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'

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• 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

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• 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

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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 25.10 - Mode of a Data

In the previous section we completed the discussion on Median. In this section we will see Mode.

Mode

Mode is that value in the data that occurs the most. 
It can be explained as follows:
• In a data set, all the values may not be different. Some values occur more than once. That is.,
    ♦ Some values may occur twice. 
    ♦ Some values may occur thrice. 
    ♦ Some other values may occur four times . . . and so on. 
We can not specify a limit. It depends on the situation
• So, from the data set, we take out the value that occur the most number of times. This value is the mode.

■ This can be put in another way:
We just mentioned that:
    ♦ Some values may occur twice. 
    ♦ Some values may occur thrice. 
    ♦ Some other values may occur four times . . . and so on. 
• When such multiple occurrences happen, we know that, we must convert the raw data into a frequency table. Such a frequency table will directly give us the number of times each value occurs. So the frequency table makes our task of finding the mode easy. All we have to do is:
• Take out that value which has the largest frequency.
■ From the above discussion, we get another information:
Unlike mean and median, mode need not be a 'number'. It can be any item like chocolate ice cream, table, chair, cricket, soccer etc.,
■ Also note that, the 'number of times' or 'frequency' is not the mode. An observation in the data set is the mode.

---

The ready made garment and shoe industries make great use of this measure of central tendency. Using the knowledge of mode, these industries decide which size of the product should be produced in large numbers.

Let us illustrate this with the help of an example:
Solved example 25.24 
Find the mode of the following marks (out of 10) obtained by 20 students:

Solution:
The first step is to arrange the raw data in ascending or descending order. The ascending order is shown below:

From the above sorted data, we can easily form the frequency table. It is shown below:

From the above table, we find that the score '9' has the maximum frequency 4. So 9 is the mode

Solved example 25.25
In a small unit of a factory, there are 5 employees : a supervisor and four labourers. The labourers draw a salary of Rs. 5,000 per month each while the supervisor gets Rs. 15,000 per month. Calculate the mean, median and mode of the salaries of this unit of the factory.
Solution:
In this problem, there are only 5 observations. So we do not need to form a frequency table. The data set is:
5000, 5000, 5000, 5000, 15000
Calculation of mean:
• Sum of all the values = 5000 + 5000 + 5000 + 5000 + 15000 = 35000
• No. of observations = 5
• Mean = 35000/5 = Rs. 7000 
Calculation of median:
There are 5 values. So n = 5. It is an odd number. We can use the procedure that we wrote earlier:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁽ⁿ⁺¹⁾⁄₂

3. This value is the median
• The list sorted in ascending order is:
5000, 5000, 5000, 5000, 15000

• i = ⁽ⁿ⁺¹⁾⁄₂ = ⁽⁵⁺¹⁾⁄₂ = ⁶⁄₂ = 3
• So the 3^rd value 5000 is the median
Calculation of mode:

From the data, we find that the salary '5000' has the maximum frequency 4. So Rs. 5000 is the mode.

---

We have completed a basic discussion about mean, median and mode. But the details that we have seen so far is not sufficient to get an accurate result on the central tendency of a data set. We need to gain more knowledge. Then only we will be able to do many problems encountered in science, engineering, business administration, social sciences etc., We will learn more in higher classes.

---

Now we will see some solved examples

Solved example 25.26
The following number of goals were scored by a team in a series of 10 matches:

Find the mean, median and mode of these scores.
Solution:
Calculation of mean:
• Sum of all the values = 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3 = 28
• No. of observations = 10
• Mean = 28/10 = 2.8
Calculation of median:

There are 10 values. So n = 10. It is an even number. We can use the procedure that we wrote earlier:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁿ⁄₂

_{3. Take out the value whose i = ⁽ⁿ⁺²⁾⁄2}

4. Calculate the mean of the values in (2) and (3). This mean is the median of the whole list
• The list sorted in ascending order is:

The numbers in yellow colour shows the sequence 'i'
• i = ⁿ⁄₂ = ¹⁰⁄₂ = 5. So the first member of the middle pair = 3
_• (ⁿ⁄₂  + 1) = 6. So the second member of the middle pair = 3
• Median of the list = Mean of 3 and 3 =  ⁽³⁺³⁾⁄₂ = ⁶⁄₂ = 3
Calculation of mode:

From the data, we find that the value '3' has the maximum frequency 4. So '3' is the mode.

