Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1

Last Updated : 05 Sep, 2024

Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10 textbook focuses on the fundamental statistical concepts and their applications. Exercise 7.5 | Set 1 contains problems designed to reinforce students' understanding of these concepts.

Statistics

Statistics involves the collection, organization, and analysis of the data to make informed decisions. It encompasses various methods to summarize and interpret numerical information. Key concepts include mean, median, mode, variance, and standard deviation. Mastery of statistics enables individuals to understand data trends make predictions and support decision-making processes based on empirical evidence.

Question 1. Find the mode of the following data:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Solution:

(i)
Value (x) 3 4 5 6 7 8 9
Frequency (f) 4 2 5 2 2 1 2
Therefore, mode = 5 because 5 occurs the maximum number of times.
(ii)
Value (x) 3 4 5 6 7 8 9
Frequency (f) 5 2 4 2 2 1 2
Therefore, mode = 3 because 3 occurs the maximum number of times.
(iii)
Value (x) 8 15 18 19 20 24 25
Frequency (f) 1 4 1 1 1 2 1
Therefore, mode = 15 because 15 occurs the maximum number of times.

Question 2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size:	37	38	39	40	41	42	43	44
Number of persons:	15	25	39	41	36	17	15	12

Find the model shirt size worn by the group.

Solution:

From the data present in the table we conclude that
Model shirt size = 40
Because shirt size 40 occurred for the maximum number of times.

Question 3. Find the mode of the following distribution.

(i)

Class interval:	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency:	5	8	7	12	28	20	10	10

Solution:

From the given table we conclude that
The maximum frequency = 28
So, the model class = 40 – 50
and,
l = 40, h = 50 40 = 10, f = 28, f1 = 12, f2 = 20
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =40+\frac{28-12}{2\times28-12-20}\times10
= 40 + 160/ 24
= 40 + 6.67
= 46.67
Hence, the mode = 46.67

(ii)

Class interval	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
Frequency	30	45	75	35	25	15

Solution:

From the given table we conclude that
The maximum frequency = 75
So, the modal class = 20 – 25
And,
l = 20, h = 25 – 20 = 5, f = 75, f₁ = 45, f₂ = 35
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =20+\frac{75-45}{2\times75-45-35}\times5
= 20 + 150/70
= 20 + 2.14
= 22.14
Hence, the mode = 22.14

(iii)

Class interval	25 - 30	30 - 35	35 - 40	40 - 45	45 - 50	50 - 55
Frequency	25	34	50	42	38	14

Solution:

From the given table we conclude that
The maximum frequency = 50
So, the modal class = 35 – 40
And,
l = 35, h = 40 – 35 = 5, f = 50, f₁ = 34, f₂ = 42
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =35+\frac{50-34}{2\times50-34-42}\times5
= 35 + 80/24
= 35 + 3.33
= 38.33
Hence, the mode = 38.33

Question 4. Compare the modal ages of two groups of students appearing for an entrance test:

Age in years	16 - 18	18 - 20	20 - 22	22 - 24	24 - 26
Group A	50	78	46	28	23
Group B	54	89	40	25	17

Solution:

For Group A:
From the given table we conclude that
The maximum frequency = 78.
So, the model class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 78, f₁ = 50, f₂ = 46
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =18+\frac{78-50}{2\times78-50-46}\times2
= 18 + 56/60
= 18 + 0.93
= 18.93 years
For Group B:
From the given table we conclude that
The maximum frequency = 89
The modal class = 18 – 20
And,
l = 18, h = 20 – 18 = 2, f = 89, f₁ = 54, f₂ = 40
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =18+\frac{89-54}{2\times89-54-40}\times2
= 18 + 70/84
= 18 + 0.83
= 18.83 years
After finding the mode of both A and B group we conclude that
the modal age of the Group A is greater than Group B.

