Find the variance of 11, 12, 13, 14 and 15.
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Correct Answer: Option A
Explanation:
\(Variance (\sigma^{2}) = \frac{\sum (x - \mu)^2}{n}\)The mean \((\mu)\) of the data = \(\frac{11 + 12 + 13 + 14 + 15}{5} = \frac{65}{5} = 13\)
\(\sigma^{2} = \frac{10}{5} = 2\)
\(Variance (\sigma^{2}) = \frac{\sum (x - \mu)^2}{n}\)The mean \((\mu)\) of the data = \(\frac{11 + 12 + 13 + 14 + 15}{5} = \frac{65}{5} = 13\)
| \(x\) | \((x - \mu)\) | \((x - \mu)^{2}\) |
| 11 | -2 | 4 |
| 12 | -1 | 1 |
| 13 | 0 | 0 |
| 14 | 1 | 1 |
| 15 | 2 | 4 |
| Total | 10 |
\(\sigma^{2} = \frac{10}{5} = 2\)