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Question
Find the mean of the following frequency distribution, using the assumed-mean method:
| Class | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
| Frequency | 10 | 20 | 30 | 15 | 5 |
Solution
| Class | Frequency `(f_i)` | Mid values `(x_i)` | Deviation `(d_i) d_i = (x_i – 150)` |
`(f_i × d_i)` |
| 100 – 120 | 10 | 110 | -40 | -400 |
| 120 – 140 | 20 | 130 | -20 | -400 |
| 140 – 160 | 30 | 150 = A | 0 | 0 |
| 160 – 180 | 15 | 170 | 20 | 300 |
| 180 – 200 | 5 | 190 | 40 | 200 |
| `Ʃ f_i` = 80 | `Ʃ (f_i × d_i)` = -300 |
Let A = 150 be the assumed mean. Then we have:
Mean, x = ` A + (Ʃ (f_i × d_i))/(Ʃ f_i)`
=150 + `(-300)/80`
`=(1200-30)/8`
=`1170/8 = 146.24`
∴ = x = 146 .24
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