RS Aggarwal Solutions: Exercise 14C - Statistics Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RS Aggarwal Solutions: Exercise 14C - Statistics Class 10 Notes | EduRev

 Page 1


1. Find the mode of the following distribution:
Marks 10 20 20 30 30 40 40 50 50 60
Frequency 12 35 45 25 13
Sol:
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 
30 40. So, the modal class is 30- 40.
Now,
Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = l + × h
= 30 + × 10
= 30 + × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100
Frequency 25 16 28 20 5
Sol: 
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 
40 60. So, the modal class is 40 60.
Page 2


1. Find the mode of the following distribution:
Marks 10 20 20 30 30 40 40 50 50 60
Frequency 12 35 45 25 13
Sol:
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 
30 40. So, the modal class is 30- 40.
Now,
Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = l + × h
= 30 + × 10
= 30 + × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100
Frequency 25 16 28 20 5
Sol: 
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 
40 60. So, the modal class is 40 60.
 
Now, 
Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20,
frequency (f1) of the modal class = 28,
frequency (f0) of class preceding the modal class = 16,
frequency (f2) of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = l + × h
= 40 + × 20
= 40 + × 20
= 40 + 12
= 52
Hence, the mode is 52.
3. Heights of students of class X are givee in the flowing frequency distribution
Height (in cm) 150 155 155 160 160 165 165 170 170 - 175
Number of 
students
15 8 20 12 5
Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.
Sol: 
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 
160 165. So, the modal class is 160 165.
Now, 
Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5,
frequency (f1) of the modal class = 20,
frequency (f0) of class preceding the modal class = 8,
frequency (f2) of class succeeding the modal class = 12
Now, let us substitute these values in the formula:
Mode = l + × h
= 160 + × 5
= 160 + × 5
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163cm.
Now, to find the mean let us put the data in the table given below:
Page 3


1. Find the mode of the following distribution:
Marks 10 20 20 30 30 40 40 50 50 60
Frequency 12 35 45 25 13
Sol:
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 
30 40. So, the modal class is 30- 40.
Now,
Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = l + × h
= 30 + × 10
= 30 + × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100
Frequency 25 16 28 20 5
Sol: 
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 
40 60. So, the modal class is 40 60.
 
Now, 
Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20,
frequency (f1) of the modal class = 28,
frequency (f0) of class preceding the modal class = 16,
frequency (f2) of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = l + × h
= 40 + × 20
= 40 + × 20
= 40 + 12
= 52
Hence, the mode is 52.
3. Heights of students of class X are givee in the flowing frequency distribution
Height (in cm) 150 155 155 160 160 165 165 170 170 - 175
Number of 
students
15 8 20 12 5
Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.
Sol: 
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 
160 165. So, the modal class is 160 165.
Now, 
Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5,
frequency (f1) of the modal class = 20,
frequency (f0) of class preceding the modal class = 8,
frequency (f2) of class succeeding the modal class = 12
Now, let us substitute these values in the formula:
Mode = l + × h
= 160 + × 5
= 160 + × 5
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163cm.
Now, to find the mean let us put the data in the table given below:
 
Height (in cm) Number of students (fi) Class mark (xi) fi xi
150 155 15 152.5 2287.5
155 160 8 157.5 1260
160 165 20 162.5 3250
165 170 12 167.5 2010
170 175 5 172.5 862.5
Total
fi = 60 fi xi = 9670
Mean = 
i i
i
i
i
f x
f
=
= 161.17
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17cm.
4. Find the mode of the following distribution:
Class 
interval
10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42
Frequency 8 6 11 20 25 22 10 4
Sol: 
As the class 26 30 has the maximum frequency, it is the modal class.
Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22
Mode, M0 = xk +
= 26 + 
= 26 + 
= (26 + 2.5)
= 28.5
5. Given below is the distribution of total household expenditure of 200 manual workers in a 
city:
Expenditure (in Rs) Number of manual 
workers
1000 1500 24
1500 2000 40
2000 2500 31
2500 3000 28
3000 3500 32
3500 4000 23
4000 4500 17
4500 5000 5
Page 4


1. Find the mode of the following distribution:
Marks 10 20 20 30 30 40 40 50 50 60
Frequency 12 35 45 25 13
Sol:
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 
30 40. So, the modal class is 30- 40.
Now,
Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = l + × h
= 30 + × 10
= 30 + × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100
Frequency 25 16 28 20 5
Sol: 
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 
40 60. So, the modal class is 40 60.
 
