
The marks obtained out of $50$ by $102$ students in a physics test are given in the frequency table below:
Marks ${\text{ }}15{\text{ 20 22 24 25 30 33 38 45}}$
Frequency $5{\text{ 8 11 20 23 18 13 3 1}}$
Find the average number of the marks.
Answer
561.9k+ views
Hint:
Here, we are given that the marks of the students and the number of students represent the frequency then the arithmetic mean or average will be given by the formula:
$\dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }}$ and here ${x_i}$ represent the marks of the students and ${f_i}$ as the frequency.
Complete step by step solution:
Here we are given the range of marks which are obtained by the students as given below
Marks ${\text{ }}15{\text{ 20 22 24 25 30 33 38 45}}$
Frequency $5{\text{ 8 11 20 23 18 13 3 1}}$
So here the frequency of $15$ marks is $5$ that means there are five students whose marks are $15$ and similarly we are given the frequency of all the marks obtained, that is 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 ${x_i}$ represent the marks of the students and ${f_i}$ as the frequency.
So ${x_1} = 15$
$
{x_2} = 20 \\
{x_3} = 22 \\
{x_4} = 24 \\
{x_5} = 25 \\
{x_6} = 30 \\
{x_7} = 33 \\
{x_8} = 38 \\
{x_9} = 45 \\
$
$\sum {{f_i}} = 5 + 8 + 11 + 20 + 23 + 18 + 13 + 3 + 1 = 102$
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} + {x_7}{f_7} + {x_8}{f_8} + {x_9}{f_9}}}{{{\text{ total number of students}}}} \\
= \dfrac{{15(5) + 20(8) + 22(11) + 24(20) + 25(23) + 30(18) + 33(13) + 38(3) + 45(1)}}{{102}} \\
= \dfrac{{2660}}{{102}} = 26.07 \\
$
Therefore mean or average mark is $26.07$
Note:
If the question asks for the mode then we need to find which mark is obtained by the maximum number of students and that will be our mode. For example here we have maximum students achieving a particular mark as $23$ so the mode will be $25$
Here, we are given that the marks of the students and the number of students represent the frequency then the arithmetic mean or average will be given by the formula:
$\dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{f_i}} }}$ and here ${x_i}$ represent the marks of the students and ${f_i}$ as the frequency.
Complete step by step solution:
Here we are given the range of marks which are obtained by the students as given below
Marks ${\text{ }}15{\text{ 20 22 24 25 30 33 38 45}}$
Frequency $5{\text{ 8 11 20 23 18 13 3 1}}$
So here the frequency of $15$ marks is $5$ that means there are five students whose marks are $15$ and similarly we are given the frequency of all the marks obtained, that is 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 ${x_i}$ represent the marks of the students and ${f_i}$ as the frequency.
So ${x_1} = 15$
$
{x_2} = 20 \\
{x_3} = 22 \\
{x_4} = 24 \\
{x_5} = 25 \\
{x_6} = 30 \\
{x_7} = 33 \\
{x_8} = 38 \\
{x_9} = 45 \\
$
$\sum {{f_i}} = 5 + 8 + 11 + 20 + 23 + 18 + 13 + 3 + 1 = 102$
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} + {x_7}{f_7} + {x_8}{f_8} + {x_9}{f_9}}}{{{\text{ total number of students}}}} \\
= \dfrac{{15(5) + 20(8) + 22(11) + 24(20) + 25(23) + 30(18) + 33(13) + 38(3) + 45(1)}}{{102}} \\
= \dfrac{{2660}}{{102}} = 26.07 \\
$
Therefore mean or average mark is $26.07$
Note:
If the question asks for the mode then we need to find which mark is obtained by the maximum number of students and that will be our mode. For example here we have maximum students achieving a particular mark as $23$ so the mode will be $25$
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

