CBSE Class 10 · Chapter 13 · Statistics and Probability · 2026-27

Statistics Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceMean of grouped data

The class mark of the class $12.5$ – $17.5$ is

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Answer: (a)

15

Why: Class mark = (lower limit + upper limit)/2.

$\frac{12.5 + 17.5}{2} = \frac{30}{2} = 15$ .

Q2

·1 mark·Multiple choiceRelationship between mean, median and mode

A grouped distribution has mean $30$ and mode $24$ . From $3 Median = Mode + 2 Mean$ , its median is

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Answer: (b)

28

Why: 3 Median = Mode + 2 Mean = 24 + 60.

$3 Median = 24 + 2 (30) = 84 \Rightarrow Median = 28$ .

Q3

·1 mark·Multiple choiceMean of grouped data

Using the step-deviation method with assumed mean $a = 25$ , class width $h = 5$ , $\sum f_{i} u_{i} = - 10$ and $\sum f_{i} = 50$ , the mean is

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Answer: (c)

24

Why: Mean = a + h × (Σfu/Σf) = 25 + 5 × (−1/5).

$\bar{x} = 25 + 5 \times \frac{- 10}{50} = 25 - 1 = 24$ .

Q4

·1 mark·Multiple choiceMedian of grouped data

For the distribution below, the median class is

Class	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	12	9	6

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Answer: (b)

20

–

30

Why: n = 40, n/2 = 20; cumulative frequencies 5, 13, 25, … first reach 20 in 20–30.

$n = 40$ , $\frac{n}{2} = 20$ . Cumulative frequencies: $5, 13, 25, 34, 40$ . The first one $\geq 20$ is $25$ , in the class $20$ – $30$ .

Q5

·1 mark·Multiple choiceMode of grouped data

The mode of the following data is

Class	10–20	20–30	30–40	40–50	50–60
Frequency	4	6	15	9	4

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Answer: (d)

36

Why: Modal class 30–40: Mode = l + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h.

Modal class $30$ – $40$ : $l = 30$ , $f_{1} = 15$ , $f_{0} = 6$ , $f_{2} = 9$ , $h = 10$ . Mode $= 30 + \frac{15 - 6}{30 - 6 - 9} \times 10 = 30 + \frac{9}{15} \times 10 = 36$ .

Q6

·1 mark·Multiple choiceMean of grouped data

The mean of the data with classes $0$ – $10$ , $10$ – $20$ , $20$ – $30$ and frequencies $4$ , $10$ , $6$ is

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Answer: (b)

16

Why: Use class marks 5, 15, 25: Σfx = 20 + 150 + 150 = 320 over Σf = 20.

$\bar{x} = \frac{4 (5) + 10 (15) + 6 (25)}{20} = \frac{320}{20} = 16$ .

Q7

·1 mark·Multiple choiceMedian of grouped data

In the formula $Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$ , the symbol $c f$ stands for the cumulative frequency

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Answer: (c) of the class before the median class

Why: cf counts all observations below the median class.

$c f$ is the cumulative frequency of the class preceding the median class: the number of observations below $l$ .

Q8

·1 mark·Multiple choiceMedian of grouped data

The median of the distribution below is

Class	0–10	10–20	20–30	30–40	40–50
Frequency	5	8	12	9	6

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Answer: (a)

\frac{155}{6}

Why: Median class 20–30 with l = 20, cf = 13, f = 12, h = 10.

$Median = 20 + \frac{20 - 13}{12} \times 10 = 20 + \frac{70}{12} = \frac{155}{6}$ (about $25.8$ ).

Q9

·1 mark·Multiple choiceMean of grouped data

If every observation of a data set is multiplied by $3$ , then the mean

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Answer: (c) is multiplied by

3

Why: Σ(3x)/n = 3 × Σx/n.

$\frac{\sum f_{i} (3 x_{i})}{\sum f_{i}} = 3 \times \frac{\sum f_{i} x_{i}}{\sum f_{i}}$ : the mean is multiplied by $3$ .

Q10

·1 mark·Multiple choiceMean of grouped data

In the assumed-mean method, $a = 50$ , $\sum f_{i} d_{i} = - 40$ and $\sum f_{i} = 20$ . The mean is

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Answer: (b)

48

Why: Mean = a + Σfd/Σf = 50 − 2.

$\bar{x} = 50 + \frac{- 40}{20} = 48$ .

Q11

·1 mark·Multiple choiceRelationship between mean, median and mode

For a distribution, the mean and the median are both $30$ . By the empirical relation, the mode is

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Answer: (a)

30

Why: Mode = 3 Median − 2 Mean = 90 − 60.

