What does Exercise 13.1 of Class 10 Maths Chapter 13 cover?

Exercise 13.1 focuses on data representation and organization. It involves converting raw data into a grouped frequency distribution table and understanding its components, such as class intervals and frequency.

Why is Exercise 13.1 important for students?

This exercise helps students organize and interpret data systematically. It builds a foundation for understanding mean, median, mode, and other statistical measures in later exercises.

How does studying Statistics in Class 10 help in real life?

Statistics is used in various fields such as business, economics, research, and even sports analysis. Understanding data helps in decision-making, data interpretation, and predicting trends.

What are common mistakes to avoid in Exercise 13.1?

Students often misplace frequency values in the table, confuse class intervals (inclusive vs. exclusive), make errors in totaling frequency values, and skip steps in data organization, leading to incorrect conclusions.

What types of questions are asked in Exercise 13.1?

The questions involve creating grouped frequency distribution tables, identifying class width, and interpreting data. Some problems require analyzing given tables to extract meaningful insights.

NCERT Solutions

Class 10

Maths

Chapter 13 Statistics

Exercise 13.1

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1

NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.1 helps students understand an important aspect of statistics, the mean of certain data. This exercise will specifically focus on different calculation methods of the mean of grouped data and the representation of this data into specialised tables. The topic is important for understanding basic statistical problems and forming a solid base for more complex real-world scenarios. So, let’s get a deeper insight into this crucial topic of statistics through the solutions of exercise 13.1.

1.0Download NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1 : Free PDF

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1

2.0Introduction to the Mean

The mean, or average, is one of the most commonly used measures of central tendency in statistics. Central tendency refers to a value that represents the center of a dataset and is close to most values in the data. Mean is the typical or central value of a dataset. In simple words, the mean provides us with an idea of the "centre" of the data by taking into account all the values in the dataset and computing a balanced average out of this dataset.

Types of Dataset:

This dataset can be divided into two types, which include:

Ungrouped Data: Ungrouped data is raw data that is not grouped into any intervals or groups. It is a set of individual values or observations. Every data point is enumerated separately, and no class intervals are utilised to aggregate the data.
Grouped Data: The exercise is mainly based on the mean of this type of data. Grouped data is data grouped or arranged into intervals or classes known as class intervals. The data is classified, and the frequency or number of occurrences for each class is recorded.

3.0Class 10 Maths Chapter 13 Statistics : Key Concepts

Exercise 13.1 involves various formulas for calculating the mean of grouped or ungrouped data. Let’s explore these formulas to easily solve questions in the exercise:

Mean for Ungrouped Data:

For ungrouped data, the mean is obtained by adding up all the individual values and then dividing the total sum by the number of values. The formula for finding the mean in ungrouped data is:

$\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$

Here,

$\sum x_{i}$ is the sum of all observations of a dataset.
n is the total number of observations in the set.

Mean for Grouped Data:

For grouped data, the process is a little different and more complex than the ungrouped data. Since the data is classified into class intervals, the mean is found by determining the class marks (the middle of the class intervals) using different methods, including:

Direct Method:

This is the easiest and most used method for finding the mean of grouped data. In this, we simply multiply each class mark(x₁) by its corresponding frequency(f_i) and then sum up these products(f₁x_i). Divide this sum by the sum of all the frequencies to get the final answer. Mathematically, we can express this with the following formula:

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{1}}$

Here, x_i is the class mark of a certain class interval of the data set. This can be calculated as:

$x_{i} = \frac{U pp er L imi t + L o w er L imi t}{2}$

This method, though easy, can sometimes result in large calculations; this is why its alternative methods are used for such large datasets.

Assumed Mean Method:

This method is used when the class mark and frequencies are large, and the chances of the calculation getting complicated are high. The method simply includes assuming a mean denoted by a from the class mark, calculating the deviation d_i for each class mark, and then finally putting these values in the following formula:

$\overset{x}{ˉ} = a + \frac{\sum f _{i} d _{i}}{\sum f _{1}}$

Here,

a is the assumed mean.
d_i is the deviation for each class mark, calculated as d_i = x_i – a.

