Answer: The pooled estimate of the common population standard deviation is \(0.550\text{ kg}\), to 3 significant figures.

Since the two populations are assumed to have the same standard deviation, use the pooled variance estimate.

For the boys, the sample size is \(10\), and the estimated standard deviation is \(0.6\). Therefore the boys' contribution to the sum of squared deviations is

\((10-1)(0.6)^2=9\times 0.36=3.24\).

For the girls, the data are \(3.9,3.1,2.9,3.1,3.6\). First find

\(\sum x=3.9+3.1+2.9+3.1+3.6=16.6\),

and

\(\sum x^2=3.9^2+3.1^2+2.9^2+3.1^2+3.6^2=55.8\).

So the girls' sum of squared deviations is

\(\sum (x-\bar{x})^2=\sum x^2-\frac{(\sum x)^2}{n}=55.8-\frac{16.6^2}{5}=0.688\).

The pooled variance estimate is therefore

\(s_p^2=\frac{3.24+0.688}{10+5-2}=\frac{3.928}{13}=0.302153\ldots\).

Hence the pooled estimate of \(\sigma\) is

\(s_p=\sqrt{0.302153\ldots}=0.5497\ldots\).

Thus \(\sigma\) is estimated as \(0.550\text{ kg}\), to 3 significant figures.