RD Sharma class 10 Solutions Chapter 7 Statistics Ex 7.5

RD Sharma Class 10 Solutions 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:

We see that 5 occurs in maximum times which is 5
∴ Mode = 5

we see that 3 occurs in maximum times i.e. 5
∴ Mode = 3

Here we see that 15 occurs in maximum times i.e. 54
∴ Mode = 15

Question 2.
The shirt sizes 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:

We see that frequency of 40 is maximum which is 41
∴Mode = 40

Question 3.
Find the mode of the following distribution.

Solution:

Question 4.
Compare the modal ages of two groups of students appearing for an entrance test :

Solution:
(i) For group A

We see that class 18-20 has the maximum frequency
∴ It is a modal class

(ii) For group B

We see that class 18-20 has the maximum frequency
∴ It is a modal class

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:

We see that class 50-60 has the maximum frequency 20
∴ It is a modal class

Question 6.
The following is the distribution of height of students of a certain class in a certain city :

Find the average height of maximum number of students.
Solution:
Writing the classes in exclusive form,

Here model class is 165.5 – 168.5
and l = 165.5, h = 3, f= 142, f₁= 118, f₂= 127

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:

(i) We see that class 35-45 has the maximum frequency 23
∴ It is a modal class

= 35.375 = 35.37 years
We see that mean is less than its mode

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:

We see that class 60-80 has the maximum frequency 61
∴ It is the modal class
Here l = 60, f= 6, f₁ = 52, f₂ = 38,h = 20
∴ Modal of life time (in hrs.)

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. (C.B.S.E. 2009)
Solution:

Here i = 20 and AM = 150

(ii) Max frequency = 14
∴ Model class = 120 – 140

Question 10.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

Solution:
Let assumed mean (A) = 32.5

(i) We see that the class 30-35 has the maximum frequency
∴ It is the modal class
Here l = 30,f = 10, f₁ = 9, f₂ = 3, h = 5

Question 11.
Find the mean, median and mode of the following data:

Solution:
Let assumed mean (A) =175

(i) Here N = 25, \(\frac { 5 }{ 3 }\) = \(\frac { 25 }{ 2 }\) = 12.5 or 13 which lies in the class 150-200
l= 150, F= 10, f= 6, h = 50

Question 12.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Solution:

We see that class 40-50 has the maximum frequency
∴ It is a modal class

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:
Let assumed mean (A) = 135

Here N = 68 , \(\frac { N }{ 2 }\) = \(\frac { 68 }{ 2 }\) = 34

Question 14.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the 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, And the modal size of the surnames.
Solution:
Let Assumed mean (A) = 8.5

(i) Here N = 100, \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50 which lies in the class 7-10
Here l = 7, F = 36, f= 40, h = 3

Question 15.
Find the mean, median and mode of the following data (C.B.S.E. 2008)

Solution:
Let assumed mean A = 70

(i) Here N = 50, then \(\frac { N }{ 2 }\) = \(\frac { 50 }{ 2 }\) = 25 which lies in the class 60-80
∴ l= 60, F= 24, f = 12 , h = 20

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:

(i) We see that the c ass 1500-2000 has maximum frequency 40
∴ It is a modal class
Here l = 1500, f = 40 , f₁ = 24 , f₂ = 33 , h =500

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:

We see that class 4000-5000 has the maximum frequency 18
∴It is a modal class

Question 18.
The frequency distribution table of agriculture holdings in a village is given below:

Find the modal agriculture holdings of the village.
Solution:
Here the maximum class frequency is 80,
and the class corresponding to this frequency is 5-7.
So, the modal class is 5-7.
l (lower limit of modal class) = 5
f₁ (frequency of the modal class) = 80
f₀ (frequency of the class preceding the modal class) = 45
f₂(frequency of the class succeeding the modal class) = 55
h (class size) = 2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

Question 19.
The monthly income of 100 families are given as below:

Solution:
In a given data, the highest frequency is 41, which lies in the interval 10000-15000.
Here, l = 10000,f₁ = 41, f₀ = 26,f₂ = 16 and h = 5000