NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

Updated on 30 Apr 2025, 05:14 PM IST

Statistics enables us to understand real-life data through interpretation of collected information. This field in mathematics handles the process of gathering numerical data followed by visual or analytical interpretation to reveal statistical patterns. The exercise works with grouped data together with frequency distribution patterns to analyze large statistical information. The procedure of grouping data followed by counting the number of occurrences in each category enables us to find significant patterns.

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NCERT Solutions Class 10 Maths Chapter 13: Exercise 13.1
Topics covered in Chapter 13 Statistics: Exercise 13.1
NCERT Solutions for Class 10 Subject Wise
NCERT Exemplar Solutions of Class 10 Subject Wise

exercise 14.1

The NCERT Solutions for Exercise 13.1 represent unprocessed data through grouped frequency tables. The frequency tables demonstrate to students which values occur in which intervals. The exercise challenges students to convert raw data into structured forms which lays a foundation for computing mean, median and mode values in future assignments. The exercise functions as an essential tool for knowledge described in NCERT Books which leads to complete understanding of problems and their practical applications.

NCERT Solutions Class 10 Maths Chapter 13: Exercise 13.1

Q1 A survey was conducted by a group of students as a part of their environment awareness program, 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.

Which method did you use for finding the mean, and why?

Answer:

Number of plants	Number of houses $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	$\sum f_{i}$ =20		$\sum f_{i} x_{i}$ =162

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{162}{20} = 8.1$

We used the direct method in this as the values of $x_{i} a n d f_{i}$ are small.

Q2 Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:

Let the assumed mean be a = 550

Daily Wages	Number of workers $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
500-520	12	510	-40	-480
520-540	14	530	-20	-280
540-560	8	550	0	0
560-580	6	570	20	120
580-600	10	590	40	400
	$\sum f_{i}$ = 50			$\sum f_{i} x_{i}$ = -240

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 550 + \frac{- 240}{50} = 550 - 4.8 = 545.20$

Therefore, the mean daily wages of the workers of the factory is Rs. 545.20

Q3 Following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Answer:

Daily pocket allowance	Number of children $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
	$\sum f_{i}$ =44+f		$\sum f_{i} x_{i}$ =752+20f

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$⟹ 18 = \frac{752 + 20 f}{44 + f}$

$⟹ 18 (44 + f) = (752 + 20 f) ⟹ 2 f = 40 ⟹ f = 20$

Therefore the missing f = 20

Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

1745918959444

Answer:

Let the assumed mean be a = 75.5

No. of heartbeats per minute	Number of women $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
65-68	2	66.5	-9	-18
68-71	4	69.5	-6	-24
71-74	3	72.5	-3	-9
74-77	8	75.5	0	0
77-80	7	78.5	3	21
80-83	4	81.5	6	24
83-86	2	84.5	9	18
	$\sum f_{i}$ =30			$\sum f_{i} x_{i}$ =12

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$

Therefore, the mean heartbeats per minute of these women are 75.9

Q5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

Let the assumed mean be a = 57

Number of mangoes	Number of boxes $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$f_{i} d_{i}$
50-52	15	51	-6	-90
53-55	110	54	-3	-330
56-58	135	57	0	0
59-61	115	60	3	345
62-64	25	63	6	150
	$\sum f_{i}$ =400			$\sum f_{i} x_{i}$ =75

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$= 57 + \frac{75}{400} = 57 + 0.1875 = 57.1875 \approx 57.19$

Therefore, the mean number of mangoes kept in a packing box is approx 57.19

Q6 The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily expenditure in rupees	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Answer:

Let the assumed mean be a = 225 and h = 50

Daily Expenditure	Number of households $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$u_{i} = \frac{d_{i}}{h}$	$f_{i} u_{i}$
100-150	4	125	-100	-2	-8
150-200	5	175	-50	-1	-5
200-250	12	225	0	0	0
250-300	2	275	50	1	2
300-350	2	325	100	2	4
	$\sum f_{i}$ =25				$\sum f_{i} x_{i}$ = -7

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$

$= 225 + \frac{- 7}{25} \times 50 = 225 - 14 = 211$

Therefore, the mean daily expenditure on food is Rs. 211

Q7 To find out the concentration of $S O_{2}$ 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:

Find the mean concentration of $S O_{2}$ in the air.

Answer:

Class Interval	Frequency $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
	$\sum f_{i}$ =30		$\sum f_{i} x_{i}$ =2.96

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{2.96}{30} = 0.099$

Therefore, the mean concentration of $S O_{2}$ in the air is 0.099 ppm

Q8 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.

1636093576677

Answer:

Number of days	Number of Students $f_{i}$	Classmark $x_{i}$	$f_{i} x_{i}$
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39

	$\sum f_{i}$ =40		$\sum f_{i} x_{i}$ =499

Mean,

$\overset{―}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{499}{40} = 12.475$ $= \frac{499}{40} = 12.475 \approx 12.48$

Therefore, the mean number of days a student was absent is 12.48 days.

Q9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

Let the assumed mean be a = 75 and h = 10

Literacy rates	Number of cities $f_{i}$	Classmark $x_{i}$	$d_{i} = x_{i} - a$	$u_{i} = \frac{d_{i}}{h}$	$f_{i} u_{i}$
45-55	3	50	-20	-2	-6
55-65	10	60	-10	-1	-10
65-75	11	70	0	0	0
75-85	8	80	10	1	8
85-95	3	90	20	2	6
	$\sum f_{i}$ = 35				$\sum f_{i} x_{i}$ = -2

Mean,

$\overset{―}{x} = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h$

$= 70 + \frac{- 2}{35} \times 10 = 70 - 0.57 = 69.43$

Therefore, the mean mean literacy rate is 69.43%

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Topics covered in Chapter 13 Statistics: Exercise 13.1

1. Grouped Frequency Distribution: A person must learn to convert extensive raw data into classification intervals combined with frequency counts that produce better reading metrics.

2. Mid-point Method: The average calculation for grouped data requires using the midpoints from class intervals to estimate the central value.

3. Mean of Grouped Data: The formula Mean = (∑fx) / ∑f applies to calculate grouped data average while recognizing f as the frequency and x as the class mark (midpoint).

4. Data Interpretation: The skill involves understanding real-world data followed by tabular structure implementation for better analysis purposes.

5. Practical Understanding of Statistics: Data handling abilities prove essential in practical conditions like budgeting surveys and planning and budget processes.

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Frequently Asked Questions (FAQs)

Q: What is the main concept of Class 10 Maths chapter 14 exercise 14.1?

To find what is the basics of statistic. Definition of mean and its formula. To learn about different methods of finding mean like Assumed mean method, step deviation method etc. Go through the ex 14.1 class 10 to command these concepts.

Q: What do you understand by the mean of a set of data?

The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. Practice class 10 maths ex 14.1 to command these concepts.

Q: What is class mark?

The class midpoint (or class mark) is a specific point in the centre of the bins (categories) in a frequency distribution table. Go through exercise 14.1 class 10 maths to get deeper understanding of concepts.

Q: Find the mean of the data?

This class 10 ex 14.1 discusses the concept of mean in detail. to find mean of data use formula discussed in this exercise. mean = sum of total data / number of data

Q: The centre of a bar in a histogram is known as?

The centre of a bar in a histogram is known as class mark.

Q: What will happen to the mean of the data if every data set has increased by the value of 5?

The mean of the data will also increase by 5.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

NCERT Solutions Class 10 Maths Chapter 13: Exercise 13.1

Answer:

Answer:

Answer:

Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

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