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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 |
Solution
The given data is shown as follows:
To find the class mark (xi) for each interval, the following relation is used.
Given this, average pocket allowance, `barx`= Rs 18
Taking 18 as the assured mean (A), di and fidi 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.
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