Find the missing frequencies f1 and f2 in the following frequency distribution, if it is known that the mean of the distribution is 1.46
| 1 | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| f1 | 46 | f1 | f2 | 25 | 10 | 5 | 200 |
| xi | fi | fixi |
|---|---|---|
| 0 | 46 | 0 |
| 1 | f1 | f1 |
| 2 | f2 | 2f2 |
| 3 | 25 | 75 |
| 4 | 10 | 40 |
| 5 | 5 | 25 |
N = 86 + f1 + f2
We have N = 200
∴ 200 = 86 + f1 + f2
Thus, f1 = 114 - f2 …(i)
Mean = 1.46
292 = 140 + f1 + 2f2
⇒ f1 + 2f2 = 152 …(ii)
Solving (i) and (ii), we get
114 - f2 + 2f2 = 152
Thus, f2 = 38
Hence, f1 = 114 - 38 = 76