(b) We know that 72% of Eastern bluebirds have a length greater than 17.1 cm, which implies \(P(Z > \frac{17.1 - 18.4}{\sigma}) = 0.72\). This corresponds to a standard normal variable \(Z\) such that \(Z = -0.583\) (since \(P(Z > -0.583) = 0.72\)).

Thus, \(\frac{17.1 - 18.4}{\sigma} = -0.583\).

Solving for \(\sigma\), we get \(\sigma = \frac{17.1 - 18.4}{-0.583} = 2.23\).

(c) For a sample of 120 bluebirds, the mean number with length greater than 17.1 cm is \(120 \times 0.72 = 86.4\).

The variance is \(120 \times 0.72 \times 0.28 = 24.192\).

Using a normal approximation, \(P(X < 80) = P(Z < \frac{79.5 - 86.4}{\sqrt{24.192}})\) (using continuity correction).

\(Z = \frac{79.5 - 86.4}{\sqrt{24.192}} = -1.4029\).

Thus, \(P(Z < -1.4029) = 1 - \Phi(1.403) = 1 - 0.9196 = 0.0804\).