$Mode = 3 (30) - 2 (30) = 30$ .

Q12

·1 mark·Multiple choiceMedian of grouped data

For the distribution below, the sum of the lower limits of the median class and the modal class is

Class	10–20	20–30	30–40	40–50	50–60
Frequency	4	6	15	9	4

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Answer: (a)

60

Why: n = 38, n/2 = 19 falls in 30–40 (cf 25), and 30–40 also has the highest frequency.

Cumulative frequencies: $4, 10, 25, 34, 38$ ; $\frac{n}{2} = 19$ lies in $30$ – $40$ (median class). Highest frequency $15$ is also in $30$ – $40$ (modal class). Sum $= 30 + 30 = 60$ .

Q13

·1 mark·Multiple choiceMean of grouped data

The mean of the distribution below is $20$ . The value of $p$ is

Class	0–10	10–20	20–30	30–40
Frequency	6	$p$	10	4

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Answer: (c)

4

Why: Σfx = 420 + 15p must equal 20(20 + p).

Class marks $5, 15, 25, 35$ : $\sum f_{i} x_{i} = 30 + 15 p + 250 + 140 = 420 + 15 p$ , $\sum f_{i} = 20 + p$ . $420 + 15 p = 400 + 20 p \Rightarrow p = 4$ .

Q14

·1 mark·Multiple choiceMedian of grouped data

A cumulative frequency table is used mainly to find the

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Answer: (d) median

Why: The median formula needs the cumulative frequency before the median class.

To locate the median class (where the cumulative frequency first reaches $\frac{n}{2}$ ) and apply the median formula, we need cumulative frequencies.

Q15

·1 mark·Multiple choiceMode of grouped data

For a grouped distribution the modal class is $60$ – $80$ , with frequency $20$ ; the classes just before and after it have frequencies $10$ and $14$ . The mode is

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Answer: (d)

72.5

Why: Mode = 60 + (20 − 10)/(40 − 10 − 14) × 20.

$Mode = 60 + \frac{20 - 10}{2 (20) - 10 - 14} \times 20 = 60 + \frac{10}{16} \times 20 = 60 + 12.5 = 72.5$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Electricity use in a housing society (4 marks)

The residents' association of a housing society recorded the electricity used last month by $40$ homes:

Units used	100–150	150–200	200–250	250–300	300–350
Number of homes	4	9	14	9	4

(i) Write the modal class. [1 mark]

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Answer:

200

–

250

The highest frequency,

14

, is in

200

–

250

. A1

(ii) Write the median class. [1 mark]

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Answer:

200

–

250

Cumulative frequencies

4, 13, 27, 35, 40

;

\frac{n}{2} = 20

first reached in

200

–

250

. A1

(iii) Find the mode of the data. [2 marks]

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Answer:

225

units

Mode = 200 + \frac{14 - 9}{28 - 9 - 9} \times 50

= 200 + 25 = 225

units. A1

OR Find the mean electricity use, using the step-deviation method. [2 marks]

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Answer:

225

units

a = 225

h = 50

u_{i} = - 2, - 1, 0, 1, 2

\sum f_{i} u_{i} = - 8 - 9 + 0 + 9 + 8 = 0

M1; mean

= 225 + 50 \times \frac{0}{40} = 225

units. A1

Case study 2: Unit-test marks (4 marks)

A mathematics teacher records the marks (out of $50$ ) scored by the $40$ students of her class in a unit test:

Marks	0–10	10–20	20–30	30–40	40–50
Number of students	3	7	12	10	8

(i) How many students scored less than

30

marks? [1 mark]

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Answer:

22

3 + 7 + 12 = 22

. A1

(ii) Write the median class. [1 mark]

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Answer:

20

–

30

\frac{n}{2} = 20

; cumulative frequencies

3, 10, 22, \dots

first reach

20

20

–

30

. A1

(iii) Find the median marks. [2 marks]

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Answer:

\frac{85}{3}

(about

28.3

)

Median = 20 + \frac{20 - 10}{12} \times 10

= 20 + \frac{25}{3} = \frac{85}{3}

. A1

OR Find the mean marks. [2 marks]

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Answer:

28.25

Class marks

5, 15, 25, 35, 45

\sum f_{i} x_{i} = 15 + 105 + 300 + 350 + 360 = 1130

M1; mean

= \frac{1130}{40} = 28.25

. A1

Next steps for Statistics

This free set is separate from the chapter's question bank. On the Statistics chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 41 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

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Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.