Question 5. The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
Frequency	3	5	16	12	13	20	5	4	1	1

Solution:

From the given table we conclude that
The maximum frequency = 20
The modal class = 50 – 60
And,
l = 50, h = 60 – 50 = 10, f = 20, f₁ = 13, f₂ = 5
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =50+\frac{20-13}{2\times20-13-5}\times10
= 50 + 70/22
= 50 + 3.18
= 53.18
Hence, the mode = 53.18

Question 6. The following is the distribution of height of students of a certain class in a city:

Height (in cm)	160 - 162	163 - 165	166 - 168	169 - 171	172 - 174
No. of students:	15	118	142	127	18

Find the average height of maximum number of students.

Solution:

Heights (exclusive)	160 - 162	163 - 165	166 - 168	169 - 171	172 - 174
Heights (inclusive)	159.5 - 162.5	162.5 - 165.5	165.5 - 168.5	168.5 - 171.5	171.5 - 174.5
No of students	15	118	142	127	18

From the given table we conclude that
The maximum frequency = 142
The modal class = 165.5 – 168.5
And,
l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f₁ = 118, f₂ = 127
Using the formula of mode
Mode = l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =165.5+\frac{142-118}{2\times142-118-127}\times3
= 165.5 + 72/39
= 165.5 + 1.85
= 167.35 cm
Hence, the average height of maximum number of students = 167.35 cm

Question 7. The following table shows the ages of the patients admitted in a hospital during a year:

Ages (in years):	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
No of students:	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.

Solution:

For mean:
Let us considered mean (A) = 30
Age (in years) Number of patients f_i Class marks x_i d_i = x_i - 275 f_id_i
5 - 15 6 10 -20 -120
15 - 25 11 20 -10 -110
25 - 35 21 30 0 0
35 - 45 23 40 10 230
45 - 55 14 50 20 280
55 - 65 5 60 30 150
N = 80 \sum f_id_i=430
From the table we get
Σf_i = N = 80 and Σf_i d_i = 430.
Using the formula of mean
Mean\ (\overline{x})=A+\frac{\sum f_id_i}{\sum f_i}
= 30 + 430/80
= 30 + 5.375
= 35.375
= 35.38
Therefore, the mean = 35.38. It represents the average age of the patients = 35.38 years.
For mode:
From the given table we conclude that
The maximum class frequency = 23
So, modal class = 35 – 45
and
l = 35, f = 23, h = 10, f₁= 21, f₂= 14
Using the formula of mode
Mode =l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =35+\frac{23-21}{2\times23-21-14}\times10\\ =35+\frac{2}{46-35}\times10
= 35 + 1.81 = 36.8
Hence, the mode = 36.8. It represents the maximum number of patients admitted in hospital of age 36.8 years.
Therefore, mode is greater than mean

Question 8. 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
No. of components:	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution:

From the given table we conclude that
The maximum class frequency = 61
So, modal class = 60 – 80
and
l = 60, f = 61, h = 20, f₁= 52, f₂= 38
Using the formula of mode
Mode =l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =60+\frac{61-52}{2\times61-52-38}\times20\\ =60+\frac{9}{112-90}\times20\\ =60+\frac{9\times20}{32}\\ =60+\frac{90}{16}
= 60 + 5.625 = 65.625
Hence, the modal lifetime of electrical components = 65.625 hours

Question 9. The following table gives the daily income of 50 workers of a factory:

Daily income	100 - 120	120 - 140	140 - 160	160 - 180	180 - 200
Number of workers	12	14	8	6	10

Find the mean, mode, and median of the above data.

