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Question
The frequency distribution:
| `x` | A | 2A | 3A | 4A | 5A | 6A |
| `f` | 2 | 1 | 1 | 1 | 1 | 1 |
where A is a positive integer, has a variance of 160. Determine the value of A.
Solution
| `x` | `f_i` | `f_ix_i` | `f_ix_i^2` |
| A | 2 | 2A | 2A |
| 2A | 1 | 2A | 4A2 |
| 3A | 1 | 3A | 9A2 |
| 4A | 1 | 4A | 16A2 |
| 5A | 1 | 5A | 25A2 |
| 6A | 1 | 6A | 36A2 |
| n = 7 | `sumf_ix_i` = 22A | `sumf_ix_i^2` = 92A2 |
∴ Variance `sigma^2 = (sumf_ix_i^2)/n - ((sumf_ix_i)/n)^2`
⇒ 160 = `(92A^2)/7 - ((22A)/7)^2`
⇒ 160 = `(92A^2)/7 - (484A^2)/49`
⇒ 160 = `(644A^2 - 484A^2)/49`
⇒ 160 = `(160A^2)/49`
⇒ A2 = 49
⇒ A = 7
Hence, the value of A is 7.
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