NCERT Solutions for Class 10 Maths Chapter 13 – Statistics

Statistics helps students organise data, interpret it clearly, and make meaningful conclusions. Chapter 13: Statistics focuses on key concepts like mean, median, and mode for grouped data, along with understanding frequency distribution tables and real-life data-based problems.

Statistics
Medium
Q.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants
0-2
2-4
4-6
6-8
8-10
10-12
12-14
Number of houses
1
2
1
5
6
2
3
Which method did you use for finding the mean, and why?
Statistics
Medium
Q.
If the median of the distribution given below is 28.5, find the values of x and y.
Class interval
Frequency
0 – 10

10– 20

20– 30

30–40

40–50

50–60
5

x

20

15

y

5
Total
60
Statistics
Medium
Q.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)
Number of students
Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52
0

3

5

9

14

28

32

35
Draw a less than type graph for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Statistics
Medium
Q.
The following distribution gives the daily income of 50 workers of a factory.
Daily income
 (in ₹)
100-120
120-140
140-160
160-180
180-200
Number of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution, and draw its graph.
Statistics
Medium
Q.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight
(in kg)
40-45
45-50
50-55
55-60
60-65
65-70
70-75
Number of students
2
3
8
6
6
3
2
Statistics
Medium
Q.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters
1-4
4-7
7-10
10-13
13-16
16-19
Number of surnames
6
30
40
16
4
4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Statistics
Medium
Q.
The following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours)
Number of lamps
1500– 2000
2000– 2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000
14
56
60
86
74
62
48
Find the median life time of a lamp.
Statistics
Medium
Q.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Length (in mm)
Number of leaves
118 – 126

127– 135

136– 144

145–153

154–162

163–171

172–180
3

5

9

12

5

4

2
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 – 180)
Statistics
Medium
Q.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Class interval
Frequency
Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60
2

6

24

45

78

89

92

98

100
Statistics
Medium
Q.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units)
Number of consumers
65 – 85

85– 105

105– 125

125–145

145–165

165–185

185–205
4

5

13

20

14

8

4
Statistics
Medium
Q.
Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages     (in ₹)
500-120
520-140
540-160
560-180
580-200
Number of workers
12
14
8
6
10
Find the mean daily wages of the workers of the factory by using an appropriate method.
Statistics
Medium
Q.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
Number
 of cars
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Frequency
7
14
13
12
20
11
15
8
Statistics
Medium
Q.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored
Number of batsmen
3000 – 4000

4000– 5000

5000– 6000

6000–7000

7000–8000

8000–9000

9000–10000

10000–11000
4

18

9

7

6

3

1

1
Find the mode of the data.
Statistics
Medium
Q.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students per teacher
Number of states / U .T.
15– 20

20– 25

25–30

30–35

35–40

40–45

45–50

50–55
3

8

9

10

3

0

0

2
Statistics
Medium
Q.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
Expenditure (in ₹)
Number of families
1000 – 1500
1500– 2000
2000– 2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000
24
40
33
28
30
22
16
7
Statistics
Medium
Q.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components.
Lifetimes  (in hours)
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
10
35
52
61
38
29
Determine the modal lifetimes of the components.
Statistics
Medium
Q.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes
50-52
53-55
56-58
59-61
62-64
Number of boxes
15
110
135
115
25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Statistics
Medium
Q.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Number of heartbeats per minute
65-68
68-71
71-74
74-77
77-80
80-83
83-86
Number of women
2
4
3
8
7
4
2
Statistics
Medium
Q.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.
Daily pocket allowance
(in ₹)
11-13
13-15
15-17
17-19
18-21
21-23
23-25
Number of children
7
6
9
13
f
5
4
Statistics
Medium
Q.
The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield
 (in kg/ha)
50-55
55-60
60-65
65-70
70-75
75-80
Number of farms
2
8
12
24
38
16
Change the distribution to a more than type distribution, and draw its graph.

These NCERT Solutions for Class 10 Maths Chapter 13 include important CBSE board questions asked between 2019 and 2025. All solutions are explained step by step in simple language to help students understand concepts clearly and score well in board exams.

NCERT Solutions for Class 10 Maths Chapter 13 – Statistics

Q1. Determine the modal lifetimes of the components.

Given: Observed lifetimes (in hours) of 225 electrical components.

