In the previous section we completed a discussion on mean. We also saw some solved examples demonstrating all three methods to find mean. In this section we will discuss about mode.
We have seen some basics about mode in part II (Details here). Let us see a new example:
Example 5:
The wickets taken by a bowler in 10 cricket matches are as follows:
5, 3, 4, 2, 0, 2, 1, 6, 2, 3
Find the mode of this data
Solution:
1. Let us analyse the data:
• In the first match he took 2 wickets
• In the second match he took 3 wickets
• In the third match he took 4 wickets
_ _ _
_ _ _
• In the tenth match he took 3 wickets
2. We have to find the mode
• That is., the item which occur the most number of times
• Is '5' the mode?
♦ '5' occurs once. If there are any other items which occur more than once, '5' cannot be the mode
♦ The next item '3' is occurs 2 times. So '5' cannot be the mode
• Then is '3' the mode?
♦ '3' occurs twice. If there are any other items which occur more than twice, '3' cannot be the mode
_ _ _
_ _ _
3. Instead of going on like this, we can speed up the work by writing the frequency of each item (or making an 'ungrouped frequency distribution table'):
We get:
Frequency of 5 = 1
Frequency of 3 = 2
Frequency of 4 = 1
Frequency of 2 = 3
Frequency of 0 = 1
Frequency of 6 = 1
• The ungrouped frequency distribution table of the above data will be as shown below:
• So the maximum frequency is 3
♦ Item having this maximum frequency is '2'
• Thus the mode of the given data is '2'
• What we saw above is an 'ungrouped frequency distribution table'.
♦ We know that, such a table is prepared when the data is small.
• If the data is large, we will be given a 'grouped frequency distribution table'.
• In example 1 above, if we are given the data as a grouped frequency distribution table (with width of class intervals 2), it will look like as in table 37.26 below:
• In this table, the maximum frequency is 5
♦ This maximum frequency is possessed by the class interval: '2 - 4'
■ So which item has the maximum frequency?
• It is not possible to answer this question
• Within the class interval '2 - 4', the items possible are '2' and '3'
• But there is no way to find the frequency of each of them.
♦ This is because, we are given a 'grouped frequency distribution table'.
• So we have to develop a new method to find the mode when 'grouped frequency distribution tables' are given to us
• When a 'grouped frequency distribution table' is given to us, we can immediately write the class interval which has the 'largest frequency'.
♦ This class interval is called the modal class
• The 'actual item' which has the largest frequency is hidden inside the modal class.
♦ It can be calculated using the formula:
Where:
l = lower limit of the modal class
h = width of the class interval (assuming all classes are of the same width)
f1 = frequency of the modal class
f0 = frequency of the class preceding the modal class
f2 = frequency of the class succeeding the modal class
Example 6:
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
Find the mode of this data
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '3 - 5' is the modal class. It has the highest frequency of '8'
2. Now we can calculate the mode using the formula:
l = lower limit of the modal class = 3
h = width of the class interval (assuming all classes are of the same width) = 2
f1 = frequency of the modal class = 8
f0 = frequency of the class preceding the modal class = 7
f2 = frequency of the class succeeding the modal class = 2
Substituting all the values, we get:
mode = 3 + (8-7⁄2×8-7-2)×2 = 3 + (1⁄16-9)×2 = 3 + (2⁄7) = 3.286
Example 7:
Consider the first example on mean that we did at the beginning of this chapter. The data was given in table 37.3. It is the marks distribution of 30 students in a mathematics examination. The mean was calculated as 62. Now find the mode of this data. Also compare and interpret the mode and the mean.