Now, 
Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20,
frequency (f1) of the modal class = 28,
frequency (f0) of class preceding the modal class = 16,
frequency (f2) of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = l + × h
= 40 + × 20
= 40 + × 20
= 40 + 12
= 52
Hence, the mode is 52.
3. Heights of students of class X are givee in the flowing frequency distribution
Height (in cm) 150 155 155 160 160 165 165 170 170 - 175
Number of 
students
15 8 20 12 5
Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.
Sol: 
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 
160 165. So, the modal class is 160 165.
Now, 
Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5,
frequency (f1) of the modal class = 20,
frequency (f0) of class preceding the modal class = 8,
frequency (f2) of class succeeding the modal class = 12
Now, let us substitute these values in the formula:
Mode = l + × h
= 160 + × 5
= 160 + × 5
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163cm.
Now, to find the mean let us put the data in the table given below:
 
Height (in cm) Number of students (fi) Class mark (xi) fi xi
150 155 15 152.5 2287.5
155 160 8 157.5 1260
160 165 20 162.5 3250
165 170 12 167.5 2010
170 175 5 172.5 862.5
Total
fi = 60 fi xi = 9670
Mean = 
i i
i
i
i
f x
f
=
= 161.17
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17cm.
4. Find the mode of the following distribution:
Class 
interval
10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42
Frequency 8 6 11 20 25 22 10 4
Sol: 
As the class 26 30 has the maximum frequency, it is the modal class.
Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22
Mode, M0 = xk +
= 26 + 
= 26 + 
= (26 + 2.5)
= 28.5
5. Given below is the distribution of total household expenditure of 200 manual workers in a 
city:
Expenditure (in Rs) Number of manual 
workers
1000 1500 24
1500 2000 40
2000 2500 31
2500 3000 28
3000 3500 32
3500 4000 23
4000 4500 17
4500 5000 5
 
Find the average expenditure done by maximum number of manual workers.
Sol: 
As the class 1500-2000 has the maximum frequency, it is the modal class.
Now, xk = 1500, h = 500, fk = 40, fk-1 = 24, fk+1 = 31
Mode, M0 = xk +
= 1500 + 
= 1500 + 
= (1500 + 320)
= 1820
Hence, mode = Rs 1820
6. Calculate the mode from the following data:
Monthly salary (in 
Rs)
No of employees
0 5000 90
5000 10000 150
10000 15000 100
15000 20000 80
20000 25000 70
25000 30000 10
Sol: 
As the class 5000-10000 has the maximum frequency, it is the modal class.
Now, xk = 5000, h = 5000, fk = 150, fk-1 = 90, fk+1 = 100
Mode, M0 = xk +
= 5000 + 
= 5000 + 
= (5000 + 2727.27)
= 7727.27
Hence, mode = Rs 7727.27
7. Compute the mode from the following data:
Age (in
years)
0 5 5 10 10 15 15 20 20 25 25 30 30 - 35
No of 
patients
6 11 18 24 17 13 5
Sol: 
As the class 15 20 has the maximum frequency, it is the modal class.
Now, xk = 15, h = 5, fk = 24, fk-1 = 18, fk+1 = 17
Page 5


1. Find the mode of the following distribution:
Marks 10 20 20 30 30 40 40 50 50 60
Frequency 12 35 45 25 13
Sol:
Here, the maximum class frequency is 45, and the class corresponding to this frequency is 
30 40. So, the modal class is 30- 40.
Now,
Modal class = 30 40, lower limit ( l) of modal class = 30, class size ( h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Mode = l + × h
= 30 + × 10
= 30 + × 10
= 30 + 3.33
= 33.33
Hence, the mode is 33.33.
2. Compute the mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100
Frequency 25 16 28 20 5
Sol: 
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 
40 60. So, the modal class is 40 60.
 
