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

The marks in the science of 8080 students of class X are given below: Find the mode of the marks obtained by the students in science.

Marks:0100 - 10102010 - 20203020 - 30304030 - 40405040 - 50506050 - 60607060 - 70708070 - 80809080 - 909010090 - 100
Frequency:3355161612121313202055441111
Explanation

Solution

Here we will be using the formula of calculating mode as shown below:
Mode=l+(f1f02f1f0f2)h{\text{Mode}} = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right)h , where
ll is the lower limit of the modal class,
hh is the size of the class interval,
f1{f_1} is known as the frequency of the modal class,
f0{f_0} is known as the frequency of the class preceding the modal class, and
f2{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 (xi{x_i}):0100 - 10102010 - 20203020 - 30304030 - 40405040 - 50506050 - 60607060 - 70708070 - 80809080 - 909010090 - 100
Frequency (fi{f_i}):3355161612121313202055441111

\RightarrowThe model class of the above-given data is 506050 - 60 with the highest frequency. And the lower limit of the modal class is l=50l = 50.
f0\Rightarrow {f_0} will be the frequency of the class preceding the modal class which is 1313having a class interval 405040 - 50.
\Rightarrow $$$${f_1} will be equal to the frequency of the modal class (5060)\left( {50 - 60} \right) which is 2020.
\Rightarrow $$$${f_2} will be the frequency of the class succeeding the modal class which is 55 having a class interval 607060 - 70.
h\Rightarrow h will be the size of the class interval which is 100=1010 - 0 = 10.
Step 2: By using the formula of mode we get:
Mode=50+(20132(20)135)×10\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:
Mode=50+(740135)×10\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:
Mode=50+(722)×10\Rightarrow {\text{Mode}} = 50 + \left( {\dfrac{7}{{22}}} \right) \times 10
Now by doing the multiplication in the above expression we get:
Mode=50+7022\Rightarrow {\text{Mode}} = 50 + \dfrac{{70}}{{22}}
By doing the addition in the above expression we get:
Mode=53.17\Rightarrow {\text{Mode}} = 53.17
The Mode of the class is 53.1753.17.

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.