Step-Deviation Method:

To further shorten the calculation of the mean, we use the step-deviation method. The method is helpful when the deviation d_i has a common factor. The method starts by dividing each deviation by class height or width, calculating the step deviation and then putting all these values in the below-mentioned formula to get the final answer:

$\overset{x}{ˉ} = a + h \times \frac{\sum f _{i} u _{i}}{\sum f _{1}}$

Here,

u_i is the step deviation, such that $u_{i} = \frac{d _{i}}{h}$
h is the height of the class interval.

To understand better and work efficiently on the concepts mentioned above, go through the NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.1 for a thorough list of questions and solutions.

4.0NCERT Class 10 Maths Chapter 13 Statistics Exercise 13.1 : Detailed Solutions

1. 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	Number of houses
0−20−2	1
2−42−4	2
4−64−6	1
6−86−8	5
8−108−10	6
10−1210−12	2
12−1412−14	3

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

Sol.

Number of plants) Class	(Number of Houses Frequency) (fi)(fi)	Marks (xi)(xi)	fixifixi
0−20−2	1	1	1
2−42−4	2	3	6
4−64−6	1	5	5
6−86−8	5	7	35
8−108−10	6	9	54
10−1210−12	2	11	22
12−1412−14	3	13	39
Total	N=20N=20		162162

We have, N=Σfi=20 and Σfi=162.
Then mean of the data is
x= N1×ΣfiXi=201×162=8.1
Hence, the required mean of the data is 8.1 plants.
We find the mean of the data by direct method because the figures are small.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)	Number of workers
500−520500−520	12
520−540520−540	14
540−560540−560	8
560−580560−580	6
580−500580−500	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Sol.

Daily wages (In Rs.)	No. of workers (fi)(fi)	Class marks (xi)(xi)	fixifixi
500−520500−520	12	510	6120
520−540520−540	14	530	7420
540−560540−560	8	550	4400
560−580560−580	6	570	3420
580−600580−600	10	590	5900
Total	N=50N=50		2726027260

We have ∑fi=50 and ∑fi=27260
Mean =∑fi∑fixi=5027260
=545.2

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

Daily pocket Allowance (in Rs.)	Number of children
11−1311−13	7
13−1513−15	6
15−1715−17	9
17−1917−19	13
19−2119−21	ff
21−2321−23	5
23−2523−25	4

Sol.

We may prepare the table as given below :

Daily pocket allowance (in Rs.)	Number of children (fi)(fi)	Class marks (xi)(xi)	didi =xi−18=xi−18	fidifidi
11−1311−13	7	12	-6	-42
13−1513−15	6	14	-4	-24
15−1715−17	9	16	-2	-18
17−1917−19	13	18=a18=a	0	0
19−2119−21	f	20	2	2 f
21−2321−23	5	22	4	20
23−2523−25	4	24	6	24
	∑fi=44+f∑fi=44+f			2f−202f−20

It is given that mean =18.
From the table, we have
a=18, N=44+f and ∑fidi=2f−40
Now, mean =a+ N1×Σfidi
Then substituting the values as given above, we have
18=18+44+f1×(2f−40)
⇒0=44+f2f−40⇒f=20.

4. 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 heart beats per minute	Number of women
65−6865−68	2
68−7168−71	4
71−7471−74	3
74−7774−77	8
77−8077−80	7
80−8380−83	4
83−8683−86	2

Sol.

No. of heart beats per min	No. of (fi)(fi)	Class (xi)(xi)
65−6865−68	2	66.5	133
68−7168−71	4	69.5	278
71−7471−74	3	72.5	217.5
74−7774−77	8	75.5	604
77−8077−80	7	78.5	549.5
80−8380−83	4	81.5	326
83−8683−86	2	84.5	169
Total	N=30N=30		2277

Mean =∑fi∑fixi=302277=75.9.

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

No. of mangoes	Number of Boxes
50−5250−52	15
53−5553−55	110
56−5856−58	135
59−6159−61	115
62−5462−54	25

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

Sol.