Now, 
Modal class = 40 60, lower limit ( l) of modal class = 40, class size ( h) = 20,
frequency (f1) of the modal class = 28,
frequency (f0) of class preceding the modal class = 16,
frequency (f2) of class succeeding the modal class = 20
Now, let us substitute these values in the formula:
Mode = l + × h
= 40 + × 20
= 40 + × 20
= 40 + 12
= 52
Hence, the mode is 52.
3. Heights of students of class X are givee in the flowing frequency distribution
Height (in cm) 150 155 155 160 160 165 165 170 170 - 175
Number of 
students
15 8 20 12 5
Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.
Sol: 
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 
160 165. So, the modal class is 160 165.
Now, 
Modal class = 160 165, lower limit ( l) of modal class = 160, class size ( h) = 5,
frequency (f1) of the modal class = 20,
frequency (f0) of class preceding the modal class = 8,
frequency (f2) of class succeeding the modal class = 12
Now, let us substitute these values in the formula:
Mode = l + × h
= 160 + × 5
= 160 + × 5
= 160 + 3
= 163
Hence, the mode is 163.
It represents that the height of maximum number of students is 163cm.
Now, to find the mean let us put the data in the table given below:
 
Height (in cm) Number of students (fi) Class mark (xi) fi xi
150 155 15 152.5 2287.5
155 160 8 157.5 1260
160 165 20 162.5 3250
165 170 12 167.5 2010
170 175 5 172.5 862.5
Total
fi = 60 fi xi = 9670
Mean = 
i i
i
i
i
f x
f
=
= 161.17
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17cm.
4. Find the mode of the following distribution:
Class 
interval
10 14 14 18 18 22 22 26 26 30 30 34 34 38 38 42
Frequency 8 6 11 20 25 22 10 4
Sol: 
As the class 26 30 has the maximum frequency, it is the modal class.
Now, xk = 26, h = 4, fk = 25, fk-1 = 20, fk+1 = 22
Mode, M0 = xk +
= 26 + 
= 26 + 
= (26 + 2.5)
= 28.5
5. Given below is the distribution of total household expenditure of 200 manual workers in a 
city:
Expenditure (in Rs) Number of manual 
workers
1000 1500 24
1500 2000 40
2000 2500 31
2500 3000 28
3000 3500 32
3500 4000 23
4000 4500 17
4500 5000 5
 
Find the average expenditure done by maximum number of manual workers.
Sol: 
As the class 1500-2000 has the maximum frequency, it is the modal class.
Now, xk = 1500, h = 500, fk = 40, fk-1 = 24, fk+1 = 31
Mode, M0 = xk +
= 1500 + 
= 1500 + 
= (1500 + 320)
= 1820
Hence, mode = Rs 1820
6. Calculate the mode from the following data:
Monthly salary (in 
Rs)
No of employees
0 5000 90
5000 10000 150
10000 15000 100
15000 20000 80
20000 25000 70
25000 30000 10
Sol: 
As the class 5000-10000 has the maximum frequency, it is the modal class.
Now, xk = 5000, h = 5000, fk = 150, fk-1 = 90, fk+1 = 100
Mode, M0 = xk +
= 5000 + 
= 5000 + 
= (5000 + 2727.27)
= 7727.27
Hence, mode = Rs 7727.27
7. Compute the mode from the following data:
Age (in
years)
0 5 5 10 10 15 15 20 20 25 25 30 30 - 35
No of 
patients
6 11 18 24 17 13 5
Sol: 
As the class 15 20 has the maximum frequency, it is the modal class.
Now, xk = 15, h = 5, fk = 24, fk-1 = 18, fk+1 = 17
 
Mode, M0 = xk +
= 15 + 
= 15 + 
= (15 + 2.3)
= 17.3
Hence, mode = 17.3 years
8. Compute the mode from the following series:
Sol: 
As the class 85 95 has the maximum frequency, it is the modal class.
Now, xk = 85, h = 10, fk = 32, fk-1 = 30, fk+1 = 6
Mode, M0 = xk +
= 85 + 
= 85 + 
= (85 + 0.71)
= 85.71
Hence, mode = 85.71
9. Compute the mode from the following data:
Class 
interval
1
5
6
10
11
15
16
20
21
25
26
30
31
35
36
40
41
45
46
50
Frequency 3 8 13 18 28 20 13 8 6 4
Sol: 
Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we 
will convert it to an exclusive form as given below:
Class 
interval
0.5
5.5
5.5
10.5
10.5
15.5
15.5
20.5
20.5
25.5
25.5
30.5
30.5
35.5
35.5
40.5
40.5
45.5
45.5
50.5
Frequency 3 8 13 18 28 20 13 8 6 4
As the class 20.5 25.5 has the maximum frequency, it is the modal class.
Now, xk = 20.5, h = 5, fk = 28, fk-1 = 18, fk+1 = 20
Mode, M0 = xk +
= 20.5 + 
= 20.5 + 
= (20.5 + 2.78)
Size 45 55 55 65 65 75 75 85 85 95 95 105 105 - 115
Frequency 7 12 17 30 32 6 10
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