Chapter 37.3 - Mode of the Data

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:

Table.37.25

• 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:

Table.37.26

• 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:

Table.37.27

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:

Table.37.3

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:

Table.37.27

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:

Table.37.28

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:

Table.37.29

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. 

Table.37.30

Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

1. First we will find the mean:

Table.37.31

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. 

Table.37.32

Find the mode and mean of this data. Interpret the two measures.
Solution:
1. First we will find the mean:

Table.37.33

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

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In the next section, we will discuss about median.

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Chapter 37.2 - Calculation of mean - Solved examples

In the previous section we saw Step deviation method. We also saw some solved examples demonstrating all three methods. In this section we will see a few more solved examples.

Solved example 37.2
Consider the following distribution of daily wages of 50 workers of a factory.

Table.37.14

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.15
• a is taken as 150, since it is the middle x_i
• h is taken as 20, since it is the class width

Table.37.15

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 7260
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get x = ⁷²⁶⁰⁄₅₀ = 145.2
2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is -240
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get d = ^-240⁄₅₀ = - 4.8
• Thus x = a + d = 150 - 4.8 = 145.2
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is -12
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get u = ^-12⁄₅₀ = - 0.24
• Thus x = a + hu = 150 + (20 × -0.24) = 150 - 4.8 = 145.2
■ This is the same value obtained by methods 1 & 2
■ In this problem, the values in the x_i column are not small. They are hundreds
• Because of these high values, we get thousands in the sixth column. It is not convenient to find all those thousands and then do calculations with them. 
• So we must reduce x_i by subtracting 'a' from it. Thus we get d_i. But then, we get hundreds in the seventh column. It is not convenient to deal with hundreds either
• So we must reduce d_i by dividing it with 'h'. We get small numbers in the eighth column
■ Upon seeing hundreds in the x_i column, we can immediately decide to use the step deviation method. We need not fill up the sixth and seventh columns 
■ The significance of the result is clear in this problem:
The average (or mean) wages of workers in that factory is 145.2 per day
• The factory owner has to deal with two items: Daily Income and Daily Expense 
• Under Daily expense, he has two items:
    ♦ Total cost of material per day
    ♦ Total cost of labour per day
• There are 50 workers. So the 'total cost of labour per day' will be approximately equal to (50×145.2) = 7260
• To obtain accurate values, we will have to apply more advanced principles of statistics. We will see them in higher classes

Solved example 37.3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Table.37.16

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.17
• a is taken as 18, since it is the middle x_i
• h is taken as 2, since it is the class width

Table.37.17

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is (752+20f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get x = ^(752+20f)⁄_(44+f) 
• But this mean is given to us as 18. So we can write:
^(752+20f)⁄_(44+f) = 18 ⟹ (752+20f) = 18(44+f) ⟹ (752+20f) = (792+18f) ⟹ 2f = 40 ⟹ f = 20

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is (-40+2f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get d = ^(-40+2f)⁄_(44+f)
• Thus x = a + d = [18 + ^(-40+2f)⁄_(44+f)]
• But this mean is given to us as 18. So we can write:
[18 + ^(-40+2f)⁄_(44+f)] = 18 ⟹ [^(-40+2f)⁄_(44+f)] = 0 ⟹ (-40+2f) = 0 ⟹ 2f = 40 ⟹ f = 20 
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is (-20+f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get u = ^(-20+f)⁄_(44+f)
• Thus x = a + hu = [18 + (2 × ^(-20+f)⁄_(44+f))]
• But this mean is given to us as 18. So we can write:

[18 + (2 × ^(-20+f)⁄_(44+f))] = 18 ⟹ [(2 × ^(-20+f)⁄_(44+f))] = 0 ⟹ [^(-20+f)⁄_(44+f)] = 0 
⟹ (-20+f) = 0 ⟹ f = 20 
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Table.37.18

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.19
• a is taken as 75.5, since it is the middle x_i
• h is taken as 3, since it is the class width

Table.37.19

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 2277
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get x = ²²⁷⁷⁄₃₀ = 75.9

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is 12
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get d = ¹²⁄₃₀ = 0.4
• Thus x = a + d = 75.5 + 0.4 = 75.9
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is 4
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get u = ⁴⁄₃₀
• Thus x = a + hu = [75.5 + (3 × ⁴⁄₃₀)] = 75.9
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Table.37.20

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:
• The class intervals given to us are not continuous. There is a gap of '1' between all the class intervals
    ♦ For example, the gap between the first and second intervals is (53-52) = 1
• We know the method to make the intervals continuous. (Details here)
• The new table 37.21 is given below:

Table.37.21

• Now we can begin to solve the problem. We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.22
• a is taken as 57, since it is the middle x_i
• h is taken as 3, since it is the class width

Table.37.22

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 22875
• The denominator is calculated at the bottom end of second column. It's value is 400
• So we get x = ²²⁸⁷⁵⁄₄₀₀ = 57.1875

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is 12
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get d = ⁷⁵⁄₄₀₀ = 0.1875
• Thus x = a + d = 57 + 0.1875 = 57.1875
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is 4
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get u = ²⁵⁄₄₀₀
• Thus x = a + hu = [57 + (3 × ²⁵⁄₄₀₀)] = 57.1875
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.6
The mean of the following distribution is 50. Also Σf = 120. Find the missing frequencies f1 and f2.

Table.37.23

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.24
• a is taken as 50, since it is the middle x_i
• h is taken as 20, since it is the class width

Table.37.24

• From the table, it is clear that, the step deviation method is appropriate because, it involves smaller numbers in the eighth column. So we will use it:
Step deviation method:
• u is given by the formula:

• The numerator will be the value at the bottom end of the sixth column. It is (4-f1+f2)
• The denominator will be the value at the bottom end of the second column. It is (68+f1+f2)
• So we get u = ^(4-f1+f2)⁄_(68+f1+f2) 
• Thus x = a + hu = [50 + (20 × ^(4-f1+f2)⁄_(68+f1+f2))]
• But this mean is given to us as 50. So we can write:
[50 + (20 × ^(4-f1+f2)⁄_(68+f1+f2))] = 50 ⟹ [(20 × ^(4-f1+f2)⁄_(68+f1+f2))] = 0
⟹ ^(4-f1+f2)⁄_(68+f1+f2) = 0 ⟹ (4-f1+f2) = 0 ⟹ f1- f2 = 4
• Now consider the denominator. In the denominator, there will be Σf. But Σf is given as 120
So we can write: Σf = (68+f1+f2) = 120 ⟹ (f1+f2) = 52
• Thus we have two equations:
(i) f1 + f2 = 52
(ii) f1 - f2 = 4
• From (ii) we get: f2 = (f1-4)
• substituting this in (i), we get: 
f1 + f1 - 4 = 52 ⟹ 2f1 = 56 ⟹ f1 = 28
• Substituting this value of f1 in (ii) we get:
f2 = 28 - 4 = 24

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In the next section, we will discuss about mode.

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