Solved example 25.27

In a mathematics test given to 15 students, the following marks (out of 100) are recorded:

Find the mean, median and mode of this data.

Solution:
Calculation of mean:
• Sum of all the values = 41 +39 +48 +52 +46 +62 +54 +40 +96 +52 +98 +40 +42 +52 +60 = 822
• No. of observations = 15
• Mean = 822/15 = 54.8
Calculation of median:
There are 15 values. So n = 15. It is an odd number. We can use the procedure that we wrote earlier:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁽ⁿ⁺¹⁾⁄₂

3. This value is the median
• The list sorted in ascending order is:

• i = ⁽ⁿ⁺¹⁾⁄₂ = ⁽¹⁵⁺¹⁾⁄₂ = ¹⁶⁄₂ = 8

• So the 8^th value 52 is the median
Calculation of mode:

From the data, we find that the value '52' has the maximum frequency 3. So '52' is the mode.

Solved example 25.28

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

Solution:

The given data is already arranged in ascending order. 

There are 10 values. So n = 10. It is an even number. We can use the procedure that we wrote earlier:

• i = ⁿ⁄₂ = ¹⁰⁄₂ = 5. So the first member of the middle pair = x
_• (ⁿ⁄₂  + 1) = 6. So the second member of the middle pair = x+2
• Median of the list = Mean of x and (x+2) =  ^(x+x+2)⁄₂ = ^(2x+2)⁄₂ = x+1
• But the median is given as 63. So we can write:
x+1 = 63 ⇒ x = 63-1 = 62

Solved example 25.29
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution:
Arrange the given data in ascending order:

From the data, we find that the value '14' has the maximum frequency 4. So '14' is the mode.

Solved example 25.30

Find the mean salary of 60 workers of a factory from the following table:

Solution:

1. We have:

2. Table below is prepared by expanding the given table

• The numerator in (1) is calculated at the bottom end of third column in the above table. It's value is 305000
• The denominator is calculated at the bottom end of second column in the above table. It's value is 60
3. So we get x = 305000/60 = 5083.33

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We have completed this discussion on Mean, Median and Mode. Part III of this discussion can be seen in chapter 37.
In the next Chapter we will see Arithmetic progressions.

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Chapter 25.9 - Median of a Data

In the previous section we completed the discussion on Mean. In this section we will see Median.

Median

We are going to learn about median and it's significance. But first we will learn how to calculate the median:
■ Any list of numbers will have a median. It is the 'middle number'.
■ If the list contains an odd number of numbers, one of the number already in the list will be the median. 
• This number will be situated at the exact middle, 
      ♦ when the whole list is sorted in ascending or descending order.
■ If the list contains an even number of numbers, the median will not be already present in the list. We will have to do some calculations to find it.
• This is because, there will be two numbers at the exact middle,
    ♦ when the whole list is sorted in ascending or descending order
• The median is the mean of those two numbers at the middle

---

Let us elaborate on the above two cases with the help of examples: fig.25.22

Case 1: When the list has an odd number of numbers

1. Consider fig. 25.22(a) below. It has 13 (an odd number) values sorted in ascending order. 

Fig.25.22

2. Each can be denoted as x_i. So we have x₁, x₂, x₃, . . . , upto x_n. where n = 13

3. 'i' denotes the position of each value in the sorted list

4. We want the value of 'i' of the median. For calculating that value of 'i', we adopt the following procedure:

(a) There are equal number of values on either side of the median. Let this equal numbers be 'x'

(b) So we get: 

x + 1 ( the median) + x = n 

⇒ 2x + 1 = n. ⇒ 2x = n-1 ⇒ x = ^(n-1)⁄₂

(c) So there are ^(n-1)⁄₂ numbers on the left side of the median

(d) That means, the 'i' value of the number just to the left of the median = ^(n-1)⁄₂

(e) So the 'i' value of the median = ^(n-1)⁄₂  + 1 = ^(n-1+2)⁄₂  = ⁽ⁿ⁺¹⁾⁄₂.

