Question

The following table gives the marks scored by students in an exam.

$\begin{array}{cc}\text{Marks}& \text{No. of students}\\ \text{Above}0& 80\\ \text{Above}10& 77\\ \text{Above}20& 72\\ \text{Above}30& 65\\ \text{Above}40& 55\\ \text{Above}50& 43\\ \text{Above}60& 23\\ \text{Above}70& 16\\ \text{Above}80& 10\\ \text{Above}90& 8\\ \text{Above}100& 0\end{array}$

Find the mean for the following data using assumed mean method.

Answer 1

Let the assumed mean 'a' be 45

Number of students $f_i$ of the class interval (0 -10) = 80 - 77 = 3

Number of students $f_i$ of the class interval (10 -20) = 77 - 72 = 5 and similary for the other class.

$\begin{array}{ccccc}\text{Class interval}& x_i& f_i& d_i=x_i-a& f_i{d}_i\\ 0-10& 5& 3& -40& -120\\ 10-20& 15& 5& -30& -150\\ 20-30& 25& 7& -20& -140\\ 30-40& 35& 10& -10& -100\\ 40-50& 45& 12& 0& 0\\ 50-60& 55& 20& 10& 200\\ 60-70& 65& 7& 20& 140\\ 70-80& 75& 6& 30& 180\\ 80-90& 85& 2& 40& 80\\ 90-100& 95& 8& 50& 400\end{array}$

Here,
$\sum f_i=80,\sum f_i{d}_i=490$
$Mean=a+\frac{\sum f_i{d}_i}{\sum f_i}$
$=45+\frac{490}{80}=45+6.125=51.25$