NCERT Solutions · Class 10 · Chapter 13: Statistics

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2

Exercise 13.2: Mode of grouped data. The modal class has the highest frequency. Mode \(= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\), where \(f_0, f_2\) are the frequencies of the classes before and after it. Mode is best for 'most common', mean for 'average'.

6 questions, 9 parts
Every answer checked by computer algebra
Free, no sign-in

Try each question first, then open its solution. Our own step-by-step solutions, set out for step marks.

Exercise 13.2 questions and solutions

Exercise 13.2, Question 1

Ages (years) of patients admitted to a hospital in a year.

Class	Frequency
5–15	6
15–25	11
25–35	21
35–45	23
45–55	14
55–65	5

(i) Find the mode.

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 35–45: \[35 + \dfrac{23 - 21}{46 - 21 - 14} \times 10 = 35 + \dfrac{20}{11}\]

Answer: About \(36.8\) years

(ii) Find the mean and compare the two.

Show solution

Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
\(\sum f_ix_i = 2830\), \(\sum f_i = 80\): \(\bar x = 35.375\).
The most common age (about 36.8) is a little above the average age (about 35.4).

Answer: Mean about \(35.37\) years; mode is higher

Practise this: Step 2, Board standard →

Exercise 13.2, Question 2

Lifetimes (hours) of 225 electrical components. Find the modal lifetime.

Class	Frequency
0–20	10
20–40	35
40–60	52
60–80	61
80–100	38
100–120	29

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 60–80: \[60 + \dfrac{61 - 52}{122 - 52 - 38} \times 20 = 60 + \dfrac{9}{32} \times 20\]

Answer: \(65.625\) hours

Practise this: Step 2, Board standard →

Exercise 13.2, Question 3

Monthly household expenditure (₹) of 200 families.

Class	Frequency
1000–1500	24
1500–2000	40
2000–2500	33
2500–3000	28
3000–3500	30
3500–4000	22
4000–4500	16
4500–5000	7

(i) Find the modal expenditure.

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 1500–2000: \[1500 + \dfrac{40 - 24}{80 - 24 - 33} \times 500 = 1500 + \dfrac{16}{23} \times 500\]

Answer: About ₹1847.83

(ii) Find the mean expenditure.

Show solution

Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
\(a = 2750\), \(h = 500\): \(u_i = -3, -2, \ldots, 4\) and \[\begin{aligned}\sum f_iu_i &= -72 - 80 - 33 + 0 + 30 + 44 + 48 + 28 \\ &= -35\end{aligned}\]
\[\begin{aligned}\bar x &= 2750 + 500 \times \dfrac{-35}{200} \\ &= 2662.5\end{aligned}\]

Answer: ₹2662.50

Practise this: Step 3, Full marks on long answers →

Exercise 13.2, Question 4

Teacher–student ratio in higher secondary schools across states (number of students per teacher).

Class	Frequency
15–20	3
20–25	8
25–30	9
30–35	10
35–40	3
40–45	0
45–50	0
50–55	2

(i) Find the mode.

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 30–35: \[30 + \dfrac{10 - 9}{20 - 9 - 3} \times 5 = 30.625\]

Answer: About \(30.6\)

(ii) Find the mean and interpret both.

Show solution

Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
\(\sum f_ix_i = 1022.5\), \(\sum f_i = 35\): \(\bar x \approx 29.2\).
Most states have about 30.6 students per teacher; the average is about 29.2.

Answer: Mean about \(29.2\)

Practise this: Step 3, Full marks on long answers →

Exercise 13.2, Question 5

Runs scored by leading batsmen in one-day internationals. Find the mode.

Class	Frequency
3000–4000	4
4000–5000	18
5000–6000	9
6000–7000	7
7000–8000	6
8000–9000	3
9000–10000	1
10000–11000	1

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 4000–5000: \[4000 + \dfrac{18 - 4}{36 - 4 - 9} \times 1000 = 4000 + \dfrac{14}{23} \times 1000\]

Answer: About \(4608.7\) runs

Practise this: Step 2, Board standard →

Exercise 13.2, Question 6

Number of cars passing a spot in 3-minute periods, observed 100 times. Find the mode.

Class	Frequency
0–10	7
10–20	14
20–30	13
30–40	12
40–50	20
50–60	11
60–70	15
70–80	8

Show solution

Mode \[= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h\] for the modal class (highest frequency).
Modal class 40–50: \[40 + \dfrac{20 - 12}{40 - 12 - 11} \times 10 = 40 + \dfrac{80}{17}\]

Answer: About \(44.7\) cars

Practise this: Step 2, Board standard →

Done the NCERT exercises? The board paper asks more

Statistics has 37 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Statistics: notes and questions Create a free account Teachers: set it as homework

Statistics in our sample papers: Sample paper 1 (questions 15, 16, 35) · Sample paper 2 (questions 17, 35) · Sample paper 3 (questions 17, 31, 36) · Sample paper 4 (questions 17, 35) · Sample paper 5 (questions 31, 35).

Also useful: free MCQs and case studies for Statistics · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.