Question: consider the table given below:

Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
students	$12$	$18$	$27$	$20$	$17$	$6$

The arithmetic mean of the marks given above is:
A. $18$
B. $28$
C. $27$
D. $6$

Hint:
Here we are given that the marks are as the class interval and the number of students represent the frequency then the arithmetic mean will be given by the formula:
$\dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }}$ and here ${x_i}$ is the average of the upper and lower limit of the class interval.

Complete step by step solution:
Here we are given the range of marks which are obtained by the students as given below

So here the frequency of $0 - 10$ marks is $12$ that means there are twelve students whose marks are between $0,10$ and similarly we are given the frequency of all the marks obtained that is the marks are obtained by how many number of students and the ranges are given that in which range the marks are obtained by how many number of students?
The mean is given by the formula $\dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }}$ and here ${x_i}$ is the average of the upper and lower limit of the class interval
So ${x_1} = \dfrac{{0 + 10}}{2} = 5$
$
  {x_2} = \dfrac{{10 + 20}}{2} = 15 \\
  {x_3} = \dfrac{{20 + 30}}{2} = 25 \\
  {x_4} = \dfrac{{30 + 40}}{2} = 35 \\
  {x_5} = \dfrac{{40 + 50}}{2} = 45 \\
  {x_6} = \dfrac{{50 + 60}}{2} = 55 \\
 $
So here mean$ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }}$
$
   = \dfrac{{{x_1}{f_1} + {x_2}{f_2} + {x_3}{f_3} + {x_4}{f_4} + {x_5}{f_5} + {x_6}{f_6}}}{{{f_1} + {f_2} + {f_3} + {f_4} + {f_5} + {f_6}}} \\
   = \dfrac{{5(12) + 18(15) + 27(25) + 20(35) + 17(45) + 6(55)}}{{12 + 18 + 27 + 20 + 17 + 6}} \\
   = \dfrac{{60 + 270 + 675 + 700 + 765 + 330}}{{100}} = 28 \\
 $
Therefore mean is $28$.

Note:
Arithmetic mean of the two numbers a and b is given by the formula $\dfrac{{a + b}}{2}$ and the geometric mean is given by the formula $\sqrt {ab} $
Harmonic mean is given by the formula $\dfrac{{2ab}}{{a + b}}$