Lifetimes (in hours) 0–20 20–40 40–60 60–80 80–100 100–120
Frequency 10 35 52 61 38 29

Solution

Modal class = 60–80 (highest frequency 61).
Using mode formula for grouped data:

Mode = l + [(f1 − f0) / (2f1 − f0 − f2)] × h

  • l = 60, h = 20
  • f1 = 61, f0 = 52, f2 = 38

Mode = 60 + [(61 − 52) / (2×61 − 52 − 38)] × 20
= 60 + (9/32) × 20
= 60 + 5.625
= 65.625 hours ≈ 65.6 hours

Q2. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure.

Expenditure (₹) No. of families
1000–1500
1500–2000
2000–2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000		 24
40
33
28
30
22
16
7

Solution (Mode)

Highest frequency = 40 ⇒ Modal class = 1500–2000
l = 1500, h = 500, f1 = 40, f0 = 24, f2 = 33

Mode = 1500 + [(40−24)/(2×40−24−33)] × 500
= 1500 + (16/23) × 500
= 1500 + 347.826…
= 1847.83 ≈ 1848

Solution (Mean)

Class marks: 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750
Mean = Σ(f×x)/Σf = 2662.5

Q3. Find the mode and mean of this data. Interpret the two measures.

Students per teacher No. of states/UT
15–20
20–25
25–30
30–35
35–40
40–45
45–50
50–55		 3
8
9
10
3
0
0
2

Solution (Mode)

Highest frequency = 10 ⇒ Modal class = 30–35
l = 30, h = 5, f1 = 10, f0 = 9, f2 = 3

Mode = 30 + [(10−9)/(2×10−9−3)] × 5
= 30 + (1/8)×5
= 30.625 ≈ 30.6 students/teacher

Solution (Mean)

Mean = Σ(f×x)/Σf = 29.21 students/teacher (approx.)

Interpretation

  • Mode (~30.6): Most common teacher-student ratio lies around 30–35 students per teacher.
  • Mean (~29.2): Overall average ratio is about 29 students per teacher across all states/UTs.

Q4. Find the mode of the data (runs scored by batsmen).

Runs scored No. of batsmen
3000–4000
4000–5000
5000–6000
6000–7000
7000–8000
8000–9000
9000–10000
10000–11000		 4
18
9
7
6
3
1
1

Solution

Modal class = 4000–5000
l = 4000, h = 1000, f1 = 18, f0 = 4, f2 = 9

Mode = 4000 + [(18−4)/(2×18−4−9)]×1000
= 4000 + (14/23)×1000
= 4608.70 ≈ 4609 runs

Q5. Find the mode (cars passing through a spot).

No. of cars 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80
Frequency 7 14 13 12 20 11 15 8

Solution

Modal class = 40–50
l = 40, h = 10, f1 = 20, f0 = 12, f2 = 11

Mode = 40 + [(20−12)/(2×20−12−11)]×10
= 40 + (8/17)×10
= 44.71 ≈ 44.7 cars

Q6. Find the median, mean and mode. Compare them.

Monthly consumption (units) No. of consumers
65–85
85–105
105–125
125–145
145–165
165–185
185–205		 4
5
13
20
14
8
4

Solution (Median)

Total N = 68 ⇒ N/2 = 34
Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68
Median class = 125–145
l = 125, h = 20, f = 20, cf(previous) = 22

Median = 125 + [(34−22)/20]×20
= 125 + 12
= 137 units

Solution (Mean)

Mean = Σ(f×x)/Σf = 137.06 units (approx.)

Solution (Mode)

Modal class = 125–145
l = 125, h = 20, f1 = 20, f0 = 13, f2 = 14

Mode = 125 + [(20−13)/(2×20−13−14)]×20
= 125 + (7/13)×20
= 135.77 ≈ 135.8 units

Comparison

Mean (≈137.06) > Median (137) > Mode (≈135.8) ⇒ data is slightly positively skewed.

Q7. If the median is 28.5, find x and y.

Class interval Frequency
0–10
10–20
20–30
30–40
40–50
50–60		 5
x
20
15
y
5
Total 60

Solution

Total frequency: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15
N/2 = 30 and median = 28.5 ⇒ Median class is 20–30
l = 20, h = 10, f = 20, cf(previous) = 5 + x

28.5 = 20 + [(30 − (5 + x))/20]×10
8.5 = (25 − x)/2 ⇒ 17 = 25 − x ⇒ x = 8
Then y = 15 − 8 = 7

Q8. Calculate the median age of 100 policy holders.

