Question

Consider the following distribution of the number of mangoes being packed in cardboard boxes where these boxes contain varying number of mangoes. Find the mean number of mangoes kept in a packing box using assumed mean method?

$$\begin{array}{cccccc}Numberofmangoes& 50-52& 53-55& 56-58& 59-61& 62-64\\ Numberofboxes& 15& 110& 135& 115& 25\end{array}$$

• A. 53.87
• B. 59.95
• C. 57.18
• D. 54.99

Answer 1

The correct option is C

57.18

The class intervals are not continuous. There is a gap of 1 between 2 intervals. So 0.5 has to be added to the upper class limit of each interval and 0.5 has to be subtracted from the lower class limit of each interval to make class intervals continuous.

$$\begin{array}{ccccc}ClassInterval& NumberofMangoes\left(f_i\right)& Classmark\left(x_i\right)& d_i=x_i-a& f_i{d}_i\\ 49.5-52.5& 15& 51& -6& -90\\ 52.5-55.5& 110& 54& -3& -330\\ 55.5-58.5& 135& 57& 0& 0\\ 58.5-61.5& 115& 60& 3& 345\\ 61.5-64.5& 25& 63& 6& 150\\ Total& \sum f_i=400& & & \sum f_i{d}_i=75\end{array}$$

Class mark = Mean($\overline{x}$) = x + $\frac{\sum f_i{d}_i}{\sum f_i}$ = 57 + $\frac{75}{400}$ = 57.1875

Let the assumed mean(a) be 57 (Choose the number in order to get minimum deviation(d) in table shown below)

$$\begin{array}{ccccc}ClassInterval& NumberofMangoes\left(f_i\right)& Classmark\left(x_i\right)& d_i=x_i-a& f_i{d}_i\\ 49.5-52.5& 15& 51& -6& -90\\ 52.5-55.5& 110& 54& -3& -330\\ 55.5-58.5& 135& 57& 0& 0\\ 58.5-61.5& 115& 60& 3& 345\\ 61.5-64.5& 25& 63& 6& 150\\ Total& \sum f_i=400& & & \sum f_i{d}_i=75\end{array}$$

Mean($\overline{x}$) = x + $\frac{\sum f_i{d}_i}{\sum f_i}$ = 57 + $\frac{75}{400}$ = 57.1875