Eduprobe
Question:
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.
Daily pocket allowance (in Rs.) | Number of children
Daily pocket allowance (in Rs.) | Number of children
11-13 | 7
13-15 | 6
15-17 | 9
17-19 | 13
19-21 | f
21-23 | 5
23-25 | 4
Solution:
Lets take assumed mean, A as 16
Class interval = 2
∴ ui = (xi - A)/h = (xi - 16)/2
Now, we have A = 16, ̄x = 18 and h = 2
We know that, Mean, ̄x = A + h(Σfiui/N)
⇒ 18 = 16 + 2(Σfiui/44+f)
⇒ 2 = 2(Σfiui/44+f)
⇒ 44 + f = Σfiui
We have:
Class interval | fi | ui | fiui
11-13 | 7 | -2.5 | -17.5
13-15 | 6 | -1.5 | -9
15-17 | 9 | -0.5 | -4.5
17-19 | 13 | 0.5 | 6.5
19-21 | f | 1.5 | 1.5f
21-23 | 5 | 2.5 | 12.5
23-25 | 4 | 3.5 | 14
Total | 44+f | | 1.5f + 20.5
Σfiui = 1.5f + 20.5
44 + f = 1.5f + 20.5
44 - 20.5 = 1.5f - f
23.5 = 0.5f
f = 23.5/0.5 = 47
Hence, the value of missing frequency, f = 47