Statistics

NCERT Solutions for Class 10 Maths Chapter 13

Chapter 13 of NCERT Maths Class 10 delves into the concepts of statistics, focusing on the collection, organization, analysis, and interpretation of data.

Key Concepts:

• Data: A collection of information, either numerical or non-numerical.
• Frequency Distribution: A table or graph that shows how often different values occur in a dataset.
• Mean: The average value of a dataset.
• Mode: The value that appears most frequently in a dataset. 
• Histogram: A bar graph where the bars are adjacent to each other and the width of each bar represents the class interval.
• Frequency Polygon: A line graph obtained by joining the midpoints of the tops of the bars in a histogram.
• Cumulative Frequency Curve (Ogive): A graph that shows the cumulative frequency of a dataset.

Applications:

• Analyzing and interpreting data from various sources, such as surveys, experiments, and real-world situations.
• Making informed decisions based on statistical analysis.
• Understanding the distribution and patterns within datasets.

In essence, the chapter on statistics provides a comprehensive understanding of data analysis techniques, enabling students to make sense of information and draw meaningful conclusions.

NCERT Solutions for Class 10 Maths Chapter 13

Exercise 13.1

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.

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

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 7)	f * d
0-2	1	1	-6	-6
2-4	3	2	-4	-8
4-6	5	1	-2	-2
6-8	7	5	0	0
8-10	9	6	2	12
10-12	11	2	4	8
12-14	13	3	6	18

• Σ(f * d) = -6 – 8 – 2 + 0 + 12 + 8 + 18 = 22
  
• Σf = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20
  
• Mean = 7 + (22/20) = 7 + 1.1 = 8.1

The assumed mean method was used because the class marks were relatively small and the intervals were of equal width, making the calculations simpler.

2. 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.

Ans : 

NCERT Solutions for Class 10 Maths Chapter 13

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

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

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 75.5)	f * d
65-68	66.5	2	-9	-18
68-71	69.5	4	-6	-24
71-74	72.5	3	-3	-9
74-77	75.5	8	0	0
77-80	78.5	7	3	21
80-83	81.5	4	6	24
83-86	84.5	2	9	18

• Σ(f * d) = -18 – 24 – 9 + 0 + 21 + 24 + 18 = 12
• Σf = 2 + 4 + 3 + 8 + 7 + 4 + 2 = 30
• Mean = 75.5 + (12/30) = 75.5 + 0.4 = 75.9

5. 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.

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

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 57.5)	f * d
50-52	51	15	-6.5	-97.5
53-55	54	110	-3.5	-385
56-58	57.5	135	0	0
59-61	60.5	115	3	345
62-64	63.5	25	6	150

• Σ(f * d) = -97.5 – 385 + 0 + 345 + 150 = -57.5
• Σf = 15 + 110 + 135 + 115 + 25 = 400
• Mean = 57.5 + (-57.5/400) = 57.5 – 0.14375 ≈ 57.36

Therefore, the mean number of mangoes kept in a packing box is approximately 57.36.

NCERT Solutions for Class 10 Maths Chapter 13

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

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

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 225)	f * d
100-150	125	4	-100	-400
150-200	175	5	-50	-250
200-250	225	12	0	0
250-300	275	2	50	100
300-350	325	2	100	200

• Σ(f * d) = -400 – 250 + 0 + 100 + 200 = -350
  
• Σf = 4 + 5 + 12 + 2 + 2 = 25
  
• Mean = 225 + (-350/25) = 225 – 14 = 211
  Therefore, the mean daily expenditure on food is ₹211.

7. 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:

Find the mean concentration of SO2 in the air.

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 0.08)	f * d
0.00-0.04	0.02	4	-0.06	-0.24
0.04-0.08	0.06	9	-0.02	-0.18
0.08-0.12	0.1	9	0.02	0.18
0.12-0.16	0.14	2	0.06	0.12
0.16-0.20	0.18	4	0.1	0.4
0.20-0.24	0.22	2	0.14	0.28

• Σ(f * d) = -0.24 – 0.18 + 0.18 + 0.12 + 0.40 + 0.28 = 0.26
• Σf = 4 + 9 + 9 + 2 + 4 + 2 = 30
• Mean = 0.08 + (0.26/30) ≈ 0.0887

Therefore, the mean concentration of SO₂ in the air is approximately 0.0887 ppm.

8. 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.

Ans : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 13)	f * d
0-6	3	11	-10	-110
6-10	8	10	-5	-50
10-14	12	7	-1	-7
14-20	17	4	4	16
20-28	24	4	11	44
28-38	33	3	20	60
38-40	39	1	26	26

• Σ(f * d) = -110 – 50 – 7 + 16 + 44 + 60 + 26 = -21
• Σf = 11 + 10 + 7 + 4 + 4 + 3 + 1 = 40
• Mean = 13 + (-21/40) ≈ 12.48

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

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

Ans :

NCERT Solutions for Class 10 Maths Chapter 13

Exercise 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year.

