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Question: The percentage of marks obtained by \( 100 \) students in an examination is given. Marks| 30-35|...

The percentage of marks obtained by 100100 students in an examination is given.

Marks30-3535-4040-4545-5050-5555-6060-65
No. of students141628231883

Find the mean marks of the student.

Explanation

Solution

First, we have to define what the terms we need to solve the problem are.
Since there is a set of students and their marks are given as respectively above in that table, like the marks thirty to thirty-five is gettable by the fourteen students is the lowest marks obtained.
The highest marks obtained is sixty to sixty-five marks by the student’s number is three only.
Thus, by all the information we need to find the mean marks.
The formula used: x=xff\overline x = \dfrac{{\sum {xf} }}{{\sum f }} formula of the Mean.

Complete step-by-step answer:
First, we need to find the average marks obtained by these students so that we can apply the formula of the means and find the answer to the given question.
Let us find the average with the use of given information like there are fourteen students can obtain the mark from 303530 - 35 hence the average marks are 32.532.5 similarly we can able to find for all average marks and thus we get 42.5,47.5,52.5,57.5,62.542.5,47.5,52.5,57.5,62.5 .
Thus, we are going to multiply the marks according to the students like 32.5×1432.5 \times 14 (fourteen students gets an average of thirty-two points five) and thus 32.5×14=45532.5 \times 14 = 455 .
Hence similarly we can obtain all the other values like 600,1190,1092.5,945,460,187.5600,1190,1092.5,945,460,187.5 which are the values of xfxf .
Hence applying the formula of mean we get x=xff\overline x = \dfrac{{\sum {xf} }}{{\sum f }} ( xfxf are the values of multiplying the average marks and students also f is the values of students only).
Thus, substitute the value in the addition (summation) we get x=xff4930110\overline x = \dfrac{{\sum {xf} }}{{\sum f }} \Rightarrow \dfrac{{4930}}{{110}} (adding all the know values as above) (where 445+600+1190+1092.5+945+460+187.5445 + 600 + 1190 + 1092.5 + 945 + 460 + 187.5 is the xf)
Further solving this we get x4930110=44.82\overline x \Rightarrow \dfrac{{4930}}{{110}} = 44.82 is the mean marks of the students.

Note: Since Mean can also be called the average values, and there are some other terms like mean which are median and mode.
All three formulas are used to find the problems related to the distribution method.
xfxf means adding all the multiplying of the average marks and students.