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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.5 | Set 1
  • Last Updated : 05 Mar, 2021

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)3456789
Frequency (f) 4252212

Therefore, mode = 5 because 5 occurs the maximum number of times.

(ii)

Value (x) 3456789
Frequency (f) 5242212

Therefore, mode = 3 because 3 occurs the maximum number of times.



(iii)

Value (x) 8151819202425
Frequency (f) 1411121

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: 3738394041424344
Number of persons: 1525394136171512

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 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency:5871228201010

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 interval10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40
Frequency304575352515

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 interval25 – 3030 – 3535 – 4040 – 4545 – 5050 – 55
Frequency253450423814

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 years16 – 1818 – 2020 – 2222 – 2424 – 26
Group A5078462823
Group B5489402517

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.

Marks0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100
Frequency35161213205411

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 – 162163 – 165166 – 168169 – 171 172 – 174
No. of students:1511814212718

Find the average height of maximum number of students.  

Solution:

Heights (exclusive)160 – 162163 – 165166 – 168169 – 171172 – 174
Heights (inclusive)159.5 – 162.5162.5 – 165.5165.5 – 168.5168.5 – 171.5171.5 – 174.5
No of students1511814212718

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 – 1515 – 2525 – 3535 – 4545 – 5555 – 65
No of students:6112123145

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 fiClass marks xidi = xi – 275fidi
5 – 15 610-20-120
15 – 2511 20-10-110
25 – 35 213000
35 – 45234010230
45 – 55145020280
55 – 6556030150
  N = 80   \sum f_id_i=430

From the table we get

Σfi = N = 80 and Σfi di = 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 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
No. of components:103552613829

Determine the modal lifetimes of the components.

Solution:

From the given table we conclude that

The maximum class frequency = 61 

So, modal calss = 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 income100 – 120120 – 140140 – 160160 – 180180 – 200
Number of workers12148610

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

Solution:

Class intervalMid value (x)Frequency (f)fxCumulative Frequency
100 – 12011012132012
120 – 14013014182026
140 – 1601508120034
160 – 1801706100040
180 – 20019010190050
  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 di, ui, and fiui as following

Number of students per teacherNumber of states/ U.T (fi)xidi = xi – 32.5Uifiui
15 – 20 317.5-15-3-9
20 – 25822.5-10-2-16
25 – 30927.5-5-1-9
30 – 351032.5000
35 – 40337.5513
40 – 45042.51020
45 – 50047.51020
50 – 55252.52048
Total35   -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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