(a) To find the number of rods expected to have a length less than 54.8 mm, we first standardize the value:

\(P(X < 54.8) = P\left(Z < \frac{54.8 - 55.6}{1.2}\right)\)

\(= P(Z < -0.6667) = 1 - 0.7477 = 0.2523\)

The expected number is then:

\(400 \times 0.2523 = 100.92\)

Rounding gives 100 or 101 rods.

(b) To find the probability that a rod's length is within half a standard deviation of the mean, we calculate:

\(P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = \Phi\left(\frac{1}{2}\right) - \Phi\left(-\frac{1}{2}\right)\)

\(= 2\Phi\left(\frac{1}{2}\right) - 1\)

\(= 2 \times 0.6915 - 1 = 0.383\)