The ages, in years, of 5 boys are 5, 6, 6, 8 and 10. Calculate, correct to one decimal place, the standard deviation of their ages.
| \(x\) | 5 | 6 | 6 | 8 | 10 | Total |
| \(x - \bar{x}\) | -2 | -1 | -1 | 1 | 3 | |
| \((x - \bar{x})^{2}\) | 4 | 1 | 1 | 1 | 9 | 16 |
Mean (\(\bar{x}\)) = \(\frac{5 + 6 + 6 + 8 + 10}{5} \)
= \(\frac{35}{5} = 7\)
\(SD = \sqrt{\frac{\sum (x - \bar{x})^{2}}{n}}\)
= \(\sqrt{\frac{16}{5}} \)
= \(\sqrt{3.2}\)
\(\approxeq 1.8 years\)
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