RD Sharma Class 10 Ex 7.5 Solutions Chapter 7 Statistics

In this chapter, we provide RD Sharma Class 10 Ex 7.5 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 7.5 Solutions Chapter 7 Statistics pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 7
Chapter NameStatistics
Exercise7.5
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 7 Statistics Ex 7.5 Download PDF

Chapter 7: Statistics Exercise – 7.5


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

Mode = 5 because it occurs the maximum number of times.

(ii)

Value (x)3456789
Frequency (f)5242212

Mode = 3 because it occurs maximum number of times.

(iii)

Value (x)815181920242526
Frequency (f)14111211

Mode = 15 because it occurs 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:

Shirt size3738394041424344
Number of persons1525394136171512

Model shirt size = 40 because it occurs maximum number of times.

Question: 3

Find the mode of the following distribution. (i)

Class interval:0 – 10 10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency:5871228201010

(ii)

Class interval:10 – 1515 – 20 20 – 2525 – 3030 – 3535 – 40
Frequency:304575352515

(iii)

Class interval:25 – 3030 – 3535 – 4040 – 4545 – 5050 – 55
Frequency:253450422915

Solution:

(i)

Class interval0-1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency5871228201010

Here the maximum frequency is 28 then the corresponding class 40 – 50 is the modal class

l = 40, h = 50 40 = 10, f = 28, f1 = 12, f= 20

= 40 + 160/ 24

= 40 + 6.67

= 46.67

(ii)

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

Here the maximum frequency is 75, then the corresponding class 20 – 25 is the modal class

l = 20, h = 25 – 20 = 5, f = 75, f1 = 45, f= 35

= 20 + 150/70

= 20 + 2.14

= 22.14 

(iii)

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

Here the maximum frequency is 50 then the corresponding class 35 – 40 is the modal class

l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f= 42

= 35 + 80/24

= 35 + 3.33

= 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:

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

For Group A Here the maximum frequency is 78, then the corresponding class 18 – 20 is model class l = 18, h = 20 – 18 = 2, f = 78, f1 = 50, f= 46 

= 18 + 56/60

= 18 + 0.93

= 18.93 years

For group B Here the maximum frequency is 89, then the corresponding class 18 – 20 is the modal class l = 18, h = 20 – 18 = 2, f = 89, f1 = 54, f= 40 Mode 

= 18 + 70/84

= 18 + 0.83

= 18.83 years

Hence the modal age for the Group A is higher than that for 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:

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

Here the maximum frequency is 20, then the corresponding class 50 – 60 is the modal class l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f= 5 

= 50 + 70/22

= 50 + 3.18

= 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 – 171172 – 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

Here the maximum frequency is 142, then the corresponding class 165.5 – 168.5 is the modal class l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127 

= 165.5 + 72/39

= 165.5 + 1.85

= 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:

We may compute class marks (xi) as per the relation

Now taking 30 as assumed mean (a) we may calculate dand fidas follows:

Age (in years)Number of patients fiClass marks xid= xi – 275fidi
5 – 15610– 20-120
15 – 251120– 10-110
25 – 35213000
35 – 45234010230
45 – 55145020280
55 – 6556030150
Total80  430

From the table we may observe that Σfi = 80, Σfdi = 430, 

= 30 + 430/80

= 30 + 5.375

= 35.375

= 35.38

Clearly, mean of this data is 35.38. It represents that on an average the age of a patients admitted to hospital was 35.38 years.

As we may observe that maximum class frequency is 23 belonging to class interval 35 – 45

So, modal class = 35 – 45 Lower limit (l) of modal class = 35 Frequency (f) of modal class = 23

Class size (h) = 10 Frequency (f1) of class preceding the modal class = 21 Frequency (f2) of class succeeding the modal class = 14 Mode

Clearly mode is 36.8. It represents that maximum number of patients admitted in hospital were of 36.8 years. 

Question: 8

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

Lifetimes (in hours):0 – 2020 – 40 40 – 6060 – 8080 – 100100 – 120
No of components:103552613829

Determine the modal lifetimes of the components.