Solution:
• For convenience, the table 37.3 is shown again below:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '40 - 55' is the modal class. It has the highest frequency of '7'
2. Now we can calculate the mode using the formula:
l = lower limit of the modal class = 40
h = width of the class interval (assuming all classes are of the same width) = 15
f1 = frequency of the modal class = 7
f0 = frequency of the class preceding the modal class = 3
f2 = frequency of the class succeeding the modal class = 6
Substituting all the values, we get:
mode = 40 + (7-3⁄2×7-3-6)×15 = 40 + (4⁄14-9)×15 = 40 + 12 = 52
3. We have already obtained the mean as 62. Now we get the mode as 60
• So we can write:
♦ The average mark of the class is 62
♦ The mark obtained by the largest number of students is 52
• That is., in a table showing the marks of all the 30 students, values near 52 will appear more than others
Now we will see some solved examples
Solved example 37.7
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:
1. First we will find the mean:
We will use the step deviation method:
• u is given by the formula:
• The numerator is the value at the bottom end of the eighth column. It is 43
• The denominator is the value at the bottom end of the second column. It is 80
• So we get u = 43⁄80
• Thus x = a + hu = [30 + (10 × 43⁄80)] = 35.375
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '35 - 45' is the modal class. It has the highest frequency of '23'
(ii) Now we can calculate the mode using the formula:
l = lower limit of the modal class = 35
h = width of the class interval (assuming all classes are of the same width) = 10
f1 = frequency of the modal class = 23
f0 = frequency of the class preceding the modal class = 21
f2 = frequency of the class succeeding the modal class = 14
Substituting all the values, we get:
mode = 35 + (23-21⁄2×23-21-14)×10 = 35 + (2⁄46-35)×10 = 36.82
3. So we can write:
• The average age of all the patients admitted at the hospital in a year is 35.3
• The people around an age of 36.8 are the most who are admitted in that year
♦ In other words, the number of patients around an age of 36.8 is greater than others
Solved example 37.8
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Determine the modal lifetimes of the components
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '60 - 80' is the modal class. It has the highest frequency of '61'
2. Now we can calculate the mode using the formula:
l = lower limit of the modal class = 60
h = width of the class interval (assuming all classes are of the same width) = 20
f1 = frequency of the modal class = 61
f0 = frequency of the class preceding the modal class = 52
f2 = frequency of the class succeeding the modal class = 38
Substituting all the values, we get:
mode = 60 + (61-52⁄2×61-52-38)×20 = 60 + (9⁄122-90)×20 = 60 + 5.625 = 65.625
Solved example 37.9
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:
We will use the step deviation method:
• u is given by the formula:
• The numerator is the value at the bottom end of the eighth column. It is -35
• The denominator is the value at the bottom end of the second column. It is 200
• So we get u = -35⁄200
• Thus x = a + hu = [2750 + (500 × -35⁄200)] = 2662.5
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '1500 - 2000' is the modal class. It has the highest frequency of '40'
(ii) Now we can calculate the mode using the formula:
l = lower limit of the modal class = 1500
h = width of the class interval (assuming all classes are of the same width) = 500
f1 = frequency of the modal class = 40
f0 = frequency of the class preceding the modal class = 24
f2 = frequency of the class succeeding the modal class = 33
Substituting all the values, we get:
mode = 1500 + (40-24⁄2×40-24-33)×500 = 1500 + (16⁄80-57)×500 = 1847.83
3. So we can write:
• The average expenditure of all the 200 families of the village is 2662.5
• The 'number of families having an expense of around 1847.83' is greater than others
Solved example 37.10
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:
1. First we will find the mean:
We will use the step deviation method:
• u is given by the formula:
• The numerator is the value at the bottom end of the eighth column. It is -23
• The denominator is the value at the bottom end of the second column. It is 35
• So we get u = -23⁄35
• Thus x = a + hu = [32.5 + (5 × -23⁄35)] = 29.2
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '30 - 35' is the modal class. It has the highest frequency of '10'
(ii) Now we can calculate the mode using the formula:
l = lower limit of the modal class = 30
h = width of the class interval (assuming all classes are of the same width) = 5
f1 = frequency of the modal class = 10
f0 = frequency of the class preceding the modal class = 9
f2 = frequency of the class succeeding the modal class = 3
Substituting all the values, we get:
mode = 30 + (10-9⁄2×10-9-3)×5 = 30 + (1⁄20-12)×5 = 30.6
3. So we can write the conclusion:
• We are given the 'number of students per teacher'
• If a value in the data is low, it indicates a better condition because, then the teacher is in charge of a lesser number of students, and so, each of those students will get better attention
• However, in this problem, we are not dealing with such aspects. We want the mean and the mode
• The mean value is 29.2.
