Question

The marks for a test in group of students are as follows find mean marks of these student marks 0-10,10-20,20-30,30-40,40-50,50-60,60-70,no. of students 3,5,12,16,11,5,4​

Answer 1

Given:

$\begin{array}{| c | c | c | c | c | c | c | c |}\cline{1-8}\bf Marks & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \cline{1-8} \bf Number\ of\ students & 3 & 5 & 12 & 16 & 11 & 5 &4 \\\cline{1-8}\end{array}$

To find:

The mean of the Marks of the students

So,

[tex]\begin{array}{| c | c | c |c|} \cline{1-4} \bf Marks & \bf Class\ Mark(x_{i}) &\bf No.\ of\ students(f_{i}) &\bf f_{i}x_{i}\\ \cline{1-4} 0-10 &\frac{0+10}{2}= 5& 3 &15\\ \cline{1-4} 10-20 & \frac{10+20}{2}= 15 & 5 &75\\ \cline{1-4} 20-30&\frac{20+30}{2}= 25&12&300 \\ \cline{1-4}30-40&\frac{30+40}{2}= 35&16&560\\ \cline{1-4}40--50&\frac{40+50}{2}= 45&11&495\\ \cline{1-4}50-60&\frac{50+60}{2}= 55&5&275\\ \cline{1-4}60-70&\frac{60+70}{2}= 65&4&260\\ \cline{1-4}\bf Total&& \sum f_{i} =56 &\sum f_{i}x_{i}=1980 \\ \cline{1-4} \end{array}[/tex]

The formula used to find the mean is

$\boxed{\bf Mean\ (\bar{x})=\frac{\sum f_{i}x_{i}}{\sum f_{i}}}$

So, the mean

$\bf Mean = \frac{1980}{56}$

Mean = 35.35

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More Information:

• Finding Mean by Assumed Mean Method.

$\boxed{\blue{ Mean\ (\bar{x}) = a+ \frac{ \sum f_{i}d_{i}}{ \sum f_{i}} }}$

Here,

Some value is taken from $\red{x_{i}}$ which lies in the middle of $x_{1},\ x_{2},\ .........,\ x_{n}$

And

$\boxed{\purple{d_{i}=x_{i}-a}}$

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• Finding 'Mean' By Step Deviation Method.

$\boxed{\red{Mean\ (x_{i})=\frac{\sum f_{i}u_{i}}{\sum f_{i}}}}$

Here,

$\boxed{\pink u_{i} = \frac{d_{i}}{h} = \frac{x_{i}-a}{h}}$

And here 'h' is the class size of each class interval. The value of 'h' is equal to the difference between the upper and lower limit of a class limits of a class or the difference between two consecutive class marks.