(i) To find the mean age of all 19 people, use the formula for the combined mean:

\(\bar{x} = \frac{(48.7 \times 12) + (38.1 \times 7)}{19} = \frac{584.4 + 266.7}{19} = \frac{851.1}{19} = 44.8 \text{ years}\)

(ii) First, find \(\Sigma x^2\) and \(\Sigma y^2\) using the given standard deviations and means:

For the first group:

\(7.65^2 = \frac{\Sigma x^2}{12} - 48.7^2\)

\(58.5225 = \frac{\Sigma x^2}{12} - 2371.69\)

\(\Sigma x^2 = 12 \times (58.5225 + 2371.69) = 29162.55\)

For the second group:

\(4.2^2 = \frac{\Sigma y^2}{7} - 38.1^2\)

\(17.64 = \frac{\Sigma y^2}{7} - 1451.61\)

\(\Sigma y^2 = 7 \times (17.64 + 1451.61) = 10284.75\)

\(Now, find the combined variance:\)

\(\text{Combined variance} = \frac{29162.55 + 10284.75}{19} - 44.79^2\)

\(= \frac{39447.3}{19} - 2006.0641\)

\(= 2076.1737 - 2006.0641 = 70.1096\)

Finally, the standard deviation is:

\(\sigma = \sqrt{70.1096} \approx 8.37 \text{ or } 8.36 \text{ years}\)