Age (less than) Cumulative frequency
Below 20
Below 25
Below 30
Below 35
Below 40
Below 45
Below 50
Below 55
Below 60		 2
6
24
45
78
89
92
98
100

Solution

Convert to class intervals (18–20, 20–25, 25–30, …, 55–60) by taking differences:
Frequencies: 2, 4, 18, 21, 33, 11, 3, 6, 2 (Total = 100)

N/2 = 50. Cumulative up to 30–35 is 45, up to 35–40 is 78 ⇒ Median class = 35–40
l = 35, h = 5, f = 33, cf(previous) = 45

Median = 35 + [(50−45)/33]×5
= 35 + (5/33)×5
= 35.76 years (approx.)

Q9. Find the median length of the leaves.

Length (mm) No. of leaves
118–126
127–135
136–144
145–153
154–162
163–171
172–180		 3
5
9
12
5
4
2

Solution

Convert to continuous classes (since measured to nearest mm):
117.5–126.5, 126.5–135.5, …, 144.5–153.5, … (class width h = 9)

N = 40 ⇒ N/2 = 20
Cumulative frequencies: 3, 8, 17, 29, … ⇒ Median class = 145–153 (continuous: 144.5–153.5)
l = 144.5, f = 12, cf(previous) = 17, h = 9

Median = 144.5 + [(20−17)/12]×9
= 144.5 + (3/12)×9
= 144.5 + 2.25
= 146.75 mm

Q10. Find the median life time of a neon lamp.

Life time (hours) No. of lamps
1500–2000
2000–2500
2500–3000
3000–3500
3500–4000
4000–4500
4500–5000		 14
56
60
86
74
62
48

Solution

N = 400 ⇒ N/2 = 200
Cumulative frequencies: 14, 70, 130, 216, … ⇒ Median class = 3000–3500
l = 3000, h = 500, f = 86, cf(previous) = 130

Median = 3000 + [(200−130)/86]×500
= 3000 + (70/86)×500
= 3000 + 406.98
= 3406.98 hours ≈ 3407 hours

Q11. Find the median, mean and modal size of surnames (number of letters).

No. of letters 1–4 4–7 7–10 10–13 13–16 16–19
No. of surnames 6 30 40 16 4 4

Solution (Mean)

Class marks: 2.5, 5.5, 8.5, 11.5, 14.5, 17.5 ⇒ Mean = 8.32 letters

Solution (Median)

N = 100 ⇒ N/2 = 50
Cumulative frequencies: 6, 36, 76, … ⇒ Median class = 7–10
Continuous l = 6.5, h = 3, f = 40, cf(previous) = 36

Median = 6.5 + [(50−36)/40]×3
= 6.5 + (14/40)×3
= 7.55 letters

Solution (Mode)

Modal class = 7–10
l = 6.5, h = 3, f1 = 40, f0 = 30, f2 = 16

Mode = 6.5 + [(40−30)/(2×40−30−16)]×3
= 6.5 + (10/34)×3
= 7.38 letters (approx.)
So, modal size ≈ 7–10 letters.

Q12. Find the median weight of the students.

Weight (kg) 40–45 45–50 50–55 55–60 60–65 65–70 70–75
No. of students 2 3 8 6 6 3 2

Solution

N = 30 ⇒ N/2 = 15
Cumulative: 2, 5, 13, 19, … ⇒ Median class = 55–60
l = 55, h = 5, f = 6, cf(previous) = 13

Median = 55 + [(15−13)/6]×5
= 55 + (2/6)×5
= 56.67 kg (approx.)

Q13. Convert to a less than type cumulative frequency distribution and draw its graph.

Daily income (₹) 100–120 120–140 140–160 160–180 180–200
No. of workers 12 14 8 6 10

Solution (Less than cumulative frequency)

Less than Cumulative frequency
< 120 12
< 140 26
< 160 34
< 180 40
< 200 50

Graph (Ogive)

Plot points: (100,0), (120,12), (140,26), (160,34), (180,40), (200,50) and join smoothly.

Q14. Draw a less than type graph. Hence obtain the median weight and verify using formula.

