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Question: The given distribution shows the number of runs scored by some top batsmen of the world in one day i...

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

Runs scoredNumber of batsmen
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

Find the mode of the data.

Explanation

Solution

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 Mode=l+(ff12ff1f2)×c{\rm{Mode}} = l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times c, where ll is the lower limit of the modal class, ff is the frequency of the modal class, f1{f_1} is the frequency of the class just preceding the modal class, f2{f_2} is the frequency of the class succeeding the modal class, cc is the width of the class interval.

Complete Step by Step Solution:

Runs scoredNumber of batsmen
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

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=4000l = 4000, frequency of the modal class f=18f = 18, frequency of the class just preceding the modal class f1=4{f_1} = 4, frequency of the class succeeding the modal class f2=9{f_2} = 9, Width of the class interval c=1000c = 1000.
Now we will calculate the mode of the frequency distribution.
Substituting l=4000l = 4000, f=18f = 18, f1=4{f_1} = 4, f2=9{f_2} = 9 and c=1000c = 1000 in the formula Mode=l+(ff12ff1f2)×c{\rm{Mode}} = l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times c, we get
Mode=4000+(1842(18)49)×1000{\rm{Mode}} = 4000 + \left( {\dfrac{{18 - 4}}{{2\left( {18} \right) - 4 - 9}}} \right) \times 1000
Multiplying the terms in the denominator, we get
Mode=4000+(143649)×1000\Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{36 - 4 - 9}}} \right) \times 1000
Adding the terms in the denominator, we get
Mode=4000+(143613)×1000\Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{36 - 13}}} \right) \times 1000
Mode=4000+(1423)×1000\Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14}}{{23}}} \right) \times 1000
Simplifying the expression, we get
Mode=4000+(1400023)\Rightarrow {\rm{Mode}} = 4000 + \left( {\dfrac{{14000}}{{23}}} \right)
Dividing 14000 by 23, we get
Mode=4000+608.69\Rightarrow {\rm{Mode}} = 4000 + 608.69
Adding the terms, we get
Mode=4608.69\Rightarrow {\rm{Mode}} = 4608.69

Therefore, the mode is 4608.694608.69.

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.