♦ So on an average, each teacher is in charge of 29.2 students
• The mode is 30.6
♦ So the number of teachers who are in charge of 30.6 students are the highest
In the next section, we will discuss about median.
We have seen some basics about mode in part II (Details here). Let us see a new example:
Example 5:
The wickets taken by a bowler in 10 cricket matches are as follows:
5, 3, 4, 2, 0, 2, 1, 6, 2, 3
Find the mode of this data
Solution:
1. Let us analyse the data:
• In the first match he took 2 wickets
• In the second match he took 3 wickets
• In the third match he took 4 wickets
_ _ _
_ _ _
• In the tenth match he took 3 wickets
2. We have to find the mode
• That is., the item which occur the most number of times
• Is '5' the mode?
♦ '5' occurs once. If there are any other items which occur more than once, '5' cannot be the mode
♦ The next item '3' is occurs 2 times. So '5' cannot be the mode
• Then is '3' the mode?
♦ '3' occurs twice. If there are any other items which occur more than twice, '3' cannot be the mode
_ _ _
_ _ _
3. Instead of going on like this, we can speed up the work by writing the frequency of each item (or making an 'ungrouped frequency distribution table'):
We get:
Frequency of 5 = 1
Frequency of 3 = 2
Frequency of 4 = 1
Frequency of 2 = 3
Frequency of 0 = 1
Frequency of 6 = 1
• The ungrouped frequency distribution table of the above data will be as shown below:
 |
| Table.37.25 |
♦ Item having this maximum frequency is '2'
• Thus the mode of the given data is '2'
• What we saw above is an 'ungrouped frequency distribution table'.
♦ We know that, such a table is prepared when the data is small.
• If the data is large, we will be given a 'grouped frequency distribution table'.
• In example 1 above, if we are given the data as a grouped frequency distribution table (with width of class intervals 2), it will look like as in table 37.26 below:
 |
| Table.37.26 |
♦ This maximum frequency is possessed by the class interval: '2 - 4'
■ So which item has the maximum frequency?
• It is not possible to answer this question
• Within the class interval '2 - 4', the items possible are '2' and '3'
• But there is no way to find the frequency of each of them.
♦ This is because, we are given a 'grouped frequency distribution table'.
• So we have to develop a new method to find the mode when 'grouped frequency distribution tables' are given to us
• When a 'grouped frequency distribution table' is given to us, we can immediately write the class interval which has the 'largest frequency'.
♦ This class interval is called the modal class
• The 'actual item' which has the largest frequency is hidden inside the modal class.
♦ It can be calculated using the formula:
l = lower limit of the modal class
h = width of the class interval (assuming all classes are of the same width)
f1 = frequency of the modal class
f0 = frequency of the class preceding the modal class
f2 = frequency of the class succeeding the modal class
Example 6:
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:
 |
| Table.37.27 |
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '3 - 5' is the modal class. It has the highest frequency of '8'
2. Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 2
f1 = frequency of the modal class = 8
f0 = frequency of the class preceding the modal class = 7
f2 = frequency of the class succeeding the modal class = 2
Substituting all the values, we get:
mode = 3 + (8-7⁄2×8-7-2)×2 = 3 + (1⁄16-9)×2 = 3 + (2⁄7) = 3.286
Example 7:
Consider the first example on mean that we did at the beginning of this chapter. The data was given in table 37.3. It is the marks distribution of 30 students in a mathematics examination. The mean was calculated as 62. Now find the mode of this data. Also compare and interpret the mode and the mean.
