Question

The marks in the science of \[80\] students of class X are given below: Find the mode of the marks obtained by the students in science.

Marks:	\[0 - 10\]	\[10 - 20\]	\[20 - 30\]	\[30 - 40\]	\[40 - 50\]	\[50 - 60\]	\[60 - 70\]	\[70 - 80\]	\[80 - 90\]	\[90 - 100\]
Frequency:	\[3\]	\[5\]	\[16\]	\[12\]	\[13\]	\[20\]	\[5\]	\[4\]	\[1\]	\[1\]

Answer 1

Hint: Here we will be using the formula of calculating mode as shown below:
\[{\text{Mode}} = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right)h\] , where
\[l\] is the lower limit of the modal class,
\[h\] is the size of the class interval,
\[{f_1}\] is known as the frequency of the modal class,
\[{f_0}\] is known as the frequency of the class preceding the modal class, and
\[{f_2}\] is known as the frequency of the class succeeding the modal class.

Complete step-by-step solution:
Step 1: We can write the above-given table data as below:

Marks (\[{x_i}\]):	\[0 - 10\]	\[10 - 20\]	\[20 - 30\]	\[30 - 40\]	\[40 - 50\]	\[50 - 60\]	\[60 - 70\]	\[70 - 80\]	\[80 - 90\]	\[90 - 100\]
Frequency (\[{f_i}\]):	\[3\]	\[5\]	\[16\]	\[12\]	\[13\]	\[20\]	\[5\]	\[4\]	\[1\]	\[1\]

\[ \Rightarrow \]The model class of the above-given data is \[50 - 60\] with the highest frequency. And the lower limit of the modal class is \[l = 50\].
\[ \Rightarrow {f_0}\] will be the frequency of the class preceding the modal class which is \[13\]having a class interval \[40 - 50\].
\[ \Rightarrow \]\[{f_1}\] will be equal to the frequency of the modal class \[\left( {50 - 60} \right)\] which is \[20\].
\[ \Rightarrow \]\[{f_2}\] will be the frequency of the class succeeding the modal class which is \[5\] having a class interval \[60 - 70\].
\[ \Rightarrow h\] will be the size of the class interval which is \[10 - 0 = 10\].
Step 2: By using the formula of mode we get:
\[ \Rightarrow {\text{Mode}} = 50 + \left( {\dfrac{{20 - 13}}{{2\left( {20} \right) - 13 - 5}}} \right) \times 10\]
By simplifying the terms inside the brackets of the above expression we get:
\[ \Rightarrow {\text{Mode}} = 50 + \left( {\dfrac{7}{{40 - 13 - 5}}} \right) \times 10\]
By doing the final calculation inside the brackets of the above expression we get:
\[ \Rightarrow {\text{Mode}} = 50 + \left( {\dfrac{7}{{22}}} \right) \times 10\]
Now by doing the multiplication in the above expression we get:
\[ \Rightarrow {\text{Mode}} = 50 + \dfrac{{70}}{{22}}\]
By doing the addition in the above expression we get:
\[ \Rightarrow {\text{Mode}} = 53.17\]
The Mode of the class is \[53.17\].

Note: Students need to remember that for calculating mode we need to find the modal class of the given intervals. Modal class is that class which is having the highest frequency distribution among all. If the intervals are not given then first, we need to convert the data into intervals with their respective frequency and then find the modal class.