Find the mean mark from the following frequency table of 100 students. MARKS OBTAINED 30 35 40 45 50 NUMBER O STUDENTS 45 26 12 10 7 About the author Arya
[tex]\large\underline{\sf{Solution-}}[/tex] Given data is Let marks obtained be represented as x and number of students is represented as f. So, [tex] \: \: \: \: \: \: \: \: \: \: \: \begin{gathered} \begin{array}{|c|c|} \bf{x} & \bf{f} \\ 30 & 45 \\35 & 26 \\40 & 12 \\45 & 10 \\50 & 7 \end{array}\end{gathered}[/tex] Calculation of Mean :- [tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf x&\sf \: f&\sf \:fx\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 30&\sf 45&\sf1350\\\\\sf 35&\sf 26&\sf910\\\\\sf 40 &\sf 12&\sf480\\\\\sf 45&\sf 10&\sf450 \\\\\sf 50&\sf 7&\sf350\\\frac{\qquad}{}&\frac{\qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}[/tex] So, we have [tex]\rm :\longmapsto\: \sum \: fx = 3540[/tex] [tex]\rm :\longmapsto\: \sum \: f = 100[/tex] Hence, Mean using Direct Method is [tex]\red{\rm :\longmapsto\:Mean = \dfrac{ \sum \: fx}{ \sum \: f}}[/tex] [tex] \rm \: \: = \: \dfrac{3540}{100} [/tex] [tex] \rm \: \: = \: 35.4[/tex] Additional Information :- Mean using Short Cut Method [tex]\red{\rm :\longmapsto\:Mean =A + \dfrac{ \sum \: fd}{ \sum \: f}}[/tex] Mean using Step Deviation Method [tex]\red{\rm :\longmapsto\:Mean =A + \dfrac{ \sum \: fu}{ \sum \: f} \times h}[/tex] Reply
[tex]\large\underline{\sf{Solution-}}[/tex]
Given data is
Let marks obtained be represented as x and number of students is represented as f.
So,
[tex] \: \: \: \: \: \: \: \: \: \: \: \begin{gathered} \begin{array}{|c|c|} \bf{x} & \bf{f} \\ 30 & 45 \\35 & 26 \\40 & 12 \\45 & 10 \\50 & 7 \end{array}\end{gathered}[/tex]
Calculation of Mean :-
[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf x&\sf \: f&\sf \:fx\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 30&\sf 45&\sf1350\\\\\sf 35&\sf 26&\sf910\\\\\sf 40 &\sf 12&\sf480\\\\\sf 45&\sf 10&\sf450 \\\\\sf 50&\sf 7&\sf350\\\frac{\qquad}{}&\frac{\qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}[/tex]
So, we have
[tex]\rm :\longmapsto\: \sum \: fx = 3540[/tex]
[tex]\rm :\longmapsto\: \sum \: f = 100[/tex]
Hence,
Mean using Direct Method is
[tex]\red{\rm :\longmapsto\:Mean = \dfrac{ \sum \: fx}{ \sum \: f}}[/tex]
[tex] \rm \: \: = \: \dfrac{3540}{100} [/tex]
[tex] \rm \: \: = \: 35.4[/tex]
Additional Information :-
Mean using Short Cut Method
[tex]\red{\rm :\longmapsto\:Mean =A + \dfrac{ \sum \: fd}{ \sum \: f}}[/tex]
Mean using Step Deviation Method
[tex]\red{\rm :\longmapsto\:Mean =A + \dfrac{ \sum \: fu}{ \sum \: f} \times h}[/tex]