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Mode of Grouped Data: A measure of central tendency is a single value that aims to describe a data set by recognising the central position within that set of data. As such, measures of central tendency are occasionally called measures of central location.
The mean, median and mode are all logical measures of central tendency, but under different conditions, some measures of central tendency become more suitable to use than others. The mode is the value of the observation, which occurs most frequently. In the following sections, we will look at the mode and learn how to calculate it and under what conditions it is most appropriate to use.
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Data is a collection of information. Grouped data is the data formed by arranging single observations of a variable into groups. A frequency distribution table of these groups provides a convenient way of summarising or analysing the data.
A measure of central tendency is a single value that aims to describe a data set by recognising the central position within that set of data. As such, measures of central tendency are occasionally called measures of central location.
The mean, median and mode grouped data are all logical measures of central tendency, but under different conditions, some measures of central tendency become more suitable to use than others. The mean or average of a number of observations is the sum of the values of all observations divided by the total number of observations. The median is the value of the given number of observations, which divides it into exactly two parts. The mode is the value of the observation, which occurs most frequently.
A mode is an observation that occurs most often, that is, the value of the observation having a maximum frequency.
Mode is used to find the ideal size or item from a collection of items. For example: in business forecasting, ready-made manufacturing garments, manufacturing shoes, etc.
Suppose a watch manufacturer is designing a series of watches. To know which of those is sold most and in high demand, we can use mode. The same way when a manufacturer designs garments. He can know the high demand size or colour using mode. Here, the mode is the most appropriate measure because it shows the highest number of repetitions in the series.
Let us now discuss the way of obtaining the mode formula for grouped data. Sometimes two or more values have the same frequency. In such cases, one cannot say which is modal value, and hence mode is ill-defined. Such a frequency distribution is also known as bimodal or multimodal distribution. So, finding such values which have a maximum frequency is known as the mode formula of grouped data. For such frequency distribution, a mode is computed by the grouping method.
To compute the mode by grouping, we need to prepare a grouping table and an analysis table to find the mode. These tables help us in determining the correct value of the mode. The grouping table consists of six columns which are constructed by using the following steps:
1. Obtain the discrete frequency distribution.
2. Take the column of frequencies as column \(I\) and encircle the maximum frequency in it.
3. Construct column \(II,\) containing the sum of the frequencies taken two at a time and encircle the maximum frequency.
4. Leave the first frequency and construct column\(III,\) containing the sum of the frequencies taken two at a time. Then, encircle the maximum frequency in column \(III.\)
5. Construct column \(IV,\) containing the sum of three frequencies and encircle the maximum frequency in it.
6. Exclude the first frequency and compute the sum of the frequencies taken three at a time to construct column \(V.\) Encircle the maximum frequency in this column.
7. Exclude the first two frequencies and compute the sum of the frequencies taken three at a time to construct column \(VI.\) Then, encircle the maximum frequency in this column
After preparing the grouping table, we prepare an analysis table by using the below steps:
1. Prepare a table. In the topmost row, write all the variable values and write column numbers from \(I\) to \(IV\) in the left-most column.
2. See the maximum frequency in the first column of the grouping table and obtain the corresponding value of the variable. Now, mark a bar \(I\) in the first row of the analysis table against the variable’s value having the maximum frequency. Continue the same procedure for the remaining five columns.
3. Find the total number of bars corresponding to each value of the variable. That value of the variable which has the maximum number of bars is the mode of the frequency distribution.
Let us understand this through an example.
Question: Compute the value of mode for the following frequency distribution
| Class | Frequency |
| \(100 – 110\) | 4 |
| \(110 – 120\) | 6 |
| \(120 – 130\) | 20 |
| \(130 – 140\) | 32 |
| \(140 – 150\) | 33 |
| \(150 – 160\) | 8 |
| \(160 – 170\) | 2 |
Answer: In the given frequency distribution, the difference between the maximum frequency and the preceding frequency is minimal. So, we will determine the modal class by the grouping method.
Analysis Table
| Column Number | \(100 – 110\) | \(110 – 120\) | \(120 – 130\) | \(130 – 140\) | \(140 – 150\) | \(150 – 160\) | \(160 – 170\) |
| \(1\) | \({\rm{I}}\) | ||||||
| \(2\) | \({\rm{I}}\) | \({\rm{I}}\) | |||||
| \(3\) | \({\rm{I}}\) | \({\rm{I}}\) | |||||
| \(4\) | \({\rm{I}}\) | \({\rm{I}}\) | \({\rm{I}}\) | ||||
| \(5\) | \({\rm{I}}\) | \({\rm{I}}\) | \({\rm{I}}\) | ||||
| \(6\) | \({\rm{I}}\) | \({\rm{I}}\) | \({\rm{I}}\) | ||||
| Total | \(1\) | \(3\) | \(5\) | \(4\) | \(1\) |
Clearly, class \(130-140\) has the maximum number of bars. So, \(130-140\) is the modal class.