5. When n = 13, we get 'i' value of the median = ⁽ⁿ⁺¹⁾⁄₂  = ⁽¹³⁺¹⁾⁄₂ = ¹⁴⁄₂ = 7

6. From the fig.25.22(a), we can see that the 7^th value is indeed the median

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So when the number of values (n) in the list is an odd number, we can find the median by the following procedure:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁽ⁿ⁺¹⁾⁄₂

3. This value is the median

---

Case 2: When the list has an even number of numbers

1. Consider fig. 25.22(b) above. It has 12 (an even number) values sorted in ascending order. 

2. Each can be denoted as x_i. So we have x₁, x₂, x₃, . . . , upto x_n. where n = 12

3. 'i' denotes the position of each value in the sorted list

4. There are two values at the 'exact middle'. We will call them the 'middle pair'. For any sorted list having an even number of values, there will be a 'middle pair'.  

5. We want the value of 'i' for both the members of that pair. For calculating those value of 'i', we adopt the following procedure:

(a) There are equal number of values on either side of the middle pair. Let this equal numbers be 'x'

(b) So we get: 

x + 2 ( the middle pair) + x = n 

⇒ 2x + 2 = n. ⇒ 2x = n-2 ⇒ x = ^(n-2)⁄₂

(c) So there are ^(n-2)⁄₂ numbers on the left side of the middle pair

(d) That means, the 'i' value of the number just to the left of the pair = ^(n-2)⁄₂

(e) So the 'i' value of the first member of the pair = ^(n-2)⁄₂  + 1 = ^(n-2+2)⁄₂  = ⁿ⁄₂.

(f) So the 'i' value of the second member of the pair = ⁿ⁄₂  + 1 .

6. When n = 12, we get:

• 'i' of the first member of the middle pair = ⁿ⁄₂ = ¹²⁄₂ = 6

• 'i' of the second member of the middle pair = (ⁿ⁄₂  + 1) = 7

7. So the 6^th and 7^th values form the middle pair. From fig.25.22(b), we can see that, this is indeed true. 

8. Once we get the i values of both the members of the middle pair, we can take them out from the list

9. Then we calculate the mean of the two members. This mean is the median of the whole list

---

So when the number of values (n) in the list is an even number, we can find the median by the following procedure:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁿ⁄₂

_{3. Take out the value whose i = ⁽ⁿ⁺²⁾⁄2}

4. Calculate the mean of the values in (2) and (3). This mean is the median of the whole list

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Now we will see some solved examples on the calculation of median

Solved example 25.22

The heights (in cm) of 9 students of a class are as follows:

Find the median of this data.

Solution:
There are 9 values. So n = 9. It is an odd number. We can use the procedure that we wrote earlier:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁽ⁿ⁺¹⁾⁄₂

3. This value is the median
• The list sorted in ascending order is:

The numbers in yellow colour shows the sequence 'i' • i = ⁽ⁿ⁺¹⁾⁄₂ = ⁽⁹⁺¹⁾⁄₂ = ¹⁰⁄₂ = 5
• So the 5^th value 149 is the median

Solved example 25.23
The points scored by a Kabaddi team in a series of matches are as follows:  

Find the median of the points scored by the team.
Solution:
There are 16 values. So n = 16. It is an even number. We can use the procedure that we wrote earlier:

1. Sort the list in ascending or descending order

2. Take out the value whose i = ⁿ⁄₂

_{3. Take out the value whose i =} (ⁿ⁄₂  + 1)

4. Calculate the mean of the values in (2) and (3). This mean is the median of the whole list
• The list sorted in ascending order is:

The numbers in yellow colour shows the sequence 'i'
• i = ⁿ⁄₂ = ¹⁶⁄₂ = 8. So the first member of the middle pair = 10
• (ⁿ⁄₂  + 1) = 9. So the second member of the middle pair = 14
• Median of the list = Mean of 10 and 14 =  ^(10+14)⁄₂ = ²⁴⁄₂ = 12

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We have completed the discussion on Median. In the next section we will see Mode

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