Solution:

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 – 80

So, modal class limit (l) of modal class = 60 Frequency (f) of modal class = 61 Frequency (f1) of class preceding the modal class = 52

Frequency (f2) of class succeeding the modal class = 38 Class size (h) = 20 Mode 

So, modal lifetime of electrical components is 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 = 50Σfx = 7260 

We have, N = 50 Then, N/2 = 50/2 = 25

The cumulative frequency just greater than N/2 is 26, then the median class is 120 – 140 such that l = 120, h = 140 – 120 = 20, f = 14, F = 12 

= 120 + 260/14

= 120 + 18.57

= 138.57

Here the maximum frequency is 14, then the corresponding class 120 – 140 is the modal class l = 120, h = 140 – 120 = 20, f = 14, f= 12, f= 8 

= 120 + 5

= 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 teacherNumber of states/ U.TNumber of students per teacherNumber of states/ U.T
15 – 20335 – 403
20 – 25840 – 450
25 – 30945 – 500
30 – 351050 – 552

Solution:

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 – 35.

So, modal class = 30 – 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f) of modal class = 10 Frequency (f1) of class preceding modal class = 9

Frequency (f2) of class succeeding modal class = 3

Mode = 30.6 It represents that most of states/ U.T have a teacher- student ratio as 30.6 Now we may find class marks by using the relation Class mark = (upper class limit + lower class limit) /2.

Now taking 32.5 as assumed mean (a) we may calculate di, ui, and fiuas following

Number of students per teacherNumber of states/ U.T (fi)xid= xi – 32.5Uifiui
15 – 20317.5-15– 3– 9
20 – 25822.5-10– 2– 16
25 – 30927.5-5– 1– 9
30 – 351032.5000
35 – 40337.5513
40 – 45042.51020
45 – 50047.51530
50 – 55252.52048
Total35   -23

Now,

So mean of data is 29.2 It represents that on an average teacher-student ratio was 29.2 

Question: 11

Find the mean, median and mode of the following data:

Classes:0 – 5050 – 100100 – 150150 – 200200 – 250250 – 300300 – 350
Frequency:2356531

Solution:

Class intervalMid value (x)Frequency (f)fxCumulative frequency
0 – 50352502
50 – 1007532255
100 – 150125562510
150 – 2001756105016
200 – 2502255112721
250 – 300275382524
300 – 350325132525
  N = 25Σfx = 4225 

We have, N = 25

Then, N/2

= 25/2

= 12.5

The cumulative frequency just greater than N/2 is 16, then the median class is 150 – 200 such that l = 150, h = 200 – 150 = 50, f = 6, F = 10

= 150 + 125/6

= 150 + 20.83

= 170.83

Here the maximum frequency is 6, then the corresponding class 150 – 200 is the modal class l = 150, h = 200 – 150 = 50, f = 6, f= 5, f= 5

= 150 + 50/2

= 150 + 25

= 175 

Question: 12

A students noted the number of cars pass through a spot on a road for 100 periods each of 3 minute and summarized it in the table given below. Find the mode of the data. 

Solution:

From the given data we may observe that maximum class frequency is 20 belonging to 40 – 50 class intervals.

So, modal class = 40 – 50 Lower limit (l) of modal class = 40 Frequency (f) of modal class = 20 Frequency (f1) of class preceding modal class = 12

Frequency (f2) of class succeeding modal class = 11 Class size = 10

= 40 + 80/17

= 40 + 4.7

= 44.7

So mode of this data is 44.7 cars 

Question: 13

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.

Monthly consumption:65 – 8585 – 105105  – 125125 – 145145 – 165165 – 185185 – 205
No of consumers:4513201484

Solution:

Class intervalMid value xiFrequency fifxCumulative frequency
65 – 857543004
85 – 1059554759
105 – 12511513149522
125 – 14513520270042
145 – 16515514217056
165 – 1851758140064
185 – 205195478068
  N = 68 Σfx = 9320

We have, N = 68 N/2 = 68/2 = 34

The cumulative frequency just greater than N/2 is 42 then the median class is 125 – 145 such that l = 125, h = 145 – 125 = 20, f = 20, F = 22 

= 125 + 12

= 137

Here the maximum frequency is 20, then the corresponding class 125 – 145 is the modal class l = 125, h = 145 – 125 = 20, f = 20, f1 = 13, f= 14

 = 125 + 140/13

= 135.77

Question: 14

100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letter English alphabets in the surnames was obtained as follows:

Number of letters:1 – 44 – 77 – 1010 – 1313 – 1616  – 19
Number surnames:630401644

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. 

