Question

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹ 18 , find the missing frequency f.

Daily pocket allowance (in Rs.)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

Answer 1

The given data is shown as follows:

To find the class mark (x_i) for each interval, the following relation is used.

Given this, average pocket allowance, `barx`= Rs 18

Taking 18 as the assured mean (A), d_i and f_id_i are calculated as follows.

Given this, average pocket allowance, `barx`= Rs 18

Daily pocket allowance (in Rs)	Number of children `(f_i)`	Class mark `(x_i)`	`f_i x_i`
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
Total	`sum f_i = 44+f`		`sum f_ix_i =752 + 20f`

The mean of the given data is given by,

`barx = (sum_(i) f_ix_i )/(sum_(i) f_i)` 

⇒ 18 =` (750+20f)/(44+f)`

⇒ 18 (44 + f) = 752 + 20f

⇒ 792 + 18 f = 752 -20f

⇒ 20f - 18 f = 792 - 752

⇒ 2f = 40

⇒ f = 20

Hence, the value of f is 20.