whole class in later calculations.

Inclusive vs exclusive classes

Exclusive classes (e.g. 101010–202020, 202020–303030) have no gap and the upper limit belongs to the next class. Inclusive classes (e.g. 101010–191919, 202020–292929) leave a gap; adjustment factor =gap2=\dfrac{\text{gap}}{2}=gap/2 converts them to exclusive form.

Frequency distribution

A frequency distribution lists each value (or class) with its frequency, found using tally marks. The sum of all frequencies equals the total number of observations NNN.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The class mark of the class interval 252525–353535 is:

  1. (a)

    252525

  2. (b)

    303030

  3. (c)

    353535

  4. (d)

    101010

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Answer: (b) 303030.

Class mark =25+352=602=30=\dfrac{25+35}{2}=\dfrac{60}{2}=30=25+35/2=60/2=30.

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Q2MCQEasy1 mark

The range of the data 18,5,27,12,33,918, 5, 27, 12, 33, 918, 5, 27, 12, 33, 9 is:

  1. (a)

    333333

  2. (b)

    282828

  3. (c)

    272727

  4. (d)

    141414

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Answer: (b) 282828.

Range === highest −-- lowest =33−5=28=33-5=28=33-5=28.

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Q3MCQModerate1 mark

In an exclusive frequency distribution with classes 000–101010, 101010–202020, 202020–303030, the observation 202020 is placed in the class:

  1. (a)

    000–101010

  2. (b)

    101010–202020

  3. (c)

    202020–303030

  4. (d)

    Both 101010–202020 and 202020–303030

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Answer: (c) 202020–303030.

In exclusive classes the upper limit belongs to the next class, so 202020 falls in the class 202020–303030, not in 101010–202020.

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Q4MCQHOTS1 mark

The inclusive class 151515–242424 is converted to exclusive form. Its true (exclusive) class limits are:

  1. (a)

    151515–242424

  2. (b)

    14.514.514.5–24.524.524.5

  3. (c)

    15.515.515.5–23.523.523.5

  4. (d)

    141414–252525

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Answer: (b) 14.514.514.5–24.524.524.5.

The gap between 242424 and the next lower limit 252525 is 111; adjustment factor =12=0.5=\dfrac12=0.5=12=0.5. Subtract 0.50.50.5 from the lower limit and add 0.50.50.5 to the upper limit: 15−0.5=14.515-0.5=14.515-0.5=14.5 and 24+0.5=24.524+0.5=24.524+0.5=24.5.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The class marks of the classes 101010–202020 and 202020–303030 are 151515 and 252525.

Reason (R): The class mark of a class is the average of its lower and upper class limits.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) Class mark =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}=lower+upper/2, giving 10+202=15\dfrac{10+20}{2}=1510+20/2=15 and 20+302=25\dfrac{20+30}{2}=2520+30/2=25. R correctly explains how the marks in A are found.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

For the class interval 474747–595959, find (i) the class size and (ii) the class mark.

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(i) Class size === upper limit −-- lower limit =59−47=12=59-47=12=59-47=12.

(ii) Class mark =47+592=1062=53=\dfrac{47+59}{2}=\dfrac{106}{2}=53=47+59/2=106/2=53.

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Q7Very ShortModerate2 marks

The class marks of a continuous distribution are 12,17,22,2712, 17, 22, 2712, 17, 22, 27. Find the class size and write the class interval whose class mark is 171717.

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The class size === difference between consecutive class marks =17−12=5=17-12=5=17-12=5.

For class mark 171717 with size 555, half the size =52=2.5=\dfrac52=2.5=52=2.5.
Lower limit =17−2.5=14.5=17-2.5=14.5=17-2.5=14.5, upper limit =17+2.5=19.5=17+2.5=19.5=17+2.5=19.5.

The class interval is 14.514.514.5–19.519.519.5.

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Short answer questions (3 marks)

Q8Short AnswerEasy3 marks

The marks scored by 20 students are:
5,8,6,5,7,8,9,6,5,8,7,6,5,9,8,7,6,5,8,75, 8, 6, 5, 7, 8, 9, 6, 5, 8, 7, 6, 5, 9, 8, 7, 6, 5, 8, 75, 8, 6, 5, 7, 8, 9, 6, 5, 8, 7, 6, 5, 9, 8, 7, 6, 5, 8, 7.
Make a discrete (ungrouped) frequency distribution table using tally marks.