Solution:

Class intervalMid value (x)Frequency (f)fxCumulative frequency
1 – 42.56156
4 – 75.53016536
7 – 108.54034076
10 – 1311.51618492
13 – 1614.545896
16 – 1917.5470100
  N = 100Σfx = 832 

We have, N = 100 N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 76, then the median class is 7 – 10 such that l = 7, h = 10 – 7 = 3, f = 40, F = 36

= 7 + 1.05

= 8.05

Here the maximum frequency is 40, then the corresponding class 7 – 10 is the modal class l = 7, h = 10 – 7 = 3, f = 40, f= 30, f= 16

= 7 + 30/34

= 7 + 0.88

= 7.88 

Question: 15

Find the mean, median and mode of the following data:

Class0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140
Frequency681012653

Solution:

Class intervalMid value (x)Frequency (f)fxCumulative frequency
0 – 20106606
20 – 4030824017
40 – 60501050024
60 – 80701284036
80 – 10090654042
100 – 120110555047
120 – 140130339050
  N = 50Σfx = 3120 

We have, N = 50 Then, N/2 = 50/2 = 25

The cumulative frequency just greater than N/2 is 36, then the median class is 60 – 80 such that l = 60, h = 80 – 60 = 20, f = 12, F = 24

= 60 + 20/12

= 60 + 1.67

= 61.67 

Here the maximum frequency is 12, then the corresponding class 60 – 80 is the modal class l = 60, h = 80 – 60 = 20, f = 12, f= 10, f= 6 

Question: 16

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:

Expenditure      FrequencyExpenditureFrequency
1000 – 1500243000 – 350030
1500 – 2000403500 – 400022
2000 – 2500334000 – 450016
2500 – 3000284500 – 50007

Solution:

We may observe from the given data that maximum class frequency is 40 belonging to 1500 -200 intervals. So, modal class = 1500 -2000

Lower limit (l) of modal class = 1500 Frequency (f) of modal class = 40 Frequency (f1) of class preceding modal class = 24

Frequency (f2) of class succeeding modal class = 33 Class size (h) = 500 

= 1500 + 347.826

= 1847.826

1847.83

So modal monthly expenditure was Rs. 1847.83

Now we may find class mark as Class mark 

Class size (h) of given data = 500 Now taking 2750 as assumed mean (a) we may calculate duas follows:

Expenditure (in Rs)Number of families fixid= xi – 2750Uifiui
1000 – 1500241250– 1500– 3– 72
1500 – 2000401750– 1000– 2– 80
2000 – 2500332250– 500– 1– 33
2500 – 3000282750000
3000 – 3500303250500130
3500 – 40002237501000244
4000 – 45001642501500348
4500 – 5000747502000428
Total200   -35

Now from table may observe that Σfi = 200, Σfidi = -35

So mean monthly expenditure was Rs. 2662.5 

Question: 17

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

Runs scoredNo of batsmenRuns scored No of batsmen
3000 – 400047000 – 80006
4000 – 5000188000 – 90003
5000 – 600099000 – 100001
6000 – 7000710000 – 110001

Find the mode of the data. 

Solution:

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 – 5000.

So, modal class = 4000 – 5000 Lower limit (l) of modal class = 4000

Frequency (f) of modal class = 18. Frequency (f1) of class preceding modal class = 4 Frequency (f2) of class succeeding modal class = 9 Class size (h) = 1000. Now

= 4000 + 608.695

= 4608.695

So mode of given data is 4608.7 runs 

All Chapter RD Sharma Solutions For Class10 Maths

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