Question

Question 5
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

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

$\begin{array}{cccccc}\text{Class interval}& f_i& x_i& d_i=x_i-57& u_i=\frac{x_i-57}{h}& f_i{u}_i\\ 49.5-52.5& 15& 51& -6& -2& -30\\ 52.5-55.5& 110& 54& -3& -1& -110\\ 55.5-58.5& 135& 57& 0& 0& 0\\ 58.5-61.5& 115& 60& 3& 1& 115\\ 61.5-64.5& 25& 63& 6& 2& 50\\ \text{Total}& 400& & & & 25\end{array}$

Now, we may observe that:

$\sum f_i=400\sum f_i{u}_i=25Mean\overline{x}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h=57+\frac{25}{400}\times 3=57+\frac{3}{16}=57+0.1875=57.1875=57.19$

Clearly, mean number of mangoes kept in a packing box is 57.19.

We have chosen step deviation method here as values of $f_i,d_i$ are big and also there is a common multiple between all $d_i$.