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**MODE Definition :**

Mode is the value of the variate which occurs most frequently. It represents the most frequent value of a series. In other words **Mode** is the value of the variable which has the highest frequency. When one speak of the ‘average student’, we generally mean the modal wage, the modal student. If we say that the modal wages obtained by workers in a factory are ` 70, we mean that the largest number of workers get the same amount. As high as ` 100 and as low as ` 50 as wages are much less frequented and they are non-modal.

**Calculation. **Mode cannot be determined in a series of individual observations unless it is converted to a discrete series (or continuous series). In a discrete series the value of the variate having the maximum frequency is the modal class. However, the exact location of mode is done by interpolation formula like median. Location of modal value in case of discrete series is possible if there is concentration of items at on point. If again there are two or more values having same maximum frequencies, (i.e. more concentration), it becomes difficult to determine mode. Such items are known as bimodal, tri-modal or multi-modal accordingly as the items concentrate at 2, 3 or more values.

**(A) For. Individual Observations **

The individual observations are to be first converted to discrete series (if possible). Then the variate having the maximum will be the mode. **Example 34 :** Calculate mode from the data (given) :

(Marks) : 10, 14, 24, 27, 24, 12, 11, 17.

(Individual observation are converted into a discrete series) Here marks 24 occurs maximum number of times, i.e. 2. Hence the modal marks is 24, or mode = 24 marks.

**Alternatively : **Arranging the numbers : 10, 11, 12, 14, 17, (24, 24) 27. Now 24 occurs maximum number of times, i.e. 2. ∴ Mode = 24 marks. [Note. When there are two or more values having the same maximum frequency, then mode is ill-defined. Such a sense is known as bimodal or multi-modal as the case may be.]

**Example 35 : **Compute mode from the following data. Marks obtained : 24, 14, 20, 17, 20, 14.

[Here 14 occurs 2 times (max.) and 20 occurs 2 times (max.)

∴ mode is ill-defined.]

**(B) For Simple Frequency Distribution Discrete Series**.

To Find the mode from the following Table :

Height (in inches) No. of Persons

57 3

59 5

61 7

62 10

63 20

64 22

65 24

66 5

67 2

69 2

Frequencies given below, in column (1) are grouped by two’s in column (2) and (3) and then by three’s in columns (4), (5), and (6). The maximum frequency in each column is marked by Bold Type. We do not find any fixed point having maximum frequency but changes with the change of grouping. In the following table, the sizes of maximum frequency in respect of different columns are arranged.

From the above table, we find 64 is the size of the item which is most frequented. The mode is, therefore, located at 64. [Note. At glance from column (1) one might think that 65 is the mode since it contains maximum frequency. This impression is corrected by the process of grouping . So it is not advisable to locate the mode merely by inspection.]

**(C) For continuous Series.** By inspections or by preparing Grouping Table and Analysis Table, ascertain the modal class. Then to find the exact value of mode, apply the following formula.

Where, I = lower class-boundary of modal class

f_{1} = frequency of modal class.

f_{0} = frequency of the class preceding modal class.

f_{2} = frequency of the class succeeding the modal class.

i = size of the class- interval of modal class.

**Note :** the above formula may also be expressed as follows :

Example 36 : Compute mode of the following distribution.

From the table it is clear that the maximum frequency is 16th : modal class is (40–50) Here l = 40, f_{0} = 12, f_{1} = 16, f_{2} = 10 (marked in table), i = 10 (= 50 –40)

**Calculation of Mode From discrete group frequency distribution. **

In such cases at first class boundaries are to be formed for applying formula.

**Example 37:** Compute mode from the following frequency distribution :

The class intervals which are in discrete form are first converted into class boundaries

Now modal class is (79.5 – 89.5), since this class has the highest frequency. Here l = 79.5, f_{0} = 40, f_{1} = 50, f_{2} = 30, i = 10

Calculation of mode from cumulative frequency distribution :

Example 38 : From the following cumulative frequency distribution of marks of 22 students in Accountancy, calculate mode :

Marks below 20 below 40 below 60 below 80 below 100

No. of students 3 8 17 20 22

**Solution : **

At first we are to transfer the above cumulative frequency distribution into a equal group frequency distribution and hence to calculate mode.

