(i) To find the mean age of the whole group, use the formula for the combined mean:

\(\bar{x} = \frac{(\bar{x}_w \times 10) + (\bar{x}_m \times 13)}{23} = \frac{(41.2 \times 10) + (46.3 \times 13)}{23} = 44.1\)

(ii) First, find \(\Sigma x_w^2\) and \(\Sigma x_m^2\) using the given standard deviations:

\(15.1^2 = \frac{\Sigma x_w^2}{10} - 41.2^2\)

\(\Sigma x_w^2 = 19254.5\)

\(12.7^2 = \frac{\Sigma x_m^2}{13} - 46.3^2\)

\(\Sigma x_m^2 = 29964.74\)

Calculate the total sum of squares:

\(\Sigma = 19254.5 + 29964.74 = 49219.24\)

Finally, find the standard deviation for the whole group:

\(\text{sd} = \sqrt{\frac{49219.24}{23} - 44.1^2} = 14.0\)