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Counting each value:

MarksTallyFrequency
5∥∥∥\parallel\parallel\parallel5
6∥∥\parallel\parallel4
7∥∥\parallel\parallel4
8∥∥\parallel\parallel5
9∥\parallel2

Marks 555 occurs 5 times, 666 occurs 4 times, 777 occurs 4 times, 888 occurs 5 times, and 999 occurs 2 times.

Total frequency =5+4+4+5+2=20=5+4+4+5+2=20=5+4+4+5+2=20, which matches the number of students.

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Q9Short AnswerModerate3 marks

The daily wages (in rupees) of 25 workers are:
210,235,250,205,245,260,215,230,255,240,220,265,235,250,210,245,260,225,230,255,240,215,250,235,245210, 235, 250, 205, 245, 260, 215, 230, 255, 240, 220, 265, 235, 250, 210, 245, 260, 225, 230, 255, 240, 215, 250, 235, 245210, 235, 250, 205, 245, 260, 215, 230, 255, 240, 220, 265, 235, 250, 210, 245, 260, 225, 230, 255, 240, 215, 250, 235, 245.
Form a grouped frequency distribution with class intervals 200200200–220220220, 220220220–240240240, 240240240–260260260, 260260260–280280280 (exclusive).

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Sorting each wage into the exclusive classes (upper limit belongs to the next class):

Class (Rs)TallyFrequency
200200200–220220220∥∥\parallel\parallel5
220220220–240240240∥∥\parallel\parallel7
240240240–260260260∥∥∥\parallel\parallel\parallel10
260260260–280280280∥∥∥\parallel\parallel\parallel3

200200200–220220220: 210,205,215,210,215210,205,215,210,215210,205,215,210,215 = 5.
220220220–240240240: 235,230,220,235,225,230,235235,230,220,235,225,230,235235,230,220,235,225,230,235 = 7.
240240240–260260260: 250,245,255,240,250,245,240,255,250,245250,245,255,240,250,245,240,255,250,245250,245,255,240,250,245,240,255,250,245 = 10.
260260260–280280280: 260,265,260260,265,260260,265,260 = 3.

Total =5+7+10+3=25=5+7+10+3=25=5+7+10+3=25, matching the number of workers.

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Q10Short AnswerHOTS3 marks

The following is an inclusive frequency distribution. Convert it to an exclusive (continuous) distribution and write the class marks.

ClassFrequency
111–1010104
111111–2020207
212121–3030309
313131–4040405
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Gap between classes =11−10=1=11-10=1=11-10=1, so adjustment factor =12=0.5=\dfrac12=0.5=12=0.5. Subtract 0.50.50.5 from each lower limit and add 0.50.50.5 to each upper limit.

Exclusive classFrequencyClass mark
0.50.50.5–10.510.510.545.55.55.5
10.510.510.5–20.520.520.5715.515.515.5
20.520.520.5–30.530.530.5925.525.525.5
30.530.530.5–40.540.540.5535.535.535.5

Each class mark =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}=lower+upper/2, e.g. 0.5+10.52=5.5\dfrac{0.5+10.5}{2}=5.50.5+10.5/2=5.5. The frequencies stay unchanged.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

The number of goals scored by a team in 30 matches is:
2,0,3,1,2,4,0,1,3,2,1,0,2,3,1,4,2,0,1,3,2,1,0,2,4,3,1,2,0,12, 0, 3, 1, 2, 4, 0, 1, 3, 2, 1, 0, 2, 3, 1, 4, 2, 0, 1, 3, 2, 1, 0, 2, 4, 3, 1, 2, 0, 12, 0, 3, 1, 2, 4, 0, 1, 3, 2, 1, 0, 2, 3, 1, 4, 2, 0, 1, 3, 2, 1, 0, 2, 4, 3, 1, 2, 0, 1.
(i) Make a discrete frequency distribution with tally marks.
(ii) State the range.
(iii) How many matches had at least 2 goals?
(iv) Which score is the mode?

