Given that the heights are normally distributed with mean \(\mu = 148\) cm and standard deviation \(\sigma = 8\) cm, we need to find the height \(h\) such that \(P(X < h) = 0.67\).

Using the standard normal distribution, we find the z-score corresponding to a probability of 0.67. From standard normal distribution tables, \(P(Z < 0.44) = 0.67\), so \(z = 0.44\).

The z-score formula is \(z = \frac{h - \mu}{\sigma}\).

Substitute the known values: \(0.44 = \frac{h - 148}{8}\).

Solve for \(h\):

\(h - 148 = 0.44 \times 8\)

\(h - 148 = 3.52\)

\(h = 148 + 3.52\)

\(h = 151.52\)

Rounding to the nearest whole number, \(h = 152\).