Number of mangoes	Number of boxes fifi	Mark xixi	ui=3xi−57ui=3xi−57	fiuifiui
49.5−52.549.5−52.5	15	51	-2	-30
52.5−55.552.5−55.5	110	54	-1	-110
55.5−58.555.5−58.5	135	57	0	0
58.5−61.558.5−61.5	115	60	1	115
61.5−64.561.5−64.5	25	63	2	50
Total	N=40N=40			25

a=57, h=3, N=400 and Σfiui=25.
By step deviation method,
Mean =a+h× N1×Σfiui=57+3×4001×25 =57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs.)	No. of households
100−150100−150	4
150−200150−200	5
200−250200−250	12
250−300250−300	2
300−500300−500	2

Find the mean daily expenditure on food by a suitable method.

Sol.

Daily	No. of Exp. (in Rs.) holds (fi)(fi)	Class ( xixi )	fixifixi
100−150100−150	4	125	500
150−200150−200	5	175	875
200−250200−250	12	225	2700
250−300250−300	2	275	550
300−350300−350	2	325	650
Total	25		5275

Mean =∑fi∑fii=255275=211

8. 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 So2So2 (in ppm)	Frequency
0.00−0.040.00−0.04	4
0.04−0.080.04−0.08	9
0.08−0.120.08−0.12	9
0.12−0.160.12−0.16	2
0.16−0.200.16−0.20	4
0.20−0.240.20−0.24	2

Find the mean concentration of SO2 in the air.

Sol.

Concentration of So2So2 (in ppm)	xixi	Frequency	fixifixi
0.00−0.040.00−0.04	0.02	4	0.08
0.04−0.080.04−0.08	0.06	9	0.54
0.08−0.120.08−0.12	0.10	9	0.90
0.12−0.160.12−0.16	0.14	2	0.28
0.16−0.200.16−0.20	0.18	4	0.72
0.20−0.240.20−0.24	0.22	2	0.44
		Σfi=30Σfi=30	ΣfiXi=2.96ΣfiXi=2.96

Mean =∑fi∑fixi=302.96=0.0986

9. 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.

No. of days	No. of students
0−60−6	11
6−106−10	10
10−1410−14	7
14−2014−20	4
20−2820−28	4
28−3828−38	3
38−4038−40	1

Sol.

No. of days	No. of students (fi)(fi)	Class marks (xi)(xi)	fixifixi
0−60−6	11	3	33
6−106−10	10	8	80
10−1410−14	7	12	84
14−2014−20	4	17	68
20−2820−28	4	24	96
28−3828−38	3	33	99
38−4038−40	1	39	39
Total	4040		499499

Mean =∑fi∑fixi=40499=12.475

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

Literacy rate (in %)	No. of cities
45−5545−55	3
55−6555−65	10
65−7565−75	11
75−8575−85	8
85−9585−95	3

Sol.

Literac yy rate (in %)	No. of cities (fi)(fi)	Class marks (xi)(xi)	fixifixi
45−5545−55	3	50	150
55−6555−65	10	60	600
65−7565−75	11	70	770
75−8575−85	8	80	640
85−9585−95	3	90	270
Total	35		2430

Mean =∑fi∑fixi=352430=69.43

5.0Benefits of Studying NCERT Class 10 Maths Chapter 13 Exercise 13.1

Improves analytical thinking by teaching students how to compute measures like mean, median, and mode.
Helps students learn how to organize raw data into grouped and ungrouped frequency distributions.
Encourages logical reasoning while solving different types of data-based questions.
Strengthens the ability to work with data interpretation and probability-based questions.

NCERT Class 10 Maths Ch 13 Statistics Other Exercises:

Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

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NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1

1.0Download NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.1 : Free PDF

2.0Introduction to the Mean

Types of Dataset:

3.0Class 10 Maths Chapter 13 Statistics : Key Concepts

Mean for Ungrouped Data:

Mean for Grouped Data:

Direct Method:

Assumed Mean Method:

Step-Deviation Method:

4.0NCERT Class 10 Maths Chapter 13 Statistics Exercise 13.1 : Detailed Solutions

5.0Benefits of Studying NCERT Class 10 Maths Chapter 13 Exercise 13.1

Frequently Asked Questions

What does Exercise 13.1 of Class 10 Maths Chapter 13 cover?

Why is Exercise 13.1 important for students?

How does studying Statistics in Class 10 help in real life?

What are common mistakes to avoid in Exercise 13.1?

What types of questions are asked in Exercise 13.1?

Join ALLEN!