So, \(l = 130,\;h = 10,\;f = 32,\;{f_1} = 20\) and \({f_2} = 33\)
Now, mode \(= l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
\(\Rightarrow {\rm{mode}} = 130 + \frac{{32 – 20}}{{64 – 20 – 33}} \times 10\)
\(\Rightarrow {\rm{mode}} = 130 + \frac{{12}}{9} \times 10 = 140.9\)
Therefore, the mode of the given distribution is \(140.9.\)
In a grouped or continuous frequency distribution with equal class intervals, we use the following steps to calculate the mode.
1. Obtain the continuous frequency table.
2. Determine the class of highest frequency, either by inspection or by grouping method.
3. Obtain the values of the below from the frequency distribution.
\(l = \) lower limit of the modal class
\(f = \) frequency of the modal class,
\(h=\) width of the modal class
\({f_1} = \) frequency of the class preceding the modal class
\({f_2} = \) frequency of the class succeeding the modal class
4. Substitute the values obtained in the previous step in the below formula,
Mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
Q.1. The wickets taken by a bowler in \(10\) cricket matches are as follows:
\(2,\;6,\;4,\;5,\;0,\;2,\;1,\;3,\;2,\;3\)
Find the mode of the data.
Ans: Clearly \(2\) is the number of wickets taken by the bowler in the maximum number of matches. So, mode of the data is \(2.\)
Q.2. Compute the mode for the following frequency distribution.
| Size | Frequency |
| \(0 – 4\) | \(5\) |
| \(4 – 8\) | \(7\) |
| \(8 – 12\) | \(9\) |
| \(12 – 16\) | \(17\) |
| \(16 – 20\) | \(12\) |
| \(20 – 24\) | \(10\) |
| \(24 – 28\) | \(6\) |
| \(28 – 32\) | \(3\) |
| \(32 – 36\) | \(1\) |
| \(36 – 40\) | \(0\) |
Ans: In the given data \(17\) is the maximum frequency and the corresponding class is \(12 – 16.\) So, modal class is \(12 – 16\) such that, \(f = 17,l = 12,\;h = 4,\;{f_1} = 9\) and \({f_2} = 12\)
We know that mode \(= l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
\( \Rightarrow {\rm{mode}} = 12 + \frac{{17 – 9}}{{34 – 9 – 12}} \times 4\)
\( \Rightarrow {\rm{mode}} = 12 + \frac{8}{{13}} \times 4 = 12 + \frac{{32}}{{13}} = 12 + 2.46 = 14.46\) Therefore, the mode of the given distribution is \(14.46.\)
Q.3. The following data gives the distribution of total household expenditure in ₹ of manual workers in a city:
| Expenditure in ₹ | Frequency |
| \(1000 – 1500\) | \(24\) |
| \(1500 – 2000\) | \(40\) |
| \(2000 – 2500\) | \(33\) |
| \(2500 – 3000\) | \(28\) |
| \(3000 – 3500\) | \(30\) |
| \(3500 – 4000\) | \(22\) |
| \(4000 – 4500\) | \(16\) |
| \(4500 – 5000\) | \(7\) |
Find the average expenditure which is being done by the maximum number of manual workers.
Ans: We know that the mode is the variable’s value, which occurs a maximum number of times in a frequency distribution. So, the average expenditure done by the maximum number of workers is the modal value. We observe that class \(1500 – 2000\) has a maximum frequency \(40.\) So, it is the modal class such that \(l = 1500,\;h = 500,\;f = 40,\;{f_1} = 24\) and \({f_2} = 33\)
Now, mode \(= l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
\( \Rightarrow {\rm{mode}} = 1500 + \frac{{40 – 24}}{{80 – 24 – 33}} \times 500 = 1500 + \frac{{16}}{{23}} \times 500 = 1847.826\) Therefore, the mode of the given distribution is \(1847.826.\)
Q.4. 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:
| Size | \(1 – 3\) | \(3 – 5\) | \(5 – 7\) | \(7 – 9\) | \(9 – 11\) |
| Number of families | \(7\) | \(8\) | \(2\) | \(2\) | \(1\) |
Find the mode of this data.