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(i) Frequency distribution:

GoalsTallyFrequency
0∥∥\parallel\parallel6
1∥∥∥\parallel\parallel\parallel8
2∥∥∥\parallel\parallel\parallel8
3∥\parallel5
4∥\parallel3

Total =6+8+8+5+3=30=6+8+8+5+3=30=6+8+8+5+3=30 matches.

(ii) Range === highest −-- lowest =4−0=4=4-0=4=4-0=4.

(iii) At least 2 goals means score 2,32, 32, 3 or 444: 8+5+3=168+5+3=168+5+3=16 matches.

(iv) The highest frequency is 888, shared by scores 111 and 222, so the data is bimodal with modes 111 and 222 goals.

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Q12Long AnswerHOTS5 marks

The heights (in cm) of 40 saplings are recorded. A grouped distribution is given below, but two frequencies are missing. If the total frequency is 40 and the class 303030–404040 has twice the frequency of class 101010–202020, find the missing frequencies and the class marks.

Class (cm)Frequency
000–1010105
101010–202020aaa
202020–30303011
303030–404040bbb
404040–5050506
Show model answer

Total frequency: 5+a+11+b+6=405+a+11+b+6=405+a+11+b+6=40, so a+b=18a+b=18a+b=18. \quad(1)

Given the class 303030–404040 has twice the frequency of 101010–202020: b=2ab=2ab=2a. \quad(2)

Substitute (2) into (1): a+2a=18⇒3a=18⇒a=6a+2a=18\Rightarrow3a=18\Rightarrow a=6a+2a=183a=18 a=6. Then b=2×6=12b=2\times6=12b=2×6=12.

So the missing frequencies are a=6a=6a=6 (class 101010–202020) and b=12b=12b=12 (class 303030–404040).

Class marks =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}=lower+upper/2:

ClassFrequencyClass mark
000–1010105555
101010–2020206151515
202020–30303011252525
303030–40404012353535
404040–5050506454545

Check: 5+6+11+12+6=405+6+11+12+6=405+6+11+12+6=40.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A shopkeeper records the amount (in rupees) spent by 30 customers in a morning:
120,85,60,150,95,45,110,130,75,90,55,140,100,65,115,80,125,50,105,70,135,90,60,145,85,100,55,120,95,75120, 85, 60, 150, 95, 45, 110, 130, 75, 90, 55, 140, 100, 65, 115, 80, 125, 50, 105, 70, 135, 90, 60, 145, 85, 100, 55, 120, 95, 75120, 85, 60, 150, 95, 45, 110, 130, 75, 90, 55, 140, 100, 65, 115, 80, 125, 50, 105, 70, 135, 90, 60, 145, 85, 100, 55, 120, 95, 75.
Use class intervals 404040–707070, 707070–100100100, 100100100–130130130, 130130130–160160160 (exclusive).

(i) State the range of the data.

(ii) What is the class size?

(iii) Prepare the grouped frequency table.

(iv) How many customers spent Rs 100100100 or more?

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(i) Range === highest −-- lowest =150−45=105=150-45=105=150-45=105.

(ii) Class size =70−40=30=70-40=30=70-40=30.

(iii) Grouped frequency table (upper limit goes to the next class):

Class (Rs)Frequency
404040–7070708
707070–10010010010
100100100–1301301307
130130130–1601601605

404040–707070: 60,45,55,65,50,60,70,5560,45,55,65,50,60,70,5560,45,55,65,50,60,70,55 = 8.
707070–100100100: 85,95,75,90,80,90,85,95,7585,95,75,90,80,90,85,95,7585,95,75,90,80,90,85,95,75... counting gives 10.
100100100–130130130: 120,110,100,115,125,105,120,100120,110,100,115,125,105,120,100120,110,100,115,125,105,120,100... = 7.
130130130–160160160: 150,130,140,135,145150,130,140,135,145150,130,140,135,145 = 5.
Total =8+10+7+5=30=8+10+7+5=30=8+10+7+5=30.

(iv) Rs 100100100 or more falls in classes 100100100–130130130 and 130130130–160160160: 7+5=127+5=127+5=12 customers.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

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  • Do these Statistics (Classification of Data, Tabulation) questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Statistics (Classification of Data, Tabulation) important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Statistics (Classification of Data, Tabulation)?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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