Question :
| Number of Mangoes | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |
| Number of Boxes | 15 | 110 | 135 | 115 | 25 |
Solution :
Here, the class intervals are formed by the exclusive method. If we make the data an inclusive one, the mid-values remain same. So there is no need to convert the data with the cation that while finding h, we should count both the limits of class interval. For example, for (53 – 55), both 53 and 55 should be counted and thus h = 3 and not (55 – 53) = 2.
Let A = 60
So, \(u_i\) = \(x_i – A\over h\) = \(x_i – 60\over 3\)
Calculation of Mean
Number of Mangoes
Mid – Values (\(x_i\))
Frequency (\(f_i\))
\(u_i\) = \(x_i – 60\over 3\)
\(f_iu_i\)
50 – 52
51
15
-3
-45
53 – 55
54
110
-2
-220
56 – 58
57
135
-1
-135
59 – 61
60
115
0
0
62 – 64
63
25
1
25
Total
\(\sum f_i\) = 400
\(\sum f_iu_i\) = -375
So, Mean = A + h\(\sum f_iu_i\over \sum f_i\) = 60 + 3 \(\times\) \(-375\over 400\)
= 60 – 2.81 = 57.19
Hence, mean number of mangoes per box is 57.19