Solution:
• For convenience, the table 37.3 is shown again below:
 |
| Table.37.3 |
• So in this problem, the class interval '40 - 55' is the modal class. It has the highest frequency of '7'
2. Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 15
f1 = frequency of the modal class = 7
f0 = frequency of the class preceding the modal class = 3
f2 = frequency of the class succeeding the modal class = 6
Substituting all the values, we get:
mode = 40 + (7-3⁄2×7-3-6)×15 = 40 + (4⁄14-9)×15 = 40 + 12 = 52
3. We have already obtained the mean as 62. Now we get the mode as 60
• So we can write:
♦ The average mark of the class is 62
♦ The mark obtained by the largest number of students is 52
• That is., in a table showing the marks of all the 30 students, values near 52 will appear more than others
Now we will see some solved examples
Solved example 37.7
The following table shows the ages of the patients admitted in a hospital during a year:
 |
| Table.37.27 |
Solution:
1. First we will find the mean:
 |
| Table.37.28 |
• u is given by the formula:
• The denominator is the value at the bottom end of the second column. It is 80
• So we get u = 43⁄80
• Thus x = a + hu = [30 + (10 × 43⁄80)] = 35.375
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '35 - 45' is the modal class. It has the highest frequency of '23'
(ii) Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 10
f1 = frequency of the modal class = 23
f0 = frequency of the class preceding the modal class = 21
f2 = frequency of the class succeeding the modal class = 14
Substituting all the values, we get:
mode = 35 + (23-21⁄2×23-21-14)×10 = 35 + (2⁄46-35)×10 = 36.82
3. So we can write:
• The average age of all the patients admitted at the hospital in a year is 35.3
• The people around an age of 36.8 are the most who are admitted in that year
♦ In other words, the number of patients around an age of 36.8 is greater than others
Solved example 37.8
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
 |
| Table.37.29 |
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '60 - 80' is the modal class. It has the highest frequency of '61'
2. Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 20
f1 = frequency of the modal class = 61
f0 = frequency of the class preceding the modal class = 52
f2 = frequency of the class succeeding the modal class = 38
Substituting all the values, we get:
mode = 60 + (61-52⁄2×61-52-38)×20 = 60 + (9⁄122-90)×20 = 60 + 5.625 = 65.625
Solved example 37.9
The following data gives the distribution of total monthly household expenditure of 200 families of a village.
 |
| Table.37.30 |
Solution:
1. First we will find the mean: |
| Table.37.31 |
• u is given by the formula:
• The denominator is the value at the bottom end of the second column. It is 200
• So we get u = -35⁄200
• Thus x = a + hu = [2750 + (500 × -35⁄200)] = 2662.5
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '1500 - 2000' is the modal class. It has the highest frequency of '40'
(ii) Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 500
f1 = frequency of the modal class = 40
f0 = frequency of the class preceding the modal class = 24
f2 = frequency of the class succeeding the modal class = 33
Substituting all the values, we get:
mode = 1500 + (40-24⁄2×40-24-33)×500 = 1500 + (16⁄80-57)×500 = 1847.83
3. So we can write:
• The average expenditure of all the 200 families of the village is 2662.5
• The 'number of families having an expense of around 1847.83' is greater than others
Solved example 37.10
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India.
 |
| Table.37.32 |
Solution:
1. First we will find the mean:
 |
| Table.37.33 |
• u is given by the formula:
• The denominator is the value at the bottom end of the second column. It is 35
• So we get u = -23⁄35
• Thus x = a + hu = [32.5 + (5 × -23⁄35)] = 29.2
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '30 - 35' is the modal class. It has the highest frequency of '10'
(ii) Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 5
f1 = frequency of the modal class = 10
f0 = frequency of the class preceding the modal class = 9
f2 = frequency of the class succeeding the modal class = 3
Substituting all the values, we get:
mode = 30 + (10-9⁄2×10-9-3)×5 = 30 + (1⁄20-12)×5 = 30.6
3. So we can write the conclusion:
• We are given the 'number of students per teacher'
• If a value in the data is low, it indicates a better condition because, then the teacher is in charge of a lesser number of students, and so, each of those students will get better attention
• However, in this problem, we are not dealing with such aspects. We want the mean and the mode
• The mean value is 29.2.
♦ So on an average, each teacher is in charge of 29.2 students
• The mode is 30.6
♦ So the number of teachers who are in charge of 30.6 students are the highest
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