Solution:

Class interval	Mid value (x)	Frequency (f)	fx	Cumulative Frequency
100 - 120	110	12	1320	12
120 - 140	130	14	1820	26
140 - 160	150	8	1200	34
160 - 180	170	6	1000	40
180 - 200	190	10	1900	50
		N = 50	\sum fx=7260

Finding Mean:
From the table we get
N = 50, fx = 7260
So using mean formula, we get
Mean = Σfx / N
= 7260/ 50
= 145.2
Hence, the mean = 145.2
Finding Median:
N/2 = 50/2 = 25
So, the cumulative frequency just greater than N/2 = 26,
The median class = 120 – 140
Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12
By using the formula of median we get
Median = l+\frac{\frac{N}{2}-F}{f}\times h\\ =120+\frac{25-12}{14}\times20
= 120 + 260/14
= 120 + 18.57
= 138.57
Hence, the median = 138.57
Finding Mode:
From the table we get
The maximum frequency = 14,
So the modal class = 120 – 140
And,
l = 120, h = 140 – 120 = 20, f = 14, f₁ = 12, f₂ = 8
By using the formula of mode we get
Mode = =l+\frac{f-f_1}{2f-f_1-f_2}\times h\\ =120+\frac{14-12}{2\times14-12-8\times20}\\ =120+\frac{40}{8}
= 120 + 5
= 125
Hence, the mode = 125

Question 10. The following distribution gives the state-wise teachers-students 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	3
20 - 25	8
25 - 30	9
30 - 35	10
35 - 40	3
40 - 45	0
45 - 50	0
50 - 55	2

Solution:

From the given table we conclude that
The maximum class frequency = 10
So, modal class = 30 - 35
and
l = 30, h = 5, f = 10, f₁ = 9, f₂ = 3
By using the formula of mode we get
Mode = l + f - f₁ 2f - f₁ - f₂ × hl + =\frac{f-f_1}{2f-f_1-f_2}\times h\\
= 30 + 120 - 12 × 530 + \frac{10-9}{2\times10-9-3}\times5
= 30 + 120 - 12 × 530 + \frac{1}{20-12}\times5
= 30 + 5/8
= 30.625
Hence, the mode = 30.6 and it represents that most of states/ U.T have a teacher-students ratio = 30.6
Now we are going to find class marks using the following formula
Class mark = \frac{upperclasslimit+lowerclasslimit}{2}
Let us considered mean(a) = 32.5, and now we are going to find d_i, u_i, and f_iu_i as following
Number of students per teacher Number of states/ U.T (f_i) x_i d_i = x_i - 32.5 U_i f_iu_i
15 - 20 3 17.5 -15 -3 -9
20 - 25 8 22.5 -10 -2 -16
25 - 30 9 27.5 -5 -1 -9
30 - 35 10 32.5 0 0 0
35 - 40 3 37.5 5 1 3
40 - 45 0 42.5 10 2 0
45 - 50 0 47.5 10 2 0
50 - 55 2 52.5 20 4 8
Total 35 -23
Using the mean formula, we get
Mean(\overline{x})=a+\sum f_id_i\sum f_i\times h\overline{x}=a+\frac{\sum f_id_i}{\sum f_i}\times h\\ =32.5+-2335\times532.5+\frac{-23}{35}\times5
= 32.5 - 23/7
= 32.5 - 3.28
= 29.2
Hence, the mean = 29.2 and it represents that on an average teacher-student ratio = 29.2.

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Conclusion

Understanding statistical concepts like mean, median and mode is crucial for the analyzing and interpreting data effectively. Exercise 7.5 | Set 1 from RD Sharma's Class 10 book provides the practice with these fundamental concepts reinforcing students' ability to the handle various data analysis problems.

Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2

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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1

Statistics

Question 1. Find the mode of the following data:

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Question 2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Find the model shirt size worn by the group.

Question 3. Find the mode of the following distribution.

(i)

(ii)

(iii)

Question 4. Compare the modal ages of two groups of students appearing for an entrance test:

Question 5. The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.

Question 6. The following is the distribution of height of students of a certain class in a city:

Find the average height of maximum number of students.

Question 7. 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.

Question 8. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Question 9. The following table gives the daily income of 50 workers of a factory:

Find the mean, mode, and median of the above data.

Question 10. The following distribution gives the state-wise teachers-students ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures:

Conclusion

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