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Ans : 

NCERT Solutions for Class 10 Maths Chapter 13

3. 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:

Ans : 

Mode = L + (f1 – f0) / (2f1 – f0 – f2) * h

Where:

• L is the lower limit of the modal class (1500)
• f1 is the frequency of the modal class (40)
• f0 is the frequency of the class preceding the modal class (24)
• f2 is the frequency of the class succeeding the modal class (33)
• h is the class size (500)

Mode = 1500 + (40 – 24) / (2 * 40 – 24 – 33) * 500

Mode ≈ 1500 + 16 / 23 * 500

Mode ≈ 1847.83

Mean : 

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 2750)	u = d/h	f * u
1000-1500	1250	24	-1500	-3	-72
1500-2000	1750	40	-1000	-2	-80
2000-2500	2250	33	-500	-1	-33
2500-3000	2750	28	0	0	0
3000-3500	3250	30	500	1	30
3500-4000	3750	22	1000	2	44
4000-4500	4250	16	1500	3	48
4500-5000	4750	7	2000	4	28

• Σ(f * u) = -72 – 80 – 33 + 0 + 30 + 44 + 28 = -43
• Σf = 24 + 40 + 33 + 28 + 30 + 22 + 16 + 7 = 200
• Mean = 2750 + (-43/200) * 500 ≈ 2662.5

Therefore, the modal monthly expenditure of the families is approximately ₹1847.83, and the mean monthly expenditure is approximately ₹2662.50.

4. 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.

Ans : 

• Mode: To find the exact modal value within this class, we can use the formula:
  • Mode = L + (f1 – f0) / (2f1 – f0 – f2) * h
  • Where:
    • L is the lower limit of the modal class (30)
    • f1 is the frequency of the modal class (10)
    • f0 is the frequency of the class preceding the modal class (9)
    • f2 is the frequency of the class succeeding the modal class (3)
    • h is the class size (5)
  • Mode = 30 + (10 – 9) / (2 * 10 – 9 – 3) * 5
  • Mode = 30 + 1 / 8 * 5
  • Mode ≈ 30.625

Finding the Mean:

Class	Class Marks (x)	Frequency (f)	Deviation (d = x – 27.5)	f * d
15-20	17.5	3	-10	-30
20-25	22.5	8	-5	-40
25-30	27.5	9	0	0
30-35	32.5	10	5	50
35-40	37.5	3	10	30
40-45	42.5	0	15	0
45-50	47.5	0	20	0
50-55	52.5	2	25	50

Σ(f * d) = -30 – 40 + 0 + 50 + 30 + 0 + 0 + 50 = 60

Σf = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35

Mean = 27.5 + (60/35) ≈ 29.21

NCERT Solutions for Class 10 Maths Chapter 13

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

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6. 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:

Ans :

NCERT Solutions for Class 10 Maths Chapter 13

Exercise 13.3

1. 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.

Ans :

2. If the median of the distribution given below is 28.5, find the values of x and y.

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NCERT Solutions for Class 10 Maths Chapter 13

3. 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 years.

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4. The lengths of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.

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5. The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.

Ans : 

NCERT Solutions for Class 10 Maths Chapter 13

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

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.

Ans : 

Median = L + (N/2 – cf) / f * h

Substituting the values:

Median = 7 + (100/2 – 36) / 40 * 3 

Median = 7 + (50 – 36) / 40 * 3 

Median = 7 + 14/40 * 3 Median ≈ 8.05

Number of letters	Midpoint (x)	Frequency (f)	Deviation (d = x – 8.5)	f * d
1-4	2.5	6	-6	-36
4-7	5.5	30	-3	-90
7-10	8.5	40	0	0
10-13	11.5	16	3	48
13-16	14.5	4	6	24
16-19	17.5	4	9	36

Σ(f * d) = -36 – 90 + 0 + 48 + 24 + 36 

= -48 Σf = 100

Mean = Assumed mean + (Σ(f * d)) / Σf 

Mean = 8.5 + (-48/100)

 Mean ≈ 8.32

Mode:

The modal class is the class with the highest frequency, which is 7-10.

To find the exact modal value, we can use the formula:

Mode = L + (f1 – f0) / (2f1 – f0 – f2) * h

Where:

• L is the lower limit of the modal class (7)
• f1 is the frequency of the modal class (40)
• f0 is the frequency of the class preceding the modal class (30)
• f2 is the frequency of the class succeeding the modal class (16)
• h is the class size (3)

Substituting the values:

Mode = 7 + (40 – 30) / (2 * 40 – 30 – 16) * 3

 Mode = 7 + 10 / 34 * 3 

Mode ≈ 7.88

7. The distribution below gives the weight of 30 students of a class. Find the median weight of the students.

Ans : 

NCERT Solutions for Class 10 Maths Chapter 13

FAQs

What is Chapter 13 of Class 10 Maths about?

Chapter 13, Surface Areas and Volumes, focuses on finding surface areas and volumes of different 3D shapes like cones, cylinders, spheres, and combinations of solids.

How do NCERT Solutions help in Chapter 13?

They provide step-by-step answers and methods that make solving surface area and volume problems easier to understand and remember.

What are the important topics in Chapter 13?

Key topics include surface area and volume of solids, converting one solid into another, and problems involving daily-life applications.

Is Chapter 13 important for board exams?

Yes, Surface Areas and Volumes is an important chapter that often has numerical questions in CBSE Class 10 board exams.

Where can I get free NCERT Solutions for Class 10 Maths Chapter 13?

You can access free and detailed NCERT Solutions for Class 10 Maths Chapter 13 on trusted educational websites and learning platforms.