Question

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

Runs scored	No. of batsman
\[3000 - 4000\]	\[4\]
\[4000 - 5000\]	\[18\]
\[5000 - 6000\]	\[9\]
\[6000 - 7000\]	\[7\]
\[7000 - 8000\]	\[6\]
\[8000 - 9000\]	\[3\]
\[9000 - 10000\]	\[1\]
\[10000 - 11000\]	\[1\]

Find the mode of the data
A. \[4608.695652\]
B. \[4408.695652\]
C. \[4202.695652\]
D. \[4882.695652\]

Answer 1

Hint: Find the maximum frequency from the given number of students. Conclude the group of maximum frequency, lower class boundary of the class, frequency of the group, i.e., maximum frequency, the frequency of the preceding class and the frequency of the succeeding class plus the width of the class. Substitute all these values in the formula of the mode and simplify the equation to find the mode.

Complete step by step solution:
To find the mode of the data, we need to find the details given in the formula of mode.
The number of batsman that maximum scored in a class of a run is the highest frequency.
Maximum frequency \[ = 18\]
The group of this maximum frequency \[ = 4000 - 5000\]
Now, we label the terms finding the following terms as given below;
\[L = \] lower class boundary of the modal group \[ = 4000\]
\[{f_1} = \] frequency of the modal group \[ = 18\]
\[{f_0} = \] frequency of the class preceding the modal class \[ = 4\]
\[{f_2} = \] frequency of the class succeeding the modal class \[ = 9\]
\[w = \] class width is the difference between the upper limit and the lower limit of a class.
\[ \Rightarrow w = 5000 - 4000\]
Subtracting the terms, we get;
\[w = 1000\]
Mode \[ = L + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times w\]
Substituting the values in the above equation, we get;
$\Rightarrow$Mode \[ = 4000 + \dfrac{{18 - 4}}{{2 \times 18 - 4 - 9}} \times 1000\]
Simplifying the equation, we get;
$\Rightarrow$Mode \[ = 4000 + \dfrac{{14}}{{36 - 13}} \times 1000\]
Further reducing the equation, we get;
$\Rightarrow$Mode \[ = 4000 + \dfrac{{14}}{{23}} \times 1000\]
Now, taking the second term and dividing it with the denominator and then multiplying it with the other number, we get;
$\Rightarrow$Mode \[ = 4000 + 608.69\]
Adding the terms, we get;
$\Rightarrow$Mode \[ = 4608.695652\]
Since rounding off to the nearest integer of the given options, we get;
\[ \Rightarrow \] Mode \[ = 4608.695652\]

$\therefore $ The correct option is A.

Note: Mode is the number or a set of intervals which appears most often in a data set. It is possible for a set of data values to have more than one mode as well or no mode as well. If there are two modes, it is known as bimodal set and more, mode is most useful as a measure of central tendency when examining categorical data, such as models of cars or flavours of soda, for which a mathematical average median value based on ordering can’t be calculated.