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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 | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
| Number of workers | 7 | 6 | 9 | 13 | f | 5 | 4 |
Solution
To find the class mark (xi) for each interval, the following relation is used.
Given that, mean pocket allowance, `barx `= Rs 18
Taking 18 as assured mean (a), di and fidi are calculated as follows.
|
Daily pocket allowance (in Rs) |
Number of children fi |
Class mark xi | di = xi − 18 | fidi |
| 11 −13 | 7 | 12 | -6 | -42 |
| 13 − 15 | 6 | 14 | -4 | -24 |
| 15 − 17 | 9 | 16 | -2 | -8 |
| 17 −19 | 13 | 18 | 0 | 0 |
| 19 − 21 | f | 20 | 2 | 2 f |
| 21 − 23 | 5 | 22 | 4 | 20 |
| 23 − 25 | 4 | 24 | 6 | 24 |
| Total |
`sumf_i=44+f ` |
2f − 40 |
From the table, we obtain
`sumf_i = 44+f`
`sumf_iu_i = 2f - 40`
`barx = a+(sumf_id_i)/(sumf_i)`
`= 18 = 18 + ((2f - 40)/(44+f))`
`0 = ((2f - 40)/(44+f))`
2f - 40 = 0
2f = 40
f = 20
Hence, the missing frequency, f, is 20.
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