Question

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

$\begin{array}{cccccc}\text{Number of mangoes}& 50-52& 53-55& 56-58& 59-61& 62-64\\ \text{Number of boxes}& 15& 110& 135& 115& 25\end{array}$

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer 1

$\begin{array}{cc}\text{Number of mangoes}& \text{Number of boxes}\\ 50-52& 15\\ 53-55& 110\\ 56-58& 135\\ 59-61& 115\\ 62-64& 25\end{array}$

We may observe that class intervals are not continuous. There is a gap of $1$ between two class intervals. So we have to add $\frac{1}{2}$ to upper class limit and subtract $\frac{1}{2}$ from lower class limit of each interval.
The class mark $\left(x_i\right)$ may be obtained by using the relation:

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

Class size $h$ of this data = $3$

Since, the given data is large, we will use Assumed Mean Method to find the mean. This method helps in reducing the calculations and results in small numerical values. We will choose the assumed mean $\left(a\right)$ in such a way so that the the value obtained in $d_i$,cancel out each other.

Now, taking $57$ as assumed mean $\left(a\right)$; we may calculate $d_i,u_i,f_i{u}_i$ as following –

$\begin{array}{ccccc}\text{Class interval}& f_i& x_i& d_i=x_i-57& f_i{d}_i\\ 49.5-52.5& 15& 51& -6& -90\\ 52.5-55.5& 110& 54& -3& -330\\ 55.5-58.5& 135& 57& 0& 0\\ 58.5-61.5& 115& 60& 3& 345\\ 61.5-64.5& 25& 63& 6& 150\\ \text{Total}& 400& & & 75\end{array}$

Now, we may observe that:

$\sum f_i=400\sum f_i{d}_i=75\text{Mean}\overline{x}=a+\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)=57+\frac{75}{400}=57+0.1875=57.1875=57.2$

Clearly, mean number of mangoes kept in a packing box is $57.2$.