Question

The given distribution shows the number of runs scored by some top batsmen of the world in one day international matches

Runs scored	Number of batsmen
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data.

Answer 1

Hint: Here, we will find the mode of the given data. The given data is in the form of the grouped frequency distribution. First, we will find the modal class, and then by using the formula for mode of grouped frequency distribution, we will find the mode. Mode is defined as the value with the maximum frequency for the grouped frequency distribution and for ungrouped frequency distribution. The class interval with maximum frequency is called the modal class.

Formula Used:
Mode is calculated by the given formula \[{\rm{Mode}} = l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times c\], where \[l\] is the lower limit of the modal class, \[f\] is the frequency of the modal class, \[{f_1}\] is the frequency of the class just preceding the modal class, \[{f_2}\] is the frequency of the class succeeding the modal class, \[c\] is the width of the class interval.

Complete Step by Step Solution:

Runs scored	Number of batsmen
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

From the given data,
Since the maximum frequency is 18, then the corresponding class is 4000-5000.
Therefore, the modal class is 4000-5000.
Thus, the lower limit of the class \[l = 4000\], frequency of the modal class \[f = 18\], frequency of the class just preceding the modal class \[{f_1} = 4\], frequency of the class succeeding the modal class \[{f_2} = 9\], Width of the class interval \[c = 1000\].
Now we will calculate the mode of the frequency distribution.
Substituting \[l = 4000\], \[f = 18\], \[{f_1} = 4\], \[{f_2} = 9\] and \[c = 1000\] in the formula \[{\rm{Mode}} = l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times c\], we get
\[{\rm{Mode}} = 4000 + \left( {\dfrac{{18 - 4}}{{2\left( {18} \right) - 4 - 9}}} \right) \times 1000\]
Multiplying the terms in the denominator, we get
\[ \Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{36 - 4 - 9}}} \right) \times 1000\]
Adding the terms in the denominator, we get
\[ \Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{36 - 13}}} \right) \times 1000\]
\[ \Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{23}}} \right) \times 1000\]
Simplifying the expression, we get
\[ \Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14000}}{{23}}} \right)\]
Dividing 14000 by 23, we get
\[ \Rightarrow {\rm{Mode}} = 4000 + 608.69\]
Adding the terms, we get
\[ \Rightarrow {\rm{Mode}} = 4608.69\]

Therefore, the mode is \[4608.69\].

Note:
We know that mode can be calculated easily by finding the variable that occurs many times but we need to keep in mind that this fact is only acceptable for the raw data. For a grouped frequency distribution, it is essential for us to find the highest frequency, only then we will be using the formula to calculate the mode for the grouped data.