Ans: Here, the maximum class frequency is \(8,\) and the class corresponding to this frequency is \(3-5.\) So, the modal class is \(3-5.\)
Now, the lower limit \(\left( l \right)\) of modal class \(=3\)
Class size \(\left( h \right) = 2\)
Frequency \(\left( f \right)\) of the modal class \(=8\)
Frequency \(\left( {{f_1}} \right)\) of the class preceding the modal class \(=7\)
Frequency \(\left( {{f_2}} \right)\) of the class succeeding the modal class \(=2\)
We know that mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
\(\Rightarrow {\rm{mode}} = 3 + \frac{{8 – 7}}{{16 – 7 – 2}} \times 2 = 3 + \frac{2}{7} = 3.286\) Therefore, mode \( = 3.286\)
Question 5: The marks distribution of \(30\) students in a mathematics examination is given below. Find the mode of this data.
| Class interval | Number of students |
| \(10 – 25\) | \(2\) |
| \(25 – 40\) | \(3\) |
| \(40 – 55\) | \(7\) |
| \(55 – 70\) | \(6\) |
| \(70 – 85\) | \(6\) |
| \(85 – 100\) | \(6\) |
Ans: The maximum class frequency is \(7,\) and the class corresponding to this frequency is \(40-55.\) So, the modal class is \(40-55.\)
Now, the lower limit \(\left( l \right)\) of modal class \(=40\)
Class size \(\left( h \right) = 15\)
Frequency \(\left( f \right)\) of the modal class \(=7\)
Frequency \(\left( {{f_1}} \right)\) of the class preceding the modal class \(=3\)
Frequency \(\left( {{f_2}} \right)\) of the class succeeding the modal class \(=6\)
We know that mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
\( \Rightarrow {\rm{mode}} = 40 + \frac{{7 – 3}}{{14 – 6 – 3}} \times 15 = 52\) Therefore, mode of the given data is \(52.\)
In this article, we have learned about the mode, mode of the grouped data, uses and properties of mode. We also learnt the formula to find the mode. You will now be able to solve problems based on this topic.
Q.1. What if there are two modes in grouped data?
Ans: If there are two modes in the grouped data, two values have the same maximum frequency. Example: Consider the data \(5,\;7,\;3,\;5,\;7,\;8,\;9,\;0.\) Here \(5,7\) appears twice. This data has \(2\) modes and is called bimodal data. Therefore, the mode of the given data is \(5,7.\)
Q.2. Explain the mode of grouped data?
Ans: In a frequency distribution, more than one value may have the maximum frequency. In such situations, the data is said to be multimodal. So, finding such values which have a maximum frequency using the below formula is known as the mode of the grouped data. mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
Q.3. How do you find the mode of grouped data?
Ans: We find the mode of the grouped data by using the below formula.
mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
Where, \(l = \) lower limit of the modal class
\(f=\) frequency of the modal class
\(h=\) width of the modal class
\({f_1} = \) frequency of the class preceding the modal class \({f_2} = \) frequency of the class succeeding the modal class
Q.4. What is the formula for mode?
Ans: The formula to find the mode is given by mode \(= l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
Where, \(l=\) lower limit of the modal class
\(f=\) frequency of the modal class
\(h=\) width of the modal class
\({f_1} = \) frequency of the class preceding the modal class \({f_2} = \) frequency of the class succeeding the modal class
Q.5. How do you find the mean median and mode grouped data?
Ans: We can find the mean of the grouped data in \(3\) methods.
Direct method:
\(\bar X = \frac{{{f_1}{x_1} + {f_2}{x_2} + ……… + {f_n}{x_n}}}{{{f_1} + {f_2} + …….. + {f_n}}}\)
Short-cut method:
\(\bar X = A + \frac{1}{N}\mathop \sum \limits_{i = 1}^n {f_i}{d_i}\)
Step-deviation method:
\(\bar X = A + h\left[ {\frac{1}{N}\mathop \sum \limits_{i = 1}^n {f_i}{u_i}} \right]\)
Where, \({u_i} = \frac{{{x_i} – A}}{h}\)
Median of grouped data:
Median \(= l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
Where \(l=\) lower frequency of the median class
\(f=\) frequency of the median class
\(h=\) size of the median class
\(F=\) Cumulative frequency of the class preceding the median class \(N = \mathop \sum \limits_{i = 1}^n {f_i}\)
Mode of grouped data:
mode \( = l + \frac{{f – {f_1}}}{{2f – {f_1} – {f_2}}} \times h\)
Where \(l=\) lower limit of the modal class
\(f=\) frequency of the modal class
\(h=\) width of the modal class
\({f_1} = \) frequency of the class preceding the modal class
\({f_2} = \) frequency of the class succeeding the modal class
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