Weight (kg) – Less than Cumulative frequency
<38 0
<40 3
<42 5
<44 9
<46 14
<48 28
<50 32
<52 35

Solution (Median by formula)

First convert to class intervals of width 2: 38–40, 40–42, 42–44, 44–46, 46–48, 48–50, 50–52
Frequencies: 3, 2, 4, 5, 14, 4, 3 (Total N = 35)
N/2 = 17.5 ⇒ Median class is 46–48 (since cf up to 44–46 is 14 and up to 46–48 is 28)

l = 46, h = 2, f = 14, cf(previous) = 14
Median = 46 + [(17.5−14)/14]×2
= 46 + (3.5/14)×2
= 46 + 0.5
= 46.5 kg

Ogive (Less than graph)

Plot points: (38,0), (40,3), (42,5), (44,9), (46,14), (48,28), (50,32), (52,35).
From y = 17.5 on the graph, read the corresponding x-value. It should come near 46.5.

Q15. Change the distribution to a more than type distribution and draw its graph.

Production yield (kg/ha) 50–55 55–60 60–65 65–70 70–75 75–80
No. of farms 2 8 12 24 38 16

Solution (More than cumulative frequency)

More than Cumulative frequency
> 50 100
> 55 98
> 60 90
> 65 78
> 70 54
> 75 16
> 80 0

Graph (More than ogive)

Plot points: (50,100), (55,98), (60,90), (65,78), (70,54), (75,16), (80,0) and join smoothly.

Q16. Find the mean number of plants per house. Which method did you use and why?

No. of plants 0–2 2–4 4–6 6–8 8–10 10–12 12–14
No. of houses 1 2 1 5 6 2 3

Solution

Class marks: 1, 3, 5, 7, 9, 11, 13
Σ(f×x) = 162 and Σf = 20

Mean = 162/20 = 8.1 plants per house

Method used: Step-deviation/assumed mean method is convenient in grouped data because it reduces calculation load
(especially when class intervals are equal).

Q17. Find the mean daily wages of the workers (appropriate method).

(Using the same table as daily income classes 100–120, 120–140, 140–160, 160–180, 180–200)

Solution

Class marks: 110, 130, 150, 170, 190
Σ(f×x) = 7260 and Σf = 50

Mean = 7260/50 = 145.2

Q18. The mean pocket allowance is ₹18. Find the missing frequency f.

Daily pocket allowance (₹) 11–13 13–15 15–17 17–19 19–21 21–23 23–25
No. of children 7 6 9 13 f 5 4

Solution

Class marks: 12, 14, 16, 18, 20, 22, 24
Total frequency = 44 + f

Σ(f×x) without the missing class = 752
Including missing class: 752 + 20f

Given mean = 18:
(752 + 20f)/(44 + f) = 18
752 + 20f = 792 + 18f
2f = 40 ⇒ f = 20

Q19. Find the mean number of mangoes per box. Which method did you choose?

No. of mangoes 50–52 53–55 56–58 59–61 62–64
No. of boxes 15 110 135 115 25

Solution

Class marks: 51, 54, 57, 60, 63
Total boxes = 400
Mean = Σ(f×x)/Σf = 22875/400 = 57.1875 ≈ 57.19 mangoes

Method chosen: Step-deviation/assumed mean method is preferred because values are larger and class intervals are equal.

Q20. Find the mean heartbeats per minute (choose a suitable method).

Heartbeats per minute 65–68 68–71 71–74 74–77 77–80 80–83 83–86
No. of women 2 4 3 8 7 4 2

Solution

Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5
Mean = Σ(f×x)/Σf = 2277/30 = 75.9 beats per minute

Method: Assumed mean method (faster for grouped data).

FAQs – NCERT Solutions for Class 10 Maths Chapter 5 (Arithmetic Progressions)

1) Arithmetic Progression (A.P.) kya hota hai?

A.P. ek number sequence hota hai jisme har next term, previous term se ek fixed number (common difference ‘d’) se increase/decrease hoti hai.
Example: 2, 5, 8, 11, … (d = 3)

2) A.P. ka nth term ka formula kya hai?

Agar first term a ho aur common difference d ho, toh
nth term: an = a + (n − 1)d

3) A.P. ke n terms ka sum kaise nikalte hain?

Sn = n/2 [2a + (n − 1)d]
Ya phir last term l known ho toh:
Sn = n/2 (a + l)

4) Boards me A.P. se kaunse type ke questions aate hain?

Usually questions nth term, sum of n terms, missing term find karna, aur word problems (salary increase, saving plan, seating arrangement, etc.)
par based hote hain. Step-by-step approach follow karne se easy ho jata hai.

5) A.P. questions quickly solve karne ka best tip kya hai?

Pehle identify karo: a (first term), d (common difference), n (number of terms).
Fir direct formula apply karo. Word problems me statement ko A.P. terms me convert karna sabse important step hai.