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

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:

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)

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) 

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)

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:

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.

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:

Find the average height of maximum number of students.  

Solution:

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:

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 fi	Class marks xi	di = xi - 275	fidi
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:

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:

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

Solution:

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:

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 (fi)	xi	di = xi - 32.5	Ui	fiui
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.

Read More:

• What is Mean in Statistics (Formula, Calculation, Examples & Properties)
• Statistics in Maths

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.