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.

Number of heartbeats per minute	50-52	53-55	56-58	59-61	62-64
Number of boxs	15	110	135	115	25

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

Answer 1

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, \(\frac{1}2\)  has to be added to the upper-class limit and \(\frac{1}2\)  has to be subtracted from the lower class limit of each interval. 

Class mark (\(x_i\)) can be obtained by using the following relation.

Number of mangoes	Number of boxes \(\bf{f_i }\)
50 -52	15
53 -55	110
56 - 58	135
59 - 61	115
62 - 64	25

Class mark  \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)

class size (h) of the data =  3 

Taking 57 as assumed mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) can be calculated as follows.   

Number of heart-beats per minute	Number of women (\(\bf{f_i}\))	\(\bf{x_i}\)	\(\bf{d_i = x_i -57}\)	\(\bf{u_i = \frac{d_i}{3}}\)	\(\bf{f_iu_i}\)
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
Total	400				25

From the table, it can be observed that  

\(\sum f_i = 400\)
\(\sum f_iu_i = 25\)

Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i}) \times h\) 
x = \(57 + (\frac{25 }{400}) \times 3\) 

x = 57 - \((\frac{3}{16})\)

x  = 57 + 0.1875 
x = 57.1875
x  = 57.19

Mean number of mangoes kept in a packing box is 57.19.

Step deviation method is used here as the values of \(f_i, d_i\) are big and also, there is a common multiple between all \(d_i\).