Question

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

$\begin{array}{cccccccc}\text{Number of days}& 0-6& 6-10& 10-14& 14-20& 20-28& 28-38& 38-40\\ \text{Number of students}& 11& 10& 7& 4& 4& 3& 1\end{array}$
$\left[3\right]$

Answer 1

We can find the class mark of each interval by using the relation:

$x_i=\frac{\text{Upper class limit + lower class limit}}{2}$

Now, taking 17 as assumed mean $A$, we can calculate $d_i$ and $f_i{d}_i$ as following:

$\begin{array}{ccccc}\text{Number of days}& \text{Number of students}{f}_i& x_i& d_i=x_i-17& f_i{d}_i\\ 0-6& 11& 3& -14& -154\\ 6-10& 10& 8& -9& -90\\ 10-14& 7& 12& -5& -35\\ 14-20& 4& 17& 0& 0\\ 20-28& 4& 24& 7& 28\\ 28-38& 3& 33& 16& 48\\ 38-40& 1& 39& 22& 22\\ \text{Total}& 40& & & -181\end{array}$
$\left[2\right]$

Now, we observe that:

$\sum f_i=40\sum f_i{d}_i=-181\text{mean}\overline{x}=A+\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)=17+\left(\frac{-181}{40}\right)$

$=17-4.525$

$=12.475$

$=12.48$

So, mean number of days for which a student was absent is 12.48 days.
$\left[1\right]$