# 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.

# 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)

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

(ii)

Mode = 3 because it occurs maximum number of times.

(iii)

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:

Find the model shirt size worn by the group.

### Solution:

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

### Question: 3

Find the mode of the following distribution. (i)

### Solution:

(i)

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)

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)

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:

### Solution:

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.

### Solution:

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:

Find the average height of maximum number of students.

### Solution:

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:

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:

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:

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:

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

### Solution:

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:

### 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

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:

### 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.

### Solution:

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:

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:

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:

### Solution:

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:

### 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:

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.

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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