Modal class is (40–60), as this class has highest frequency.

Here l = 40, f_{0 }= 5, f_{1} = 9, f_{2} = 3, i = 20

Calculation of missing frequency :

**Example 39 : **Mode of the given distribution is 44, find the missing frequency

Marks 10–20 20–30 30–40 40–50 50–60 60–70

No. of students 5 8 12 –– 10 8

Solution : Since mode is 44, so modal class is 40–50.

let the missing frequency be f_{1}

**Miscellaneous examples :**

1. If two variates x and y are related by 2x = 3y – 1, and mean of y be 9 ; find the mean of x.

2. If 2u = 5x is the relation between two variables x and u and harmonic mean of x is 0.4, find the harmonic mean of u

3. The relation between two variables x and y is 3y – 2x + 5 = 0 and median of y is 40, find the median of x.

From 3y – 2x + 5 = 0 we get As the median is located by position, so median of x is

4. Mode of the following frequency distribution is 24 and total frequency is 100. Find the values of f_{1} and f_{2} .

C.I : 0 –10 10–20 20–30 30– 40 40–50

Frequency : 14 f_{1} 27 f_{2} 15

Mode is 24, so modal class is (20–30). From the formula of mode we find

14 + f_{1} + 27 + f_{2} + 15 = 100

or, f_{1 }+ f_{2} = 100 – 56 = 44 …….….(1)

or, 40 = 270 – 10f_{1} or, 10f_{1} = 230 or, f_{1} = 23. From (1) , f_{2} = 44 – 23 = 11 ∴ f_{1} = 23, f_{2} = 11

5. The following are the monthly salaries in rupees of 20 employees of a firm :

130 125 110 100 80 76 98 103 122 66

145 151 65 71 118 140 116 85 95 151

The firm gives bonuses of ` 10, 15, 20, 25 and 30 for individuals in the respective salary group : exceeding ` 60 but not excedding ` 80, exceeding ` 80 but not exceeding ` 100 and so on up to exceeding ` 140 but not exceeding ` 160. Find the average bonus paid per employee.

**Solution:** From the monthly salaries of the employees, we find the number of employees lying in the salary groups mentioned as follows :

Marks obtained by 30 students in History of a Test Examination 2012 of some school are as follows :

34 36 10 21 31 32 22 43 48 36

48 22 39 26 34 39 10 17 47 38

40 51 35 52 41 32 30 35 53 23

construct a frequency table with class intervals 10–19. 20–29 etc. Calculate the median and mode from the frequency distribution.

Solution:

Median = value of N^{th}/2 i.e., 30 2 i.e. 15th term

So median class is (29.5 – 39.5)

Advantages of mode :

(i) It can often be located by inspection.

(ii) It is not affected by extreme values. It is often a really typical value.

(iii) It is simple and precise. It is an actual item of the series except in a continuous series.

(iv) Mode can be determined graphically unlike Mean.

Disadvantages of mode :

(i) It is unsuitable for algebraic treatment.

(ii) When the number of observations is small, the Mode may not exist, while the Mean and Median can be calculated.

(iii) The value of Mode is not based on each and every item of series.

(iv) It does not lead to the aggregate, if the Mode and the total number of items are given.

**Empirical Relationship among Mean, Median and Mode**

A distribution in which the values of Mean, Median and Mode coincide, is known symmetrical and if the above values are not equal, then the distribution is said asymmetrical or skewed. In a moderately skewed distribution, there is a relation amongst Mean, Median and Mode which is as follows :

Mean – Mode = 3 (Mean – Median)

If any two values are known, we can find the other

**Example 40 : **In a moderately asymmetrical distribution the mode and mean are 32.1 and 35.4 respectively. Calculate the Median. From the relation, we find

3 Median = 2 Mean + Mode

or 3 Median = 2 × 35.4 + 32.1 = 70.8 + 32.1 = 102.9

∴ Median = 34.3

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