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

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 = 

= 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 = 

= 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 = 

= 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 = 

= 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 = 

= 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 = 

= 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 = 

= 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 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

From the table we get

Σf_i = N = 80 and Σf_i d_i = 430.

Using the formula of mean

= 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 

= 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 

= 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

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 = 

= 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 = 

= 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 + 

= 30 + 120 – 12 × 530 + 

= 30 + 120 – 12 × 530 + 

= 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 = 

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

= 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.