NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 help students understand and solve questions related to grouped data and mean calculation in a clear and exam-oriented manner. This exercise builds a strong foundation in Statistics, which is an important scoring unit in Class 10 Maths.
Designed as per the latest CBSE Class 10 Maths syllabus, Exercise 14.1 focuses on calculating the mean of grouped frequency distribution using systematic steps. Regular practice of this exercise helps students improve calculation speed, accuracy, and confidence for board examinations.
NCERT Solutions Class 10 Maths Chapter 14 – Statistics Exercise 14.1
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?
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.
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
|
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
|
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?
Q.
The table below shows the daily expenditure on food of 25 households in a locality.
|
Daily Expenditure (in ₹)
|
100-150
|
150-200
|
200-250
|
250-300
|
300-350
|
|
Number of households
|
4
|
5
|
12
|
2
|
2
|
Find the mean daily expenditure on food by a suitable method.
Q.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below.
|
Concentration of SO2 (in ppm)
|
Frequency
|
|
0.00 – 0.04
0.04– 0.08
0.08 – 0.12
0.12–0.16
0.16–0.20
0.20–0.24
|
4
9
9
2
4
2
|
Find the mean concentration of SO2 in the air.
Q.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
|
Number of days
|
0-6
|
6-10
|
10-14
|
14-20
|
20-28
|
28-38
|
38-40
|
|
Number of students
|
11
|
10
|
7
|
4
|
4
|
3
|
1
|
Q.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
|
Literacy rate (in %)
|
45-55
|
55-65
|
65-75
|
75-85
|
85-95
|
|
Number of cities
|
3
|
10
|
11
|
8
|
3
|
Q.
The following table shows the ages of the patients admitted in a hospital during a year.
|
Age (in years)
|
5-15
|
15-25
|
25-35
|
35-45
|
45-55
|
55-65
|
|
Number of patients
|
6
|
11
|
21
|
23
|
14
|
5
|
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
NCERT Solutions Class 10 Maths Chapter 14 – Statistics Exercise 14.1
These solutions are prepared in a structured format so students can clearly understand how to organize data, apply formulas correctly, and avoid common mistakes in exams.
Q1. Mean number of plants per house (20 houses)
Class intervals (plants): 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14
Frequencies (houses): 1, 2, 1, 5, 6, 2, 3
Class marks (xᵢ): 1, 3, 5, 7, 9, 11, 13
Compute fᵢxᵢ:
-
0–2: f=1, x=1 ⇒ fx=1
-
2–4: f=2, x=3 ⇒ fx=6
-
4–6: f=1, x=5 ⇒ fx=5
-
6–8: f=5, x=7 ⇒ fx=35
-
8–10: f=6, x=9 ⇒ fx=54
-
10–12: f=2, x=11 ⇒ fx=22
-
12–14: f=3, x=13 ⇒ fx=39
∑f = 20, ∑fx = 162
Mean = (∑fx)/(∑f) = 162/20 = 8.1
✅ Answer: Mean plants per house = 8.1
✅ Method used: Direct method, because xᵢ and fᵢ are small, calculation is easy.
Q2. Mean daily wages (50 workers)
Classes: 500–520, 520–540, 540–560, 560–580, 580–600
f: 12, 14, 8, 6, 10
Class marks xᵢ: 510, 530, 550, 570, 590
Take assumed mean a = 550
dᵢ = xᵢ − a : -40, -20, 0, 20, 40
fᵢdᵢ:
12(-40)=-480, 14(-20)=-280, 8(0)=0, 6(20)=120, 10(40)=400
∑f = 50, ∑fd = -240
Mean = a + (∑fd / ∑f)
= 550 + (-240/50)
= 550 - 4.8
= 545.2
✅ Answer: Mean daily wages = ₹ 545.20
✅ Method: Assumed mean method (numbers are large, easier than direct).
Q3. Missing frequency f (mean pocket allowance = ₹18)
Classes: 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
f: 7, 6, 9, 13, f, 5, 4
Class marks xᵢ: 12, 14, 16, 18, 20, 22, 24
Take a = 18
dᵢ = xᵢ − a : -6, -4, -2, 0, 2, 4, 6
fᵢdᵢ:
7(-6)=-42
6(-4)=-24
9(-2)=-18
13(0)=0
f(2)=2f
5(4)=20
4(6)=24
∑f = 44 + f
∑fd = (-42-24-18) + 0 + 2f + 20 + 24
= (-84 + 44) + 2f
= -40 + 2f
Mean formula: 18 = a + (∑fd)/(∑f)
18 = 18 + (-40 + 2f)/(44 + f)
So (-40 + 2f)=0 ⇒ 2f=40 ⇒ f=20
✅ Answer: f = 20
Q4. Mean heartbeats per minute (30 women)
Classes: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86
f: 2, 4, 3, 8, 7, 4, 2
Class marks xᵢ: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5
Class size h = 3, take a = 75.5
dᵢ = xᵢ − a: -9, -6, -3, 0, 3, 6, 9
uᵢ = dᵢ/h: -3, -2, -1, 0, 1, 2, 3
fᵢuᵢ: 2(-3)=-6, 4(-2)=-8, 3(-1)=-3, 8(0)=0, 7(1)=7, 4(2)=8, 2(3)=6
∑f = 30, ∑fu = 4
Mean = a + h(∑fu/∑f)
= 75.5 + 3(4/30)
= 75.5 + 0.4
= 75.9
✅ Answer: Mean heartbeat = 75.9 per minute
✅ Method: Step-deviation (common factor, easier arithmetic).
Q5. Mean mangoes per box
Given (not continuous): 50–52, 53–55, 56–58, 59–61, 62–64
f: 15, 110, 135, 115, 25
Make continuous:
49.5–52.5, 52.5–55.5, 55.5–58.5, 58.5–61.5, 61.5–64.5
Class marks xᵢ: 51, 54, 57, 60, 63
h = 3, take a = 57
dᵢ = xᵢ − a: -6, -3, 0, 3, 6
uᵢ = dᵢ/h: -2, -1, 0, 1, 2
fᵢuᵢ: 15(-2)=-30, 110(-1)=-110, 135(0)=0, 115(1)=115, 25(2)=50
∑f = 400, ∑fu = 25
Mean = a + h(∑fu/∑f)
= 57 + 3(25/400)
= 57 + 0.1875
= 57.1875 ≈ 57.19
✅ Answer: Mean mangoes per box = 57.19
✅ Method: Step-deviation (nice class width, easier).
Q6. Mean daily expenditure (25 households)
Classes: 100–150, 150–200, 200–250, 250–300, 300–350
f: 4, 5, 12, 2, 2
Class marks xᵢ: 125, 175, 225, 275, 325
h = 50, take a = 225
dᵢ = xᵢ − a: -100, -50, 0, 50, 100
uᵢ = dᵢ/h: -2, -1, 0, 1, 2
fᵢuᵢ: 4(-2)=-8, 5(-1)=-5, 12(0)=0, 2(1)=2, 2(2)=4
∑f = 25, ∑fu = -7
Mean = a + h(∑fu/∑f)
= 225 + 50(-7/25)
= 225 - 14
= 211
✅ Answer: Mean expenditure = ₹ 211
Q7. Mean concentration of SO₂ (ppm)
Intervals: 0.00–0.04, 0.04–0.08, 0.08–0.12, 0.12–0.16, 0.16–0.20, 0.20–0.24
f: 4, 9, 9, 2, 4, 2
Class marks xᵢ: 0.02, 0.06, 0.10, 0.14, 0.18, 0.22
h = 0.04, take a = 0.14
dᵢ = xᵢ − a: -0.12, -0.08, -0.04, 0, 0.04, 0.08
uᵢ = dᵢ/h: -3, -2, -1, 0, 1, 2
fᵢuᵢ: 4(-3)=-12, 9(-2)=-18, 9(-1)=-9, 2(0)=0, 4(1)=4, 2(2)=4
∑f = 30, ∑fu = -31
Mean = a + h(∑fu/∑f)
= 0.14 + 0.04(-31/30)
= 0.14 - 0.0413...
= 0.0987 ≈ 0.099
✅ Answer: Mean SO₂ concentration = 0.099 ppm
Q8. Mean days absent (40 students)
Classes: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40
f: 11, 10, 7, 4, 4, 3, 1
Class marks xᵢ: 3, 8, 12, 17, 24, 33, 39
Take a = 17
dᵢ = xᵢ − a: -14, -9, -5, 0, 7, 16, 22
fᵢdᵢ: 11(-14)=-154, 10(-9)=-90, 7(-5)=-35, 4(0)=0, 4(7)=28, 3(16)=48, 1(22)=22
∑f = 40, ∑fd = -181
Mean = a + (∑fd/∑f)
= 17 + (-181/40)
= 17 - 4.525
= 12.475 ≈ 12.48
✅ Answer: Mean absent days = 12.48 days
Q9. Mean literacy rate (35 cities)
Classes: 45–55, 55–65, 65–75, 75–85, 85–95
f: 3, 10, 11, 8, 3
Class marks xᵢ: 50, 60, 70, 80, 90
Take a = 70
dᵢ = xᵢ − a: -20, -10, 0, 10, 20
fᵢdᵢ: 3(-20)=-60, 10(-10)=-100, 11(0)=0, 8(10)=80, 3(20)=60
∑f = 35, ∑fd = -20
Mean = a + (∑fd/∑f)
= 70 + (-20/35)
= 70 - 0.5714
= 69.4286 ≈ 69.43%
✅ Answer: Mean literacy rate = 69.43%
FAQs – Class 10 Maths Chapter 14 Exercise 14.1
Q1. What is the focus of Exercise 14.1?
Exercise 14.1 focuses on finding the mean of grouped data using proper statistical methods in an exam-friendly format.
Q2. Why is Exercise 14.1 important for board exams?
Questions based on mean calculation from grouped frequency tables are commonly asked in CBSE board exams and are considered scoring if solved accurately.
Q3. How can students prepare effectively for Exercise 14.1?
Students should practice preparing frequency tables neatly, calculate class marks correctly, and revise the mean formula thoroughly to strengthen their preparation.
Q4. How do NCERT Solutions help in scoring better marks?
NCERT Solutions provide step-by-step explanations, clear tabular presentation, and accurate calculations that